Sample MCAT Question

The Atom

➢ Atomic Structure:

o Q1. Draw and describe the Bohr model of an atom: include the relative size of the components, the charges on each species and their relative mass. Describe the aspects of the Bohr model still believed to be accurate and those that have been updated and/or replaced by modern quantum mechanics and atomic orbital theory.

o Element Symbols:

▪ Z = atomic number (i.e., the number of protons)

▪ A = mass number (i.e., the number of protons + the number of neutrons)

o Isotopes:

▪ Q2. What is an isotope? Do all isotopes have an odd mass number?

o Cations vs. Anions:

▪ Any atom or molecule with fewer electrons than protons is a cation. Any atom or molecule with more electrons than protons is an anion.

▪ Q3. Metals form , but non-metals form .

▪ Cations are smaller than their neutral counterpart and anions are larger than their neutral counterpart. Q4. Why? Explain the difference in size between anions, cations, and neutral atoms.

Sample MCAT Question

1) A student introduces a particle of unknown identity between two oppositely-charged electrodes and

notes that it accelerates toward one of the two electrodes. The particle could be any of the following,

EXCEPT a(n):

A) anion.

B) cation.

C) neutron.

D) proton.

Solution: The acceleration of the particle toward an electrode indicates that it is charged. Because the stem does not specify

which electrode the particle accelerates toward, the particle could be positively or negatively charged. Anions, cations and protons are all charged, but a neutron is not, making C the best answer.

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Periodic Table & Characteristics of the Elements

➢ Terminology:

o Q5. Identify the following referring to the periodic table: period, group/family, alkali metals, alkaline earth metals, transition metals, lanthanides, actinides, halogens, noble gases, s-block, p-block, d-block and f-block.

➢ Metals vs. Non-metals:

o METALS: THINK OF METALS AS: larger atoms with loosely held electrons. Metals “like” to lose electrons and form positive ions. They are lustrous, ductile, malleable and excellent conductors of both heat and electricity. They are involved in ionic bonds with nonmetals.

o NON-METALS: THINK OF NON-METALS AS: smaller atoms with tightly held electrons. Non-metals “like” to gain electrons and form negative ions. They have lower melting points than metals, and form covalent bonds with non-metals. Most of the O-Chem stuff involves non-metals.

➢ Periodic Table Trends:

o Size Matters: The periodic table trends are largely dependent on atomic radius, or in other words, size. Smaller atom’s nuclei are closer to their valence electrons, so according to Coulomb’s Law, F = kqq/r2, those electrons are held more tightly to the positively charged nucleus. This greater force will also cause the atom to be more electronegative, have a higher ionization energy, greater electron affinity and less metallic character than a larger atom. Larger atoms are better at stabilizing charges, don’t form pi bonds (Q6. Why?) and have d orbitals where they can “house” extra electrons.

o Families are Similar: Elements in the same family (column) tend to have similar properties, both chemical and physical (e.g., SiH4 will behave similar to CH4).

o Zeffective: Most MCAT books say way too much about this. We are only aware of one (1) MCAT question that has ever required an understanding of Zeffective. As atoms increase in size, they are surrounded by more and more electrons. The “effective nuclear charge,” felt by the valence electrons of such atoms is less than expected. This is because the electrons in between the valence shell and the nucleus “shield” the valence electrons, preventing them from feeling the full charge of the nucleus. Hydrogen’s electron feels its full nuclear charge because there is zero shielding.

o Q7. State whether the following will increase of decrease as you go across a period or down a group of the periodic table: Electron Affinity, Electronegativity, Ionization Energy, Atomic Radius & Metallic Character.

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Quantum Mechanics

➢ Overview: Every electron in an atom has a unique “address” or “location.” Electron addresses consist of four numbers. One could think of the first number as giving the street name (i.e., shell), the second as giving the house or apartment (i.e., subshell), and the third as giving the room within that apartment (i.e., orbital). Electrons can come in pairs, with two electrons sharing one orbital. They are like two siblings sharing the same room. They are differentiated by their “spin.”

➢ First Quantum Number:

o a.k.a. “n” or “the principle quantum number”

o Gives the shell (e.g., valence electrons are in the outermost “shell”) and represents the approximate relative energy of electrons in that shell.

o Q8. Which orbital is higher in energy, a 4p or a 5s? What about a 3d and a 4s?

➢ Second Quantum Number:

o a.k.a. “ℓ” or “the azimuthal quantum number” or “the angular momentum quantum number”

o Gives the subshell or orbital; has values of 0, 1, 2 or 3, and from this we know the shape:

▪ 0 = s ; 1 = p ; 2 = d ; 3 = f

➢ Third Quantum Number:

o a.k.a. “mℓ” or “the magnetic quantum number”

o Gives the orbital orientation; has a value of -ℓ to ℓ (from the azimuthal quantum number)

▪ Designates the orientation of the subshell where an electron is most likely to be found (i.e., which “dumbbell” of a p subshell)


2) A student desiring to eject an electron from a metal sample must choose between three metal elements

available to him in the lab: sodium, magnesium, or potassium. To most easily accomplish his objective,

the student should choose:

A) Na

B) K

C) Mg

D) Mg or K, because they exhibit equal metallic properties.

Solution: The periodic table trends indicate that elements to the left and down in the table will most easily give up a valence

electron. Mg is one box to the right of Na, and K is one box below Na. Thus, B is the best answer, as Na is easier than Mg

and K is easier than Na. The difference between adjacent family members is greater than the difference between adjacent periods, making B the best answer.

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➢ Fourth Quantum Number:

o a.k.a. “ms” or ”the electron spin quantum number”

o Gives the spin, which is either +½ or -½. Positive spin is represented by an up arrow in an electron configuration diagram and negative spin is represented by a down arrow.

▪ Q9. Continue the street/apartment/room analogy used above to explain how many rooms there are in an s, p, d and f subshell.

▪ Q10. What are the Heisenberg Uncertainty and Pauli Exclusion Principles?

Electron Configuration

➢ Overview: Electron configurations are a list of quantum numbers and the number of electrons in each. Energy level diagrams are the stair-step diagrams you drew in general chemistry with lines representing orbitals and up and down arrows representing electrons.

o The only remotely difficult MCAT question you’ll face regarding electron configuration is to provide the configuration for cations and anions. For cations, move back one box in the periodic table for each electron missing. For anions, move forward one box for each extra electron.

o Q11. What is the electron configuration of calcium in calcium sulfate? What element in its ground state will have the same electron configuration as a chloride ion?

Energy Levels

➢ Definition: Energy levels represent the energies of the electrons in an atom. They are quantized! In other words, they look like stair steps, and do NOT look like a ramp. Electrons can be in energy level 1 or energy level 2, but never anywhere in between.

➢ Energy Levels & Photon-Light Emission:

o Because energy levels are quantized, you cannot cause an electron to move up one energy level unless you add an amount of energy equal to the difference in energy between the current energy level and the higher energy level. If an electron is struck by a photon with an energy that is lower than the difference in energy between two energy levels in that atom the photon will pass through the atom without being absorbed. If an electron drops to a lower energy level, energy is released as a photon (i.e., as electromagnetic radiation). The energy released will be exactly equal to the difference between the two energy levels.

➢ The Work Function:

o Bombarding certain metals with energy can cause the ejection of an electron from their outermost shell (i.e., valence electron). The amount of energy required to do this is called the “work function,” and is usually given the variable φ. This may sound similar to “Ionization Energy.” However, they are NOT the same. Ionization energies are measured for lone atoms in a gaseous state. The work function refers specifically to valence electrons being ejected from the surface of a solid metal. If the energy added is less than the work function, the electron won’t be ejected. If it is greater than the work function, the excess energy will be transferred into the kinetic energy of the ejected electron.

▪ KE = E – φ ; where E is the amount of energy added and KE is the kinetic energy of the ejected electron. Because energy is usually added via bombardment with photons, E can be replaced with hf, the formula for the energy of a photon.

E = hf ; where E = the energy of a photon, h = Planck’s Constant (which is always given) and f = frequency.

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Q12. Derive a related equation that would allow you to calculate the energy of a photon knowing only velocity and wavelength.

Q13. A certain metal is known to have a work function of 500J. If a photon of 500J strikes the surface of the metal, what will be the result?

Radioactive Decay & Half Life

➢ Definition: Radioactive decay is the process by which unstable atoms change their chemical composition over time. The nucleus sometimes loses or gains electrons, lose bundles of protons and neutrons called “alpha particles,” or even transforms one subatomic particle into another.

➢ Types of Decay:

o Alpha Decay: The loss of one He nucleus, which has a mass number of 4 and atomic number of 2.

o Beta Decay: A neutron is changed into a proton with the ejection of an electron.

o Electron Capture: A proton is changed into a neutron via capture of an electron.

o Positron Emission: A proton is changed into a neutron, with expulsion of a positron. (Note: Some sources will label positron emission as a type of Beta decay. However, the MCAT has asked previous questions that clearly demonstrate reference to Beta decay as defined above, separate and distinct from positron emission.)

o Gamma Emission: Gamma rays are usually emitted as a byproduct of the types of decay outlined above. Gamma decay does not change the number of nucleons!

o Important Notes:

▪ THINK OF NEUTRONS AS: neutron = a proton + an electron ▪ THINK OF PROTONS AS: proton = neutron + a positron


3) Two unique light sources are used to bombard a single metal sample with photons (φ = 349J). The first

light source delivers photons with an energy of 700J at a rate of 1 x 105 photons per second. The

second light source delivers 350 J photons at exactly twice that rate. If each light source is shone onto

the surface of the metal for exactly one second, which of the following statements is true?

A) The second light source will eject twice as many electrons as the first.

B) The second light source will eject half as many electrons as the first.

C) Neither light source will eject electrons.

D) Electrons ejected by the second light source will have greater kinetic energy.

Solution: Answer A is correct. Because both light sources deliver photons with energies that exceed the work function, both

will eject electrons in a one-to-one ratio with photons. However, because the second light source delivers twice as many

photons per second, it will eject twice the number of electrons during the one-second period. Answer D is close to true, but it

is actually the first light source that would result in electrons with greater KE.

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➢ Half-Life Problems:

o Definition: The half-life of a substance (t1/2) is the amount of time required for exactly one-half of the mass of that substance to disappear due to radioactive decay.

o Solving Half-Life Problems: There are three variables involved in half-life problems: half-life, t1/2, time

elapsed, t, and the amount of the substance in grams, g. You are usually given two of those three and

asked to solve for the third. Try to do this conceptually in your head—do not use a formula. For example, one question may tell you that we currently have 500 g of element Z, and that the half-life of element Z is 10 years. It then asks how much of element Z will remain 40 years from now. The half-life tells us that every 10 years the substance is cut in half. Thus, in 40 years it will be cut in half four times. Write 500 g on your scratch paper, then cut it in half four times to get the correct answer: 31.25 g. Another question may give you the initial and final mass of the substance plus the amount of time elapsed, and ask you to calculate the half-life. In such a case, first decide how many times the substance had to be cut in half to go from the initial value to the final value (i.e., the number of half-lives). Divide the total time elapsed by the number of half-lives to get the length of each half-life.

o Always take notes when counting half-lives. Many students make silly errors because they miscalculate the number of half-lives. Write it down and you won’t mess up! This is yet another application of the Altius mantra:

Write it Down and Draw it Out!

Bonding

➢ Bonding and Anti-Bonding Orbitals:

o Most of this is beyond the MCAT. Just remember the following:

1) Anti-bonding orbitals are higher in energy than bonding orbitals. 2) Bonding orbitals contain electrons that are “in phase” and are said to be “attractive”; anti-bonding

orbitals contain “out of phase” electrons that are said to be “repulsive.” 3) Know what drawings of bonding and anti-bonding orbitals look like.

➢ Covalent vs. Ionic:

o Q14. Describe the differences between covalent and ionic bonds.

o Good vs. Poor Electrolytes: Covalent compounds that dissociate 100% in water, such as strong acids and strong bases, make good electrolytes. Other covalent compounds are usually poor electrolytes. Ionic compounds that are soluble in water always make good electrolytes.

o Ionic Character: All bonds that are not between two atoms of the same element have some ionic character. It is basically a measure of the polarity of the bond. Ionic species such as NaCl have close to 100% ionic character. Covalent bonds between two non-metals of nearly identical electronegativity have close to zero ionic character.

▪ Q15. Which two elements in the periodic table, if united in a bond, would create a bond with the maximum possible ionic character?

o Condosity: The “condosity” of a solution is the concentration (molarity) of a Sodium Chloride (NaCl) solution that will conduct electricity exactly as well as the solution in question. For example, for a 2.0 M KCl solution, we would expect the condosity to be something more than 2.0. Why would we expect it to be above 2.0? Because potassium is more metallic than sodium. Thus, we know that it will be a better conductor. This means the NaCl solution will have to be slightly more concentrated in order to conduct electricity exactly as well as the KCl solution.

▪ Q16. What is the expected condosity of a 3.0 M LiCl solution?

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➢ Important Terminology Regarding Covalent Bonds:

o Bond Length = distance between the nuclei of the atoms forming the bond.

o Bond Energy = the amount of energy required to break a bond, which is a measure of bond strength. This is an extremely common point of confusion. Remember the following:

**Energy is always REQUIRED to break a bond. Energy is always RELEASED when a bond is formed.**

**Stable compounds (e.g., N2) have HIGH bond energies. Unstable compounds (e.g., ATP) have LOW bond energies.**

▪ When something is said to be a “high energy” molecule that does NOT mean it has high bond energy. In fact, it means the very opposite. It is unstable and thus requires very little energy to dissociate the bond.

o Bond Dissociation Energy = (as far as the MCAT is concerned) think of this as the same thing as bond energy—the amount of energy required to break or “dissociate” the bond. If you wish to be more precise, the bond energy is the average of the gas-phase bond dissociation energies of all bonds of that same type in that same molecule.

o Heat of Combustion = the amount of energy released when a molecule is combusted with oxygen. All covalent bonds are broken and reformed in a radical reaction. The higher the energy of the molecule (i.e., less stable) the higher the heat of combustion.

▪ Q17. Why do bonds form? Is energy required or released when a bond is formed?

▪ Q18. Use a reaction coordinate diagram to explain why the heat of combustion is greatest for the most unstable molecules.

➢ Coordinate Covalent Bonds: A covalent bond in which both electrons shared in the bond are donated by one atom. In most cases, more than one of these “donor” molecules surround and bind a single “recipient” molecule. The donor molecule must have a lone pair and the recipient molecule must have an empty orbital. If a molecule does not have a lone pair of electrons it CANNOT form coordinate covalent bonds with metals or other Lewis acids. The complex formed by the metal and the molecules forming coordinate covalent bonds with that metal, is called a “coordination complex.”

Formulas

➢ Empirical vs. Molecular Formulas: Q19. Provide a conceptual definition for empirical and molecular formula.

o Calculating Percent Mass:

▪ Percent Mass = (mass of one element/total mass of the compound)(100%)

▪ Q20. What is the percent mass of carbon in glucose? What is the percent mass of hydrogen in water?

➢ Deriving a Formula from Percent Mass:

1) Change the percent mass for each element into grams (i.e., 15% = 15g). 2) Convert the grams of each element into moles by dividing by molar mass. 3) Look at the element with the lowest number of moles. Calculate approximately how many times it will

divide into each of the other molar amounts for each of the other elements—this number will be the subscript for each element in the empirical formula. If the subscripts are not at their lowest common denominator, reduce to get the empirical formula.

o An empirical formula is all you can get from percent mass alone. To get the molecular formula, you must be given the MW of the unknown compound. If you have the molecular weight of the actual compound, simply divide that MW by the MW of the empirical formula. You should get a whole number. Multiply each subscript by that number to get the molecular formula.

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Inorganic Nomenclature

➢ Inorganic nomenclature is primarily the naming of ionic compounds:

o General Ionic Compounds: Name the cation first, then the anion (i.e., calcium sulfate is CaSO4, not SO4Ca, and is not called “sulfate calcium”)

o Transition Metals: When written in words, compounds that include transition elements must have a roman numeral showing the oxidation state of the metal (e.g., iron(II)sulfate vs. iron(III)sulfate).

o Monatomic ions: Named by replacing the last syllable with “-ide.” (e.g., sulfide ion, hydride ion, chloride ion, etc.)

o The following polyatomic ions must be memorized:

▪ Q21. Provide a formula and charge for each of the following: hydroxide, nitrate, nitrite, chlorate, chlorite, hypochlorite, perchlorate, carbonate, bicarbonate, ammonium, sulfate, phosphate, manganite, permanganate, and cyanide.

o Acids: Follow the “ate-ic – ite-ous” convention. If the ion name ends in “-ate,” replace that ending with “-ic” as in: Nitrate → Nitric Acid. If the ion name ends in “-ite,” replace that ending with “-ous” as in: Nitrite → Nitrous Acid. If the parent is a single ion rather than a polyatomic ion, replace the “ide” ending with “-ic” and add “Hydro-” as a prefix, as in: Iodide → Hydroiodic Acid.

o Binary Compounds: Name the element furthest down and to the left on the periodic table first; use poly prefixes as necessary (e.g., Nitrogen Trioxide, Carbon Monoxide, Sulfur Dioxide, etc.). Some have common names such as ammonia and water.

Reactions

➢ Reaction Types:

o Combination o Decomposition o Single Displacement o Double Displacement (a.k.a. “metathesis reaction”)

➢ Step-by-Step Instructions for Balancing a Reaction:

1) Balance the number of carbons

2) Balance the number of hydrogens

3) Balance the number of oxygens

4) Balance the number of any remaining elements

5) If necessary, use fractions. For example, if you have seven oxygens on one side of the reaction and one O2 on the other side, put 7/2 in front of the O2.

6) Multiply all of the species on both sides of the reaction by the denominator of any fractions.

7) Double check your work. The number of atoms of each element found on either side of the reaction should be equal. Count them to be certain your coefficients are correct. The charges should also balance. One of the most common errors is failure to fully multiply by a coefficient. For example, you might accidentally count 2CO2 as 2 oxygens when there are actually 4 oxygens present.

• You must double-check every reaction you see on the MCAT to ensure that it is balanced. Yes, they will give you unbalanced reactions without telling you!

• Q22. Write balanced reactions for the combustion of methane, ethane, and propanol.

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The Mole, Stoichiometry & Mole-to-Mole Conversions

➢ Terminology:

o Q23. Provide a conceptual definition for each of the following: atomic weight, molecular weight, molar mass, mole, and Avogadro’s number.

o Mole-to-Mole Conversion: This is a basic skill you should have learned early on in general chemistry. However, we are surprised how many students struggle with this skill and end up missing points because of it. We have found that students perform much better on these problems if they have a map already in their head of where they can go. We do not want you to memorize the flowchart shown below. However, if you can internalize it naturally you will always be able to plan out exactly how to move from what you are given to what you need to calculate.

➢ Limiting Reagent:

o You must have a balanced equation.

o You must convert to moles first.

o Compare the number of moles you have to the number of moles required to run one cycle of the reaction, as indicated by the coefficients. The reactant you will run out of first is the limiting reagent.

o The reactant you have the least of, in either grams or moles, is NOT necessarily your limiting reagent. For example, suppose for the combustion of methane you have 1.5 moles of O2 and only 1.0 mole of methane. Because you need two moles of O2 to react with one mole of methane, you will run out of O2 first and it is therefore your limiting reagent—even though you have more moles of O2 than you do methane.

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➢ Theoretical Yield:

o Q24. Provide conceptual definitions for: theoretical yield, actual yield and percent yield.

o Remember that yield is a function of reactants and equilibrium, NOT rate. Thus, adding a catalyst may increase rate, but will NOT increase yield. For the MCAT, focus on these two ways to increase yield:

1) Start with more reactants

2) Shift the equilibrium to the right using one of the actions described by Le Chatelier’s Principle (discussed on the subsequent page).

The most common method of shifting the equilibrium in this way is to remove products as they are formed. By doing so, you force the reaction into a constant state of “catch up.” The reaction continually produces more product in an attempt to reach equilibrium.

CAVEATS: Action 1) above will increase overall quantity of yield, but NOT percent yield. Furthermore, it will only work if you add more of the limiting reagent. Adding more of any non-limiting reagent (i.e., a reagent that is in excess) will have no effect.

➢ Moles of Oxygen Needed for Combustion: You will occasionally be asked to predict the species that will require the most oxygen to combust. Here is a simple ranking system to make such predictions quickly:

o Add 1.0 for each carbon and subtract 0.5 for each oxygen. The higher the resulting number the more oxygen necessary for full combustion. This is NOT, however, the actual number of moles required—this is only a ranking system. The only way to determine the exact moles of oxygen required for a combustion reaction is to write out and balance the combustion equation.

▪ Q25. Which requires the most oxygen to combust, propane, propanol or propanoic acid?

Equilibrium

➢ Terminology:

o Q26. Define equilibrium conceptually. Picture equilibrium and describe it in words?

➢ The Law of Mass Action:

o Keq = [products]x/[reactants]y

o Keq is written with every term raised to an exponent equal to its coefficient in the balanced equation (remember, however, that you do NOT do this when writing rate laws). Pure liquids (ℓ) and pure solids (s) are never included!

o Q27. How does each of the following affect equilibrium: addition of a catalyst, increased temperature, lowering the Ea, stabilizing the transition state, addition of reactants/products?

➢ The Reaction Quotient: The Equilibrium Constant can only be calculated at equilibrium. If you make the exact same calculation using concentration values taken at any point other than equilibrium the result is called the Reaction Quotient, Q.

o If Q > K, the reaction will proceed to the left or reactants.

o If Q < K, the reaction will proceed to the right or products.

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➢ Le Chatelier’s Principle: Systems already at equilibrium, that experience change, will shift to the left or to the right to reduce the effects of that change and re-establish equilibrium.

o Q28. When you disrupt the equilibrium, creating a “shift” according to Le Chatelier’s principle, what happens to Keq? Does it change?

o Q29. Predict the effects of doing each of the following to a reaction at equilibrium: 1) adding/removing reactants, 2) adding/removing products, 3) increasing/decreasing pressure, and 4) increasing/decreasing temperature.

Kinetics

➢ Terminology:

o Q30. Provide conceptual definitions for, and clarify the difference between, kinetics and thermodynamics.

➢ Collisions Cause Reactions!

o For a reaction to occur:

1) The reactants must collide with enough energy to overcome the energy of activation

2) The reactants must be in the correct spatial orientation

o Rate is measured as the change in molarity (M) of the reactants per second (M/s)

▪ Q31. How will increasing each of the following affect reaction rate: [reactants], [products], [catalyst], energy of activation (Ea), energy of the transition state, energy of the reactants, and temperature?

▪ Q32. Draw an energy coordinate diagram for both an endothermic and an exothermic reaction. Draw an energy coordinate diagram for a two-step reaction with a fast step and a slow step. For all, label products, reactants, ∆H, energy of activation (Ea), and transition state(s).


4) In a sealed container at room temperature, a student is observing the endothermic decomposition

reaction wherein sulfuric acid is broken down into water, sulfur dioxide and diatomic oxygen. After the

reaction has reached equilibrium, the student transfers the entire contents of the reaction vessel into an

evacuated vessel with twice the volume. Which of the following is expected AFTER the transfer?

A) Less water will be produced.

B) The amount of sulfur dioxide gas present will increase.

C) Less sulfuric acid will decompose.

D) The reaction rate will increase due to the reduced pressure.

Solution: First write a balanced reaction: 2H2SO4 <=> 2H2O + 2SO2 + O2. Note that all gases are on the product side of this

equation. The increase in container volume effectively reduces pressure, shifting the equilibrium to the side with the greater

number of moles of gas (i.e., it will shift to the right). This makes Answer B correct.

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➢ Writing Rate Laws: Know how to write a rate law, how to determine one from a table of experimental values, and how to predict experimental results based on a given rate law.

o Rate laws assume the following:

1) Reactions only proceed forward (we ignore the reverse reaction)

2) We only consider the first few seconds of the reaction, when there is a high concentration of each reactant and any catalysts (e.g., enzymes).

o Rate Law Exponents: Students often get the false idea that the exponents in the rate law are given by the coefficients in the balanced equation. The exponents equal the “order” of each reactant. Only if you are specifically told that the reaction is “elementary” do the coefficients equal the exponents in the rate law. For the MCAT, always assume that they do NOT.

o To calculate the “order” of each reactant using experimental data:

1) Find two trials where the [reactant] in question changed, but all other parameters did NOT (i.e., the concentrations of the other reactants, temperature, pressure, etc., all remained constant).

2) Note the factor by which the reactant concentration changed.

3) Note the factor by which the rate changed across those same two trials.

4) Solve for Y in the following equation: XY = Z ; where X = the factor by which the [reactant]

changed, Z = the factor by which the rate changed, and Y = the order of the reactant. Recall that

any number raised to the zero power is equal to one.

o The “overall order” of a reaction = the sum of the exponents in the rate law

➢ Rate Order Graphs: These graphs will only be linear when the reaction has only a single reactant, OR when it is part of a multiple-reactant reaction where the rate is independent of ALL the other reactants (e.g., the other reactant is zero order, or is in excess).

o Zero Order: [A] vs. time is linear (i.e., yields a straight line) with slope = -k

o First Order: ln[A] vs. time is linear with slope =-k

o Second Order: 1/[A] vs. time is linear with slope = k

▪ Q33. For a reaction with two reactants, A and B, a graph of 1/[A] vs. time is non-linear. Which of the following is known? 1) The reaction cannot be second order in A and independent of B, 2) Reactant B must be involved in the rate law, 3) Reactant B cannot be in excess.

▪ Q34. For a reaction with two reactants, A and B, a graph of ln[A] vs. time is linear. How many of the following statements are true? 1) Reactant A must be first order, 2) Reactant B could be first order, 3) Reactant B cannot be impacting rate, 4) Reactant B must be in excess.

▪ Q35. Describe how you could use the rate order graphs to determine the order of each reactant in a multi-reactant reaction experimentally in the lab: What would you need to measure? What would you do with the data?

➢ Rates of Multi-Step Reactions:

o If there is a slow step, the slow step always determines the rate.

o If the slow step is first, the rate law can be written as if it were the only step.

o If the slow step is second (based on a few assumptions that are beyond the MCAT) the rate law can still be written as the rate law for the slow step, with the product of the first step being treated as a reactant in the second (slow) step.

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➢ Catalysts: A catalyst is any substance that increases reaction rate without itself being consumed in the process.

o Q36. What do catalysts change? What do they NOT change?

o Q37. What is the difference between an enzyme and a catalyst? Are all catalysts enzymes? Are all enzymes catalysts?

o Rate Laws for Catalyzed Reactions:

▪ Technically, the rate law is the sum of the rate law for the uncatalyzed reaction and the rate law for the catalyzed reaction. This fact, however, can usually be ignored, and you can assume that the rate law for the reaction is exactly equal to the rate law for the catalyzed reaction alone. Under such an assumption, write the rate law in the same way as normal, with the concentration of the

catalyst added in as a reactant. The rate constant, k, is sometimes replaced with kcat.

o Q38. Why can you usually ignore the rate law for the uncatalyzed reaction?

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Sample MCAT Questions

A student submits the following partial data table for a lab project in which students analyzed the reaction of

nitric oxide with hydrogen gas. It is known that the reaction is second order with respect to nitric oxide.

2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g)

5) The order of the above reaction with respect to hydrogen gas is:

A) zero order.

B) first order.

C) second order.

D) impossible to determine without additional information.

Solution: There are no trials across which [H2] changes, while [NO] remains constant. Normally, this is what we look for first

to predict rate order. However, we were given the information that the reaction is 2nd order with respect to NO. If we divide

the rate for Trial 1 by the rate for Trial 3 we see that the rate decreased by a factor of 64. Knowing the order of NO, we can say that when [NO] went down by a factor of 4, the rate should have gone down by a factor of 16. It actually went down by a

factor of 64, or 16*4. Thus the effect of the change in the [H2] must have accounted for this additional 4-fold decrease. Using

XY = Z we can see that if concentration went down by 4 and rate also went down by 4, the reaction must be first order with

respect to H2. It will help to recall from the manipulating equations section that if two variables are changed by factors x and y, respectively, the overall change to the equation is equal to x times y. B is the correct answer. This would be classified as a “hard” question on the MCAT. You will need to do well on these question types to earn a top score.

6) The lab instructor asks the student to estimate the expected rate for Trial 4. Compared to the rate of

Trial 3, Trial 4 should be approximately:

A) 64 times faster.

B) 128 times faster.

C) 256 times faster.

D) 1024 times faster.

Solution: Given the information in the stem and the answer to the previous question, we now know that the reaction is first

order with respect to hydrogen gas and second order with respect to nitric oxide. Between Trials 3 and 4 NO goes up by a factor of 8, which means the rate should go up by 82 or 64. Simultaneously H2, which is first order, goes up by 16. Thus, the overall reaction should go up by 16*64, or 1024. Answer D is thus correct.

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Sample MCAT Question