Tutorial #1 - Unit 1 – Reactin Kinetics...Chemistry 12 Tutorial #1 – Reaction Kinetics 1 Tutorial #1 - Unit 1 – Reaction Kinetics Multiple Choice Explanations . The answer is

Chemistry 12 Tutorial #1 – Reaction Kinetics

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Tutorial #1 - Unit 1 – Reaction Kinetics Multiple Choice Explanations

The answer is B Reaction “A” looks simple but the diatomic bonds in H2 & I2 must be broken before the reaction can take place. This means there is a higher energy requirement (activation energy), so the reaction is fairly slow at room temp. The “at room temp” is important because this reaction could be very fast at a high temp. Reaction “B” is the answer because there are no bonds broken, the Ag+ and the CrO42-(aq) simply join together (precipitation reactions are always fast!) Reaction “C” is slow because of the solid (it is heterogeneous). Reactions can take place only on the surface of the lead. Reaction “D” is slow because covalent C-H bonds in CH4 must be broken. This requires more energy.


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The answer is C The higher the Temp. and the higher the [HCl], the faster the rate, so Rx. 3 is fastest (high temp. and high [HCl]), followed by Rx. 1 (high temp. and low [HCl]) and slowest is Rx. 2 (low temp. and low [HCl])

The answer is D “D” is the definition of Activation Energy. “A” is just Kinetic Energy, “B” is the actual energy of the AC, not an energy difference on the PE Graph, “C” of course is the ∆H.


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Ea (rev) = +10 kJ

∆H (rev) = -20 kJ

The answer is A The EA for the reverse reaction is shown by the blue arrow placed on the diagram above and the ΔH for the reverse reaction is shown by the red arrow. It is –20 because it’s going down from 130 to 110. The EA for the forward reaction would be 140 – 110 = 30 and the ∆H for the forward reaction would be 130 – 110 = +20. Look at the graph and make sure you know how these are obtained!


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The answer is B Increasing the temperature increases the average kinetic energy of the molecules so there are more collisions with a high kinetic energy (II) and if they are moving faster, there is also a greater probability (chance) of collisions (I). The potential energy barrier for a successful collision is called the activation energy, which cannot be altered by changing the temp. It can only be altered by using a suitable catalyst.

The answer is D A catalyst can change only the activation energy (the energy difference between the reactants or products and the activated complex). It cannot change the energy difference between the reactants and products (∆Η) or the potential energy of the reactants or products. Looking at “B”, only a change in temperature can change the kinetic energy.


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The answer is A Since N2 & O2 are the only things added, the concentrations of these are both high. Since these are the reactants, the rate of the forward reaction starts out fast. Since there is no product to begin with, the rate of the reverse reaction is initially zero. As time goes on and the system approaches equilibrium, N2 & O2 are used up, their concentrations decrease and the rate of the forward reaction decreases. In the meantime, and as N2 & O2 were reacting, NO2 was being formed, so the [NO2] would be increasing and the chance that one NO2 molecule could collide with another is increasing, therefore the rate of the reverse reaction would be increasing as the system approaches equilibrium. Remember the graph of forward and reverse reaction rate vs. time:

Forward rate

Reverse rate

equilibrium time


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The answer is C “Dynamic” means things are continuing to happen. In other words the forward and reverse reactions continue to occur. Remember “A” is not true because maximum randomness (entropy) is NOT always achieved at equilibrium. “B” and “D” are both characteristics of equilibrium but these do not explain why it’s called dynamic.

The answer is A In reaction “A”, there is 1 mole of gas on the left side and 2 moles of gas on the right, therefore entropy is increasing as the reaction proceeds. In “B”, entropy is decreasing because gases and forming a liquid, in “C”, 3 moles of gas are forming 2 moles of gas and in “D”, (aq) is going to solid.


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The Answer is B “B” is the fastest reaction because aqueous ions are highly mobile and more concentrated than molecules in a gas. The aqueous ions have a high probability of colliding. Also, when Ag+ and I- react, there are no bonds to break. They simply collide and bond to form the solid. (The activation energy is very low.) “A” is slow at room temperature because the covalent bonds in the diatomic molecules of H2 and I2 must be broken before they can react to form HI. This takes energy (fairly high activation energy), and is therefore slow at room temperature. In answer “C”, many C—C covalent bonds (in the sugar) must be broken (and also the diatomic bond in O2 ). Covalent bonds require a high activation energy and are therefore slow to break. Answer “D” is slow at room temperature because there are 23 reacting particles, therefore the reaction would require numerous steps and hence is quite slow. (“B” is a 2 particle collision which takes place in a single step).


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The Answer is A Rate is defined as ∆amount/∆time, which in this case would be ∆Volume CO2/∆time. Looking at the graph above, so can see that ∆Volume CO2/∆time is just the slope of the graph. The slope of the graph is steepest in the time interval between 0 and 1 minute. Therefore the rate of the reaction is fastest during this interval. This makes sense because generally as a reaction proceeds (starting with reactants), the rate starts out fast and then slows down as the reactants are used up and their concentration(s) decrease.

The slope is steepest in this


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The Answer is B The only thing that can change the average kinetic energy of molecules is a change in the temperature. Adding a catalyst simply decreases the activation energy (minimum energy required for a successful collision) by providing an alternate mechanism (pathway) for the reaction. Increasing the surface area allows for more possible collisions between reactant molecules in two different phases (heterogeneous). Increasing the concentration, again increases the probability of collisions but does not affect the kinetic energy.

The Answer is D Looking at the red markers on the graph above, the ∆H for the forward reaction is the net energy difference between the products and the reactants (arrow IV). Because PE or enthalpy is decreasing as the reaction proceeds from left to right, the value of ∆Hforward would be a negative number (exothermic). The ∆Hreverse

Reactants


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would be the same amount but a positive value (endothermic). Just to review, arrow I is Ea(forward), arrow II is really nothing because the starting point of the graph is arbitrary. (Only the energy differences among reactants, products and activated complex is relevant in these PE graphs.) Arrow III is Ea(reverse).

The Answer is A A catalyst only provides an alternate reaction mechanism which has a lower activation energy. The heat of reaction (∆H) is the net energy difference between the reactants and the products. The energies of the reactants and the products cannot be changed with a catalyst, so ∆H cannot be changed with a catalyst. Since a catalyst lowers both the Ea(forward) and the Ea(reverse) by the same amount, both the forward and the reverse reactions will be speed up by the catalyst. Therefore the equilibrium will not shift and the equilibrium constant (Keq) will remain the same. Remember, the only thing that can change the value of Keq is the temperature


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The Answer is D Remember that a reaction intermediate is produced in one step and used up in a later step. (So it is on the right in a step and on the left in a step somewhere below). If you look at the question above you will see that HOOBr and HOBr are both reaction intermediates. When the reaction intermediates are cancelled out and the rest of the equations are added, you obtain the overall reaction which shows the reactants and the products. Remember that a catalyst is used up (on the left) of one step and regenerated (on the right) in a later step.

HOOBr is an intermediate

HOBr is an intermediate

Overall reaction: 4HBr + O2 2 H2O + 2 Br2

HBr and O2 are reactants

H2O and Br2 are products


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The Answer is D According to the Chemistry 12 PLO’s from the Ministry: You can see that I, II and III in the box above are contained in the characteristics of equilibrium as described by the Ministry. Recall that concentrations of reactants and products are constant but NOT usually equal to each other. Macroscopic means observable.

It is expected that students will: D4. describe chemical equilibrium as a closed system at constant temperature:

1. whose macroscopic properties are constant 2. where the forward and reverse reaction rates are equal 3. that can be achieved from either direction 4. where the concentrations of reactants and products are constant


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The Answer is C For the answers “A” and “C” in which ∆H is negative, the reactions are exothermic and the tendency toward minimum enthalpy favors the products: Reaction “A” has 1 mole of gas (mog) in the reactants and 2 mog’s in the products, so products have maximum entropy. Reaction “C” has 3 mog’s in the reactants and 2 mog’s in the products, so maximum entropy favors the reactants. So in reaction “C” minimum enthalpy favors the products and maximum entropy favors the reactants so these two driving forces are opposing each other. If you check reaction “B” you will see that because it is endothermic, minimum enthalpy favors the reactants, and because it goes from 3 mog’s to 2 mog’s, maximum entropy also favors the reactants. In reaction “D”, minimum enthalpy favors the reactants (endothermic) and maximum entropy also favors the reactants. (10 mog’s in reactants and 9 mog’s in products.)

PE

reactants

products

Reaction proceeds

An exothermic reaction

Products have minimum enthalpy


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The Answer is C The ratios of all of the species present in the reaction is constant. Therefore you are able to calculate the RATE of production EXACTLY like calculating MOLES of one species given the moles of another species. 4.0x10-4 moles x 3 mol H2 = 6.0 x10-4 moles 2 mol NH3


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The Answer is B The key to this question is the OPEN SYSTEM.

I. Balance - can measure weight – the weight of the system will decrease because a gas is formed and will leave the system because it is open.

II. Pressure Gauge – will measure pressure. The P will NOT change because it is an open system. The gas is escaping.

III. pH meter - DO NOT look at H2 Gas. This will not affect pH. The [HCl] will decrease as the reaction proceeds forward, so the pH will increase.


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The Answer is C The heat of reaction (∆H) is the net energy difference between the reactants and the products and CAN NOT be changed by temperature. Increasing the temperature increases the average kinetic energy of the molecules so there are more collisions with a high kinetic energy (III) and if they are moving faster, there is also a greater probability (chance) of collisions (II).

The Answer is D The activated complex is the INTERMEDIATE complex that is TEMPORARILY formed from the combination of the reactants (unstable) . As the reactants approach each other they slow down because the e-‘s repel so their kinetic energy decreases. As the KE decreases the PE increases (high PE) .


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The Answer is B As stated many times before, as the T is increased the KE is increased, if the KE is increased then the speed at which the molecules are traveling is increased which will increased the number of collisions. If the number of collisions is increased then the reaction rate will increase and the time for the reaction to occur will decrease.

The Answer is D


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The Answer is D

The Answer is D


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The Answer is B


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Tutorial #1 - Unit 1 – Reaction Kinetics Student handout


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The slope is steepest in this interval


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Tutorial #1 - Unit 1 – Reaction KineticsTutorial #1 - Unit 1 – Reaction Kinetics

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Tutorial #1 - Unit 1 – Reactin Kinetics...Chemistry 12 Tutorial #1 – Reaction Kinetics 1 Tutorial #1 - Unit 1 – Reaction Kinetics Multiple Choice Explanations . The answer is