Gases and Limiting Reagents Class Example: The Haber process produces ammonia, NH 3, for fertilizer manufacturing N 2 (g) + 3H 2 (g) → 2NH 3 (g) (a)Find

Gases and Limiting ReagentsClass Example: The Haber process produces ammonia, NH3, for fertilizer manufacturing

N2(g) + 3H2(g) → 2NH3(g)

(a) Find the limiting reagent when 2.22 L of N2(g) at STP is reacted with 2.22 L of H2(g) at 1150 mm Hg at 19.5 oC.

(b) If 4.00 L of both N2(g) and H2(g) at STP are reacted what volume of NH3(g) would be produced at STP?

Gases and Limiting Reagents – NH3(g)• Class example – Part (b): 4.00 L of N2(g) at STP and

4.00 L of H2(g) at STP contain the same number of moles of gas. Why? Since H2(g) is consumed three times as quickly as N2(g) in the reaction

• N2(g) + 3H2(g) → 2NH3(g)

• necessarily H2(g) is the limiting reagent. If the reaction goes to completion the number of moles of 2NH3(g) formed would be two thirds the initial number of moles of H2(g).

Gases and Limiting Reagents – NH3(g)

• If the reaction goes to completion the number of moles of NH3(g) formed would be two thirds of the initial number of moles of H2(g). The volume of NH3(g) formed would be two thirds of the initial volume of H2(g). (Avogadro’s Hypothesis).

• Volume NH3(g) = 4.00 L H2(g) x

• = 2.66 L NH3(g)

Gases and Limiting Reagents – NH3(g)• Alternate approach: We could use more

familiar calculations based on moles of reactants and products. With reactants at STP

• Moles H2(g) = 4.00 L H2(g) x

• = 0.176 mol = Initial # moles N2(g)

• To get the moles of N2(g) reacting

• 0.176 mol H2(g) x = 0.0587 mol N2(g)

Gases and Limiting Reagents – NH3(g)• Aside: The number of moles NH3(g) formed:• 0.176 mol H2(g) x = 0.117 mol NH3(g)

Moles N2(g) Moles H2(g) Moles NH3(g)

Intial 0.176 0.176 0.000

Changes -0.0587 -0.176 0.117

Final 0.117 0.000 0.117

Mass N2(g) Mass H2(g) Mass NH3(g)

Initial 4.93 0.355 0.000

Changes -1.65 -0.355 2.00

Final 3.28 0.000 2.00

Gas Stoichiometry – Text Example• Stoichiometric calculations can involve gases,

solids and solutions. A simple example from the text involves the decomposition of sodium azide – a very rapid reaction used in automobile air bags. The very rapid production of nitrogen gas quickly inflates the air bags in the car. (Subsequent reactions produce silicate glass.)

Copyright 2011 Pearson Canada Inc. 6 - 7

Stoichiometry and Dalton’s Law• Dalton’s Law can be useful when a reaction

produces a mixture of gases. It cane also be needed when a gas produced by a chemical reaction is collected over water. In such cases the gas collected will contain water vapor.

• Then: PWet Gas = PDry Gas + Pwater Vapor

• The vapor pressure of water increases with T.



Kinetic Theory of Matter• Observations:• All gases exert pressure on the walls of their

container.• The pressure exerted by a fixed amount of gas

in a rigid container increases steadily as the gas temperature rises (Charles’s Law).

• At a given T equal amounts (moles) of different gases with the same T and V exert the same pressure.

Kinetic Theory of Matter• The observations on the previous slide can be

explained if we assume that:• Gas molecules are in rapid translational

motion.• Gas molecules move more quickly as the T

increases. (Average kinetic energy α Temp.)• At a given T the average kinetic energy of He

and Ar atoms (for example) are the same. (He atoms move faster? Why?) Molecules?

Kinetic Theory of Gases• Absolute zero (0 Kelvin): molecules cease to

rotate or translate (zero kinetic energy).• Finite temperatures ( > 0 Kelvin): increase T

and gas molecules translate (and rotate) with higher and higher average energies. At a given T a distribution of molecular velocities (and, correspondingly, kinetic energies!) is seen.


Distribution of Molecular Speeds – the effect of mass and temperature

Figure 6-16

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6 Slide 15 of 41

M

3RTu rms

6-8 Gas Properties Relating to the Kinetic-Molecular Theory


•Diffusion–Net rate is proportional to molecular speed.

•Effusion–A related phenomenon.

Graham’s Lawrate of effusion of A

rate of effusion of B(urms )A

(urms )B

3RT/MA

3RT/MB MB

MA


• Only for gases at low pressure (natural escape, not a jet).• Tiny orifice (no collisions)• Does not apply to diffusion.

Ratio used can be:· Rate of effusion (as above)· Molecular speeds· Effusion times

· Distances traveled by molecules· Amounts of gas effused.

Graham’s Law Demonstration• Graham’s Law tells us that light gaseous

molecules move more quickly (higher average velocity) than heavy molecules. This is sometimes demonstrated (inadvertently) in chemistry labs when bottles of concentrated HCl(aq) and NH3(aq) are left often in close proximity. The following reaction cane be seen

• HCl(g) + NH3(g) → NH4Cl(s) Phase change!


Gas Diffusion• In the previous slide the solid ammonium

chloride, NH4Cl(s), forms closer to the opening of the HCl(aq) bottle than the NH3(aq) bottle. This tells us that the lighter ammonia molecules diffuse through the atmosphere faster than the heavier hydrogen chloride molecules.



Visualizing Molecular MotionFigure 6-14


6-7 Kinetic Molecular Theory of Gases

• Particles are point masses in constant,

random, straight line motion.

• Particles are separated by great

distances.

• Collisions are rapid and elastic.

• No force between particles.

• Total energy remains constant.

Gas Nonideality

• Class discussion of evidence for.

Documents

Gases and Limiting Reagents Class Example: The Haber process produces ammonia, NH 3, for fertilizer manufacturing N 2 (g) + 3H 2 (g) → 2NH 3 (g) (a)Find