Page 253 - Theory and Problems of BEGINNING CHEMISTRY

242 RATES AND EQUILIBRIUM [CHAP. 16

Supplementary Problems

16.23. Compare the function of a catalyst in a chemical reaction with that of a marriage broker—a matchmaker.
Ans. They both facilitate an effect without permanent change in themselves.

16.24. Identify or explain each of the following terms: (a) equilibrium, (b) rate of reaction, (c) catalyst, (d) completion,
(e)LeChˆatelier’s principle, ( f ) stress, (g) shift, (h) shift to the right or left, (i) equilibrium constant, and (j) equilibrium
constant expression.
Ans. (a) Equilibrium is a state in which two exactly opposite processes occur at equal rates. No apparent change
takes place at equilibrium. (b) Rate of reaction is the number of moles per liter of reactant that reacts per unit
time. (c) A catalyst is a substance that alters the rate of a chemical reaction without undergoing permanent
change in its own composition. (d) A reaction goes to completion when one or more of the reactants is
entirely used up. In contrast, a reaction at equilibrium has some of each reactant formed again from products,
and no reactant or product concentration goes to zero. (e)LeChˆatelier’s principle states that when a stress
is applied to a system at equilibrium, the equilibrium tends to shift to reduce the stress. ( f ) Stress is a
change of conditions imposed on a system at equilibrium. (g) Shift is a change in the concentrations of
reactants and products as a result of a stress. (h) Shift to the right is production of more products by using
up reactants; shift to the left is exactly the opposite. (i) An equilibrium constant is a set value of the ratio of
concentrations of products to concentrations of reactants, each raised to the appropriate power and multiplied
together. (j) An equilibrium constant expression is the equation that relates the equilibrium constant to the
concentration ratio.

16.25. Nitrogen and sulfur oxides undergo the following rapid reactions in oxygen. Determine the overall reaction. What is
the role of NO?

2NO + O 2 −→ 2NO 2
NO 2 + SO 2 −→ NO + SO 3
Ans. Combining the first reaction with twice the second yields the following reaction:
2SO 2 + O 2 −→ 2SO 3
That reaction will proceed only slowly in the absence of the NO and NO 2 . The NO is a catalyst. It speeds
up the conversion of SO 2 to SO 3 and is not changed permanently in the process.

16.26. State, if possible, what happens to the position of the equilibrium in each of the following cases.
−→
N 2 + 3H 2 ←− 2NH 3 + heat
(a) Heat is added and ammonia is added. (b) Nitrogen is added and the volume is reduced. (c) Ammonia is taken out
and a catalyst is added. (d) Nitrogen and ammonia are both added.
Ans. (a) Addition of ammonia and addition of heat will each cause the equilibrium to shift to the left; addition
of both will also cause shift to the left. (b) Each stress—addition of nitrogen and increase in pressure (by
reduction of the volume)—causes a shift to the right, and so the equilibrium will shift to the right when both
stresses are applied. (c) Addition of a catalyst will cause no shift, but removal of ammonia will cause a
shift to the right, and so these stresses together will cause a shift to the right. (d) You cannot tell, because
addition of the nitrogen would cause a shift to the right and addition of ammonia would cause a shift to the
left. Since no data are given and Le Chˆatelier’s principle is only qualitative, no conclusion is possible.

16.27. What is the best set of temperature and pressure conditions for the Haber process—the industrial process to convert
hydrogen and nitrogen to ammonia?
Ans. To get the equilibrium to shift as much as possible toward ammonia, you should use a high pressure and a
low temperature. However, the lower the temperature, the slower the reaction. Therefore, a high pressure,
200 atm, and an intermediate temperature, 500 C, are actually used in the industrial process. A catalyst to
◦
speed up the reaction is also used.
16.28. Consider a change in total volume of the equilibria described. In each case, calculate K for a 1.0-L total volume and
again for a 2.0-L total volume, and explain in each case why there is or is not a difference.