Chapter 10 Homework Answers

Transcription

1 Chapter 10 Homework Answers 10-5 The ratio of the equilibrium concentrations of trans--butene to cis--butene at 400 C is 1.7 according to the data in section 10.. This ratio will always have this value no matter how much material is present, so long as the system is at equilibrium at 400 C The rate law describes how the rate of the reaction is dependent upon the concentrations of the various compounds in the reaction. The rate law does not tell you anything about the equilibrium ratio of reactants and products The concentrations of N and H will decrease while the concentration of NH 3 will increase until a time when no further change will occur. Beyond this point in time, the system is at equilibrium. The concentration of H will decline three times as rapidly as that of N while the concentration of NH 3 will increase twice as rapidly as N declines. Concentration (M) Kinetic Region N H Equilibrium Region NH3 Time Sketch of changes in the concentration of reactants and products with time for the reaction of N (g) + 3 H (g) NH 3 (g). The initial concentration [N = [H = 1.0 M. No NH 3 is initially present No. At equilibrium both the forward and reverse reactions continue, but at the same rate (a) K c = [H O [N (b) K [NO [H c = [NO [Cl (c) K [NOCl c = [ NO [ NO [ O 10-4 K c (c) = K c (a) x K c (b) = [ CO 1 [ CO [ O = (1.1 x ) x (7.1 x 10-1 ) = 0.78 x [H [ O 1 [ H O = [CO [H [ CO [H O

2 [HI 10-6 K c = [H [I ( ) trial I K c = = ( ) ( ) trial II K c = trial III K c = K c is a constant. ( ) ( ) ( ) = ( ) ( ) ( ) = When Q c > K c The reaction must shift to the left to reach equilibrium. Answer (c) NH 3 (g) N (g) + 3 H (g) - C + C +3 C C = 0.34 C = (N ) = (H ) = N (g) + 3 H (g) NH 3 (g) initial M M 0 change - C -3 C C equilibrium 0.9 M (0.078) (0.078) [N eq = 0.9 M [H eq = M [NH 3 eq = M K c = [ NH 3 [ N [ H 3 = (0.156) (0.9)(0.766) 3 = N O 4 (g) NO (g) K c =5.8 x 10-5 initial M change C - C equilibrium C 1.00 M - C This reaction must proceed to the left to form N O 4. In fact, the equilibrium constant for NO (g) N O 4 (g) is 1/K c = 1.7 x Thus, most of the NO will react to produce N O 4 and [N O 4 will be about 0.50 M.

3 10-49 PCl 5 (g) PCl 3 (g) + Cl (g) K c = at 450K initial 1.00 M 0 0 change - C C C equilibrium C C C C 1.00 C = Assume C<<1.00 C = C = Check, C is 3.6% of the initial concentration. The assumption that C<<1.00 is valid. [PCl 5 eq = 0.96 M [PCl 3 eq = [Cl eq = M % decomposition of PCl 5 = x 100% = 3.6% The difference in the % decomposition of PCl 5 in this reaction carried out at 450 K and the same reaction carried out at 50 K is due to the different values of K c at the different temperatures. There is less tendency for product formation in this reaction system at the higher temperature Equilibrium constants will change with temperature depending upon the nature of the reaction involved Many different examples can be given to answer this question. The following are two examples. Consider a room filled with people. The stress of being crowded-in is alleviated when some people leave the room. Consider the fatigue and irritability associated with a lack of sleep. Increasing the amount of sleep decreases the fatigue and helps to bring the person back into equilibrium with themselves and their surroundings When the pressure changes on a reaction with gaseous components which was initially in a state of equilibrium, the reaction system shifts in the direction which will best offset the externally applied pressure change. A decrease in pressure will favor the shift in reaction direction which generates more gaseous components. An increase in pressure will cause the reaction to shift in the direction which has the smaller amounts of gaseous components. With a decrease in pressure for the reactions listed: (a) The reaction will shift to the right in the direction of formation of NO. (b) The reaction will not shift because the number of moles of gaseous products is the same as the number of moles of gaseous reactants. (c) The reaction will shift to the left in the direction of the decomposition of N O 3.

4 10-66 A decrease in concentration of the underlined component of the reactions initially at equilibrium will have the effect of shifting the reaction system to decrease the concentration of all components of the other side of the reaction equation. The concentrations of all components of the side where the concentration of a component was decreased will tend to increase. (a) Decreasing the concentration of NO(g) will cause the reaction to shift to the right, forming more NO. (b) Decreasing the concentration of O (g) will cause the reaction to shift to the left, forming more O (g). (c) Decreasing the concentration of F (g) will cause the reaction to shift to the left, forming more Cl (g) An increase in the volume of the container by a factor of would decrease the concentrations of all components of the reaction and also the pressure of the system. 4 NH 3 (g) + 5 O (g) 4 NO(g) + 6 H O(g) The decreased pressure would shift the equilibrium to the right in the direction of the greater number of moles of gaseous products.

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