CHM1 16 Answers to End-of-Chapter Problems Collection

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Answers to Ch 16 Answers to End-of-Chapter Problems

**NOT COMPLETED | HOWEVER SOME OF THE PROBLEMS HAVE AN ANSWER**

1. `K_c = ([SO_3]^2)/([O_2][SO_2]^2)`

2. `K_c = ([NO]^2)/([O_2]^(0.5) [N_2 O])`

3. `K_c = ([Cu^(2+)])/([Ag^(+)]^2)`

4. `K_c = ([CO_2])/([CaCO_3])`

5. `K_c = [H_2 O][CO_2]`

2. `K_c: 24.5` `K_p: 1.002` Atm

3. `Q_c = 83.33 > K_c` Therefore the reaction shifts to the left.

4.`H_2 (g) + I_2 (g) -> 2HI (g)`

H2(g)+I2(g)→2HI(g)H2(g)+I2(g)→2HI(g)

5. `K_p = (P_(CO_2))/(P_(O_2))`

6.a) The original equation has been multiplied by 2, so K_c must be squared

`K_c^' = (K_c)^2 = (4.0 x 10^4)^2= 1.5 x 10^9`

b) The original equation has been multiplied by ½ (divided by 2) so K_c must be raised to the ½ power. The value of K_c’’ is the square root of the original K_c value.

Reaction:	HA	A^-	H₃O⁺
I	0.650 mol	0.000 mol	0.000 mol
C	-0.400 mol	+0.400 mol	+0.400 mol
E	0.250 mol	0.400 mol	0.400 mol

10.

11. Use the following formula: `DeltaG=-RT lnK`

`= 8.314 * 298 * ln(2.81*10^(-16)) = -8.87*10^5`

`= 8.871 kJ`

17. Q = 0.96 (Q > K), so the reaction will proceed to the left, and CO and H₂O will form.

18.

(1) Removing O₂ will decrease PO2, thereby decreasing the denominator in the reaction quotient and making Q_p >K_p. The reaction will proceed to the left as written, increasing the partial pressures of SO₂ and O₂ until Q_p once again equals K_p. (2) Removing O₂ will decrease PO2 and thus increase Q_p, so the reaction will proceed to the left. The partial pressure of SO₃ will decrease.
K_p and Q_p are both equal to PCO2. (1) Removing CO₂ from the system causes more CaCO₃ to react to produce CO₂, which increases PCO2 to the partial pressure required by K_p. (2) Adding (or removing) solid CaCO₃ has no effect on PCO2 because it does not appear in the expression for K_p (or Q_p).

19. a. The formation of NH₃ is exothermic, so we can view heat as one of the products. If the temperature of the mixture is decreased, heat (one of the products) is being removed from the system, which causes the equilibrium to shift to the right. Hence the formation of ammonia is favored at lower temperatures.

b. The decomposition of calcium carbonate is endothermic, so heat can be viewed as one of the reactants. If the temperature of the mixture is decreased, heat (one of the reactants) is being removed from the system, which causes the equilibrium to shift to the left. Hence the thermal decomposition of calcium carbonate is less favored at lower temperatures.