Question

14. Consider the following reaction:
2AgCl(s)  2Ag(s) + Cl2(g) H° = 127.1 kJ; S° = 115.7 J/K at 298 K
Suppose 41.0 g of silver(I) chloride is placed in a 70.0 L vessel at 298 K. When
equilibrium is reached, what will the partial pressure of chlorine gas be?
(R = 0.0821 L · atm/(K · mol) = 8.31 J/(K · mol))
A) 0.050 atm B) 5.1  10–23 atm C) 0.95 atm
D) 0.10 atm E) 5.7  10–17 atm

Answer 1

First we can calculate the dG by using

dG = dH – TdS

dG = 127.1 – 298 * 0.115.7

dG = 92.62 kJ

now calculate the K

∆Gr° = - R∙T∙ln(K)
92.62*1000 = -8.3145 *298 lnK

ln K = -37.38

K = 5.84*10^-17

The correct answer is choice E

since, Kp = partial pessure of Cl2

Ag and AgCl would not be included since they are solids, Kp value is very low so it indicates forward reaction. also, originally there was no Cl2, so the reaction would be moving in forward direction.

First we can calculate the dG by using

dG = dH – TdS

dG = 127.1 – 298 * 0.115.7

dG = 92.62 kJ

now calculate the K

∆Gr° = - R∙T∙ln(K)
92.62*1000 = -8.3145 *298 lnK

ln K = -37.38

K = 5.84*10^-17

The correct answer is choice E

since, Kp = partial pessure of Cl2

Ag and AgCl would not be included since they are solids, Kp value is very low so it indicates forward reaction. also, originally there was no Cl2, so the reaction would be moving in forward direction.