Partial pressure required for no reaction spontaneity

Question

Consider the following reaction:

CaO(s) CO2(g) → CaCO3(s); ∆G° = –130.9 kJ at 298 K

At what partial pressure of CO2(g) will the reaction no longer be spontaneous at 298 K? (R = 0.0821 L · atm/(K · mol) = 8.31 J/(K · mol))

A. 1.00 atm

B. 8.77 × 1022 atm

C. 1.59 × 108 atm

D. 1.14 × 10–23 atm

E. 6.28 × 10–9 atm

Heres another one that got me stuck for many tens of minutes... Im studying for an ACS final

Answer 1

We know a relation between equilibrium constant, Kp, and G is

∆G° = -RTlnK, where

R = 0.008314 kJ K^(-1) mol^(-1) <-- notice units of kJ for Gibb's free energy

T = temperature in Kelvins = 298K

K = equilibrium constant = Kp in this case

Thus,

∆G° = -RTlnK

-130.9 = -(0.008314)(298)lnK

e^52.83398484 = K

K = Kp = 8.820796689 × 10^(22)

Since we know Kp = 1/P(CO2), then

8.820796689E22 = 1/P(CO2)

P(CO2) = ~1.14E-23 atm

[Answer: D]