Question
Using the Gibbs energy change, ΔG0 = +63.3 KJ for the following reaction,
Ag2CO3 (s) r2Ag+ (aq) + CO32- (aq)
the Ksp of Ag2CO3 (s) in water at 250 C is (R= 8.314 JK-1 mol-1)
-
3.2 x 10-26
-
8.0 x 10-12
-
2.9 x 10-3
-
7.9 x 10-2
Solution
B.
8.0 x 10-12
ΔG0 is related to Ksp by the equation,ΔG0 = --2.303 RT log Ksp
We have Given,
ΔG0 = +63.3 KJ = 63.3 x 103 J
Thus, substitute ΔG0 = 63.3 x 103 J
R = 8.314 JK-1 mol-1
and T =298 K [ 25 +273 K] into the above equation to get,
63.3 x 103 = -2.303 x 8.314 x 298 log Ksp
log Ksp = -11.09
Ksp = antilog (-11.09)
Ksp = 8.0 x 10-12
