Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

Using the Gibbs energy change, ΔG∘=+63.3kJ\Delta G^\circ = +63.3kJ , for the following reaction, Ag2CO3⇌2Ag(aq)++CO3(aq)2−Ag_2CO_3 \rightleftharpoons 2Ag^+_{(aq)}+CO^{2^−}_{3(aq)} the KspK_{sp} of Ag2CO3(s)Ag_2CO_3(s) in water at 25∘25^∘C is (R=8.314JK−1mol−1R=8.314JK^{−1}mol^{−1})

A

3.2×10−263.2 \times 10^{-26}

B

8.0×10−128.0 \times 10^{-12}

C

2.9×10−32.9 \times 10^{-3}

D

7.9×10−27.9 \times 10^{-2}

Answer

8.0×10−128.0 \times 10^{-12}

Explanation

Solution

ΔG∘ΔG^\circ is related to KspK_{sp} by the equation
ΔG∘=−2.303RT  log⁡  KspΔG^\circ = - 2.303RT\, \, \log\, \, K_{sp}
Given, ΔG∘=+63.3KJΔG^\circ = + 63.3 KJ
=63.3×103J= 63.3 \times 10^3 J
Thus, substitute ΔG∘=63.3×103J, ΔG^\circ = 63.3 \times 10^3 J ,
R=8.314JK−1mol−1R = 8.314 JK^{-1} mol^{-1} and T= 298 K [25 + 273 K]
from the above equation to get
63.3×103=−2.303×8.314×298  log⁡  Ksp63.3 \times 10^3 = -2.303 \times 8.314 \times 298\, \, \log\, \, K_{sp}
∴logKsp=−11.09\therefore log K_{sp} = -11.09
⇒Ksp=antilog⁡(−11.09)\Rightarrow K_{sp} = anti \log(-11.09)
    Ksp=8.0×10−12\, \, \, \, K_{sp} = 8.0 \times 10^{-12}