Using the Gibbs change, ∆ G ^ o = + 63.3kJ, for the following reaction, A g _ 2 C O _ 3 (g) ⇌ 2A g ^ + (aq) + C O _ 3 ^ 2 - (aq) the K _ sp of A g _ 2 C O _ 3 (s) in water at 25 ^ o C is (R = 8.314J K ^ - 1 mo l ^ - 1 )

Correct answer: B. 8.0 × 10 ^ - 12 ∆ G ^ o = - 2.303RTlog K _ sp 63.3 × 10 ^ 3 = - 2.303 × 8.31 × 298log K _ sp - 11.09 = log K _ sp 8 × 10 ^ - 12 = K _ sp

NEET 2014

Exam: NEET
Year: 2014
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2014 · Chemistry · Thermodynamics (C)

Using the Gibbs change, ∆ G ^ o = + 63.3kJ, for the following...

Using the Gibbs change,

∆ {G}^{o} = + 63.3kJ,

for the following reaction,

A{g}_{2}C{O}_{3}\left(g\right)

⇌ 2A{g}^{+}\left(aq\right)

+ C{O}_{3}^{2 -}\left(aq\right)

the {K}_{sp} of A{g}_{2}C{O}_{3}\left(s\right)

in water at

{25}^{o}C is \left(R = 8.314J{K}^{- 1}mo{l}^{- 1}\right)

Options

A $3.2 \times {10}^{- 26}$
B $8.0 \times {10}^{- 12}$
C $2.9 \times {10}^{- 3}$
D $7.9 \times {10}^{- 2}$

Correct answer

B. $8.0 \times {10}^{- 12}$

Step-by-step solution

$∆ {G}^{o} = - 2.303RTlog {K}_{sp}$

$63.3 \times {10}^{3} = - 2.303 \times 8.31 \times 298log {K}_{sp}$

$- 11.09 = log {K}_{sp}$

$8 \times {10}^{- 12} = {K}_{sp}$

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