Question
For the auto-ionization of water at $${25^ \circ }C,$$ $${H_2}O\left( l \right) \rightleftharpoons {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right)$$ equilibrium constant is $${10^{ - 14}}.$$
What is $$\Delta {G^ \circ }$$ for the process?
A.
$$ \simeq 8 \times {10^4}\,J\,mo{l^{ - 1}}$$
B.
$$ \simeq 3.5 \times {10^4}\,J\,mo{l^{ - 1}}$$
C.
$$ \simeq 2 \times {10^4}\,J\,mo{l^{ - 1}}$$
D.
$${\text{None of these}}$$
Answer :
$$ \simeq 8 \times {10^4}\,J\,mo{l^{ - 1}}$$
Solution :
$$\Delta {G^ \circ } = - RT\,\,{\text{ln}}K$$
$$ = - 8.314J{K^{ - 1}}\,mo{l^{ - 1}} \times $$ $$298\,K \times 2.303\,{\text{log}}\,{10^{ - 14}}$$
$$ = 7.98 \times {10^4} \simeq 8 \times {10^4}\,J\,mo{l^{ - 1}}$$