Question

Consider the auto ionization of water at 25°C:
$H_{2}O\left(l\right)\iff H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)K_w=1.00\times {10}^{-14}$
a. Calculate $\Delta G^o$for this process at 25°C.
b. At${40}^{o}C$,$K_w=2.92\times {10}^{-14}$ ,. Calculate $\Delta G^o$at ${40}^{o}C$.

Short answer

1. The$\Delta G°$ at 25°C is$79.6KJ/mol$
2. The $\Delta G°$at 40°C is$81.6KJ/mol$

Step-by-step solution

Step 1: Introduction to the Concept

The Gibbs free energy change and the equilibrium constant have the following relationship:

$\Delta G^o=-RTlnK$

$\Delta G=\Delta G^o+RTlnK$

Where,

$\Delta G=$ Gibbs free energy change

$\Delta G°=$Gibbs free energy change at standard conditions

R = Universal gas constant ()

T=Temperature

K= Equilibrium constant

Step 2: Determination of  at 25°C

a)

Here the given reaction is,

$H_{2}O\left(l\right)\iff H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)K_w=1.00\times {10}^{-14}$

From equation (1),

$\Delta G_R^o=-R\times T\times ln\times K$

$∆{G^{\circ }}_R=-0.008314KJ/mol-K\times \left(25+273\right)K\times \ln \left(1.00\times {10}^{-14}\right)$

$∆{G^{\circ }}_R=79.9KJ/mol$

Step 3: Determination of  at 40°C

b)

Here the given reaction is,

$H_{2}O\left(l\right)\iff H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)K_w=2.92\times {10}^{-14}$

From equation (1),

role="math" localid="1651131750136" $\Delta G_R^o=-R\times T\times ln\times K$

$∆{G^{\circ }}_R=-0.008314KJ/mol-K\times \left(40+273\right)K\ln \left(2.92\times {10}^{-14}\right)$

$∆{G^{\circ }}_R=81.1KJ/mol$

Therefore $\Delta G°$at 25°C is determined as $79.9KJ/mol$ and $\Delta G°$ at 40°C is determined as$81.1KJ/mol$