Chapter 17, Problem 17.125CP

(a)

Interpretation Introduction

To determine:

Calculate $\text{ΔG}$ at $25^{o}C$ and indicate whether the reaction is spontaneous in the forward or reverse direction.

Calculations:

Using equation, ${\text{ΔG=ΔG}}^{\text{o}}\text{+RTlnK}$

K= $\frac{[{\mathrm{H}}_3{O}^{+}(aq)][{\text{OH}}^{-}(\text{aq)}]}{[{\text{H}}_{\text{2}}\text{O(l)}]}$

$[{\text{H}}_{\text{2}}\text{O(l)}]=1$

So, K= ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}][{\text{OH}}^{-}(aq)]$

$$

  $$

So, $\text{ΔG}$ = $\text{79}\text{.9 KJ+8}{\text{.314×10}}^{\text{-}}{}^{\text{3}}{\text{ KJK}}^{-1}{\text{mol}}^{-1}(298)\ln (1.0\times 1.0)$

$\Delta G$ = 79.9 KJ/mole
As $\text{ΔG}$ >0, reaction is spontaneous in forward direction

(b)

Interpretation Introduction

To determine:

Calculate $\text{ΔG}$ at ${\text{25}}^{\text{o}}\text{C}$ and indicate whether the reaction is spontaneous in the forward or reverse direction.

Calculations:

Using equation, ${\text{ΔG=ΔG}}^{\text{o}}\text{+RTlnK}$

K= ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq)][OH}}^{\text{-}}\text{(aq)]}$

$$

$\text{ΔG}$ = $79.9\text{ KJ}+8.314\times {10}^{-}{}^3{\text{ KJK}}^{-1}{\text{mol}}^{-1}(298)\times \ln (1.0\times {10}^{-7}\times 1.0\times {10}^{-7})$

$\text{ΔG}$ = 0.028 KJ/mole
As $\text{ΔG}$ value is close to zero, reaction is in equilibrium(c)

Interpretation Introduction

To determine:

Calculate $\text{ΔG}$ at ${\text{25}}^{\text{o}}\text{C}$ and indicate whether the reaction is spontaneous in the forward or reverse direction.

Calculations:

Using equation, ${\text{ΔG=ΔG}}^{\text{o}}\text{+RTlnK}$

K= ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq)][OH}}^{\text{-}}\text{(aq)]}$

$$

$\text{ΔG}$ = $79.9\text{KJ}+8.314\times {10}^{-}{}^3{\text{KJK}}^{-1}{\text{mol}}^{-1}(298)\times \ln (1.0\times {10}^{-7}\times 1.0\times {10}^{-10})$

$\text{ΔG}$ = -890 KJ/mole
As value of $\text{ΔG}$ is less than 0, $\text{ΔG}$ is negative
So, reaction is spontaneous in forward direction
Equilibrium constant of the reaction:
${\text{ΔG=ΔG}}^{\text{o}}\text{+RTlnK}$

  $K={\text{e}}^{-\text{ΔG}}{}^{\text{/RT}}$

K= ${\text{e}}^{-(79.9{\text{ KJmol}}^{-1}/8.314\times {10}^{-3}\times 298)}$

K= $9.8\times {10}^{-15}$

K= $1.0\times {10}^{-14}$