Question

For the reaction${\mathbf{HCO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\rightleftharpoons }{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{"}{\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$, $\mathbf{\Delta G}\mathbf{°}\mathbf{=}\mathbf{+}\mathbf{59}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{}$at $\mathbf{298}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{K}$ . Find the value of Kfor the reaction.

Short answer

The value of K for the given reaction is $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}$.

Step-by-step solution

Equilibrium constant

Equilibrium constant (K): In the equilibrium reaction, the ratio is between the concentration of the reactant and the concentration of the product. If K's value is smaller than 1, the reaction should be moved to the left; if K is more than 1, the reaction should be moved to the right.

$\mathrm{\Delta G}°=-\mathrm{RT}\ln \mathrm{K}$

$\mathrm{K}={\mathrm{e}}^{\left(-∆\mathrm{G}°/\mathrm{RT}\right)}$

Here,

$∆\mathrm{G}°$is the change in free energy.

K is the equilibrium constant.

R is Gas constant 8.314

T is Temperature.

The value of K for the given reaction.

The given reaction is:

${\mathrm{HCO}}_3^{-}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{CO}}_3^{2-}\hspace{1em}\mathrm{\Delta G}°=+59.0\mathrm{kJ}/\mathrm{mol}$

Temperature $=298.15\mathrm{K}$

$$\begin{gathered} K=e^{\left(-∆G°/RT\right)} \\ =e^{-\left(59.0\times {10}^{3}J/mol\right)/\left(8.314472J/K.mol\right)\left(298.15K\right)} \\ =e^{-\left(59.0\times {10}^{3}J/mol\right)/2478.9598J/mol} \\ =e^{-\left(23.80\right)} \\ =\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}} \end{gathered}$$