The equilibrium constant at 25^ C for the process CO^ 3+ (aq)+6 N… - Vidyadip

Question

The equilibrium constant at $25^{\circ} \mathrm{C}$ for the process $\mathrm{CO}^{3+}(\mathrm{aq})+6 \mathrm{NH}_3(\mathrm{aq}) \rightleftharpoons\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_6\right]^{3+}(\mathrm{aq})$ is $2.5 \times 10^6$. Calculate the value of $\Delta \mathrm{G}^{\circ}$ at $25^{\circ} \mathrm{C}\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$. In which direction is the reaction spontaneous under standard conditions?

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Answer

$\Delta\text{G}^\circ=-2.303\text{ RT}\log\text{K}$$=-2.303\times8.314\times298\log2.5\times10^6$
$=-5705.8[0.3980+6.0000]$ $[\log2.5=0.3980,\log10^6=6.0000]$
$=\frac{-5705.8\times6.3980}{1000}=-36.505\text{ kJ mol}^{-1}$
The reaction is spontaneous is forward direction under standard conditions because $\Delta\text{G}=-\text{ve}.$

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