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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics5 Marks
Question
The standard Gibbs energy change for the reaction, $\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$ is -33.2kJ $mol^{-1}$ at 298K.
  1. Calculate the equilibrium constant for the above reaction.
  2. What would be the equilibrium constant if the reaction is written as $\frac{1}{2}\text{H}_2(\text{g})+\frac{3}{2}\text{H}_2(\text{g})\rightleftharpoons\text{NH}_3(\text{g})$
  3. What will be the equilibrium constant if the reaction is $\text{NH}_3(\text{g})\rightleftharpoons\frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2(\text{g})$
✓

Answer

$\log\text{K}=-\frac{\Delta\text{G}}{2.303\text{RT}}$
$\therefore\log\text{K}=-\frac{-33.2\times10^3\text{J mol}^{-1}}{2.303\times(8.314\text{JK}^{-1}\text{mol}^{-1})\times(298\text{K})}$
$=5.82\ \text{or K}=6.6\times10^5$
For the given reaction, $\Delta\text{G}^\circ=\frac{1}{2}\times(-33.2)=-16.6\text{kJ mol}^{-1}$
$\log\text{K}=-\frac{-16.6\times10^3\text{J mol}^{-1}}{2.303\times(8.314\text{JK}^{-1}\text{mol}^{-1})\times(298\text{K})}$
$=2.91\ \text{or K}=8.1\times10^2$
For the given reaction $\text{NH}_3(\text{g})\overrightarrow{\ \ \ \ \ }\ \frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2(\text{g})$
$\Delta\text{G}^\circ=-[-16.6\text{kJ mol}^{-1}]=16.6\text{kJ mol}^{-1}$
$\log\text{K}=-\frac{16.6\times10^3\text{JK}^{-1}\text{mol}^{-1}}{2.303\times(8.314\text{JK}^{-1}\text{mol}^{-1})\times(298\text{K})}$
$=-2.91\text{ or K}=1.23\times10^{-3}$
 

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