1. Using the data given below, calculate the value of equilibrium… - Vidyadip

Question

Using the data given below, calculate the value of equilibrium constant for the reaction at 298K.

$3\text{CH}\equiv\text{CH}(\text{g})\rightleftharpoons\text{C}_6\text{H}_6(\text{g})$ assuming ideal gas behaviour, $\Delta_\text{f}\text{G}^\circ[\text{HC}\equiv\text{CH(g)}]=2.09\times10^5\text{J mol}^{-1},$ $\Delta_\text{f}\text{G}^\circ[\text{C}_6\text{H}_6(\text{g})]=1.24\times10^5\text{J mol}^{-1},$ R = 8.314JK^-1 mol^-1.

Based on your calculated value, comment whether this process can be recommended as a practical method for making benzene.

✓

Answer

$\Delta\text{G}^\circ=\Delta_\text{f}\text{G}^\circ[\text{C}_6\text{H}_6(\text{g})]-3\Delta_\text{f}\text{G}^\circ[\text{CH}\equiv\text{CH(g)}]$

$=1.24\times10^5\text{J mol}^{-1}-3\times2.09\times10^5\text{J mol}^{-1}$

$=1.24\times10^5\text{J mol}^{-1}-6.27\times10^5\text{J mol}^{-1}$

$=-5.03\times10^5\text{J mol}^{-1}$

$-5.03\times10^5=-2.303\times8.314\times298\log\text{K}$

$\Rightarrow\log\text{K}=\frac{5.03\times10^5\text{J}}{5705.8\text{J}}=+88.1559$

$\Rightarrow\text{K}=1.432\times10^{88}$

Yes, this process can be recommended as a practical method for making benzene as value of K is very high.

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