JEE MAIN - Chemistry (2023 - 8th April Evening Shift - No. 18)
For complete combustion of ethene.
$$\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$
the amount of heat produced as measured in bomb calorimeter is $$1406 \mathrm{~kJ} \mathrm{~mol}^{-1}$$ at $$300 \mathrm{~K}$$. The minimum value of $$\mathrm{T} \Delta \mathrm{S}$$ needed to reach equilibrium is ($$-$$) _________ $$\mathrm{kJ}$$. (Nearest integer)
Given : $$\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$
Explanation
In the given combustion reaction, the number of moles of gaseous products ($\Delta n_g$) is -2 (since we have 2 moles of CO₂ gas on the product side and 4 moles of gaseous reactants).
The change in internal energy ($\Delta U$) is given as -1406 kJ/mol.
We can calculate the change in enthalpy ($\Delta H$) using the equation $\Delta H = \Delta U + \Delta n_gRT$, where $R$ is the ideal gas constant and $T$ is the temperature.
Substituting the given values:
$ \begin{aligned} \Delta H &= \Delta U + \Delta n_gRT \\\\ &= -1406 \mathrm{kJ/mol} + (-2) \times 8.3 \times 10^{-3} \mathrm{kJ/K/mol} \times 300 \mathrm{K} \\\\ &= -1406 \mathrm{kJ/mol} - 4.98 \mathrm{kJ/mol} \\\\ &\approx -1411 \mathrm{kJ/mol} \end{aligned} $
At equilibrium, $\Delta G = 0$, so $\Delta H = T \Delta S$, which gives $T \Delta S = \Delta H$.
Therefore, the minimum value of $T \Delta S$ needed to reach equilibrium is -1411 kJ
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