QJEEM23Cheb62: JEE Main 2023 Chemistry - Class 11 JEEM23Cheb62 Solution: For complete combustion of ethene. C 2 H 4 ( g ) + 3 O 2 ( g )…

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:

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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