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 : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

<p>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). </p> <p>The change in internal energy ($\Delta U$) is given as -1406 kJ/mol.<br/><br/> 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.</p> <p>Substituting the given values:</p> <p>$ \begin{aligned} \Delta H &amp;= \Delta U + \Delta n_gRT \\\\ &amp;= -1406 \mathrm{kJ/mol} + (-2) \times 8.3 \times 10^{-3} \mathrm{kJ/K/mol} \times 300 \mathrm{K} \\\\ &amp;= -1406 \mathrm{kJ/mol} - 4.98 \mathrm{kJ/mol} \\\\ &amp;\approx -1411 \mathrm{kJ/mol} \end{aligned} $</p> <p>At equilibrium, $\Delta G = 0$, so $\Delta H = T \Delta S$, which gives $T \Delta S = \Delta H$. <br/><br/>Therefore, the minimum value of $T \Delta S$ needed to reach equilibrium is -1411 kJ</p>