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Step-by-Step Solution
Step 1: Identify the Given Data
• Combustion reaction of ethene:
C_{2}H_{4}(g) + 3\,O_{2}(g) \rightarrow 2\,CO_{2}(g) + 2\,H_{2}O(l)
• Heat released (in a bomb calorimeter) at 300 K: -1406\,\mathrm{kJ\,mol^{-1}}
• Gas constant, R = 8.3\,\mathrm{J\,K^{-1}\,mol^{-1}} (which is 8.3\times10^{-3}\,\mathrm{kJ\,K^{-1}\,mol^{-1}} )
Step 2: Calculate the Change in Moles of Gas ( \Delta n_{g} )
On the reactant side, we have a total of 4 moles of gas (1 mole of C_{2}H_{4} + 3 moles of O_{2} ).
On the product side, there are 2 moles of gas (2 moles of CO_{2} ).
Hence,
\Delta n_{g} = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 2 - 4 = -2.
Step 3: Relate the Internal Energy Change ( \Delta U ) and Enthalpy Change ( \Delta H )
For gaseous reactions at constant temperature, the relation between enthalpy change and internal energy change is:
\Delta H = \Delta U + \Delta n_{g}\,R\,T.
We know:
\Delta U = -1406\,\mathrm{kJ\,mol^{-1}}, \quad \Delta n_{g} = -2, \quad R = 8.3\times10^{-3}\,\mathrm{kJ\,K^{-1}\,mol^{-1}}, \quad T = 300\,\mathrm{K}.
Step 4: Compute \Delta H
Substitute the known values:
\begin{aligned}
\Delta H &= \Delta U + \Delta n_{g} \, R \, T \\
&= -1406\,\mathrm{kJ\,mol^{-1}} + (-2)\times \bigl(8.3\times10^{-3}\,\mathrm{kJ\,K^{-1}\,mol^{-1}}\bigr) \times (300\,\mathrm{K}) \\
&= -1406\,\mathrm{kJ\,mol^{-1}} - 4.98\,\mathrm{kJ\,mol^{-1}} \\
&\approx -1411\,\mathrm{kJ\,mol^{-1}}.
\end{aligned}
Therefore, \Delta H \approx -1411\,\mathrm{kJ\,mol^{-1}}.
Step 5: Determine T\Delta S at Equilibrium
At equilibrium, the Gibbs free energy change \Delta G is zero. Using the relation
\Delta G = \Delta H - T\Delta S = 0,
we get
T\,\Delta S = \Delta H.
Since \Delta H \approx -1411\,\mathrm{kJ\,mol^{-1}}, we have
T\Delta S = -1411\,\mathrm{kJ\,mol^{-1}}.
Thus, the minimum value of T\Delta S needed to reach equilibrium is approximately -1411\,\mathrm{kJ\,mol^{-1}}.
