Question

For complete combustion of ethene,
\(C_2H_4(g) + 3O_2(g) \longrightarrow 2CO_2(g) + 2H_2O (l)\)
the amount of heat produced as measured in bomb calorimeter is 1406 kJ mol^-1 at 300K. The minimum value of \(T\Delta S\) needed to reach equilibrium is (-) _____ . kJ.(Nearest integer)
Given: R = 8.3 J K^-1 mol^-1

Hint:

Always include the ∆n_gRT correction term in bomb calorimetry problems when calcu lating thermodynamic equilibrium parameters

Answer 1

Correct Answer: 1411

The thermodynamic relationship is given by: 

\( \Delta G = \Delta H - T\Delta S \)

At equilibrium, \( \Delta G = 0 \), so:

\( T\Delta S = \Delta H - \Delta n_gRT \)

Given Data:

• \( \Delta H = -1406 \, \text{kJ mol}^{-1} \)
• \( R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1} = 0.0083 \, \text{kJ K}^{-1} \text{mol}^{-1} \) (converted to kJ for consistency)
• \( T = 300 \, \text{K} \)

Step 1: Calculate \( \Delta n_g \):

The reaction gives:

\( \Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} \)

From the reaction:

\( \Delta n_g = 2 - \frac{7}{2} = -\frac{3}{2} \)

Step 2: Substitute Values into \( T\Delta S = \Delta H - \Delta n_gRT \):

\( T\Delta S = -1406 + \left(-\frac{3}{2} \cdot 0.0083 \cdot 300 \right) \)

Step 3: Simplify:

\( T\Delta S = -1406 + (-3.735) \)

\( T\Delta S \approx -1409.735 \, \text{kJ} \)

Step 4: Round to the Nearest Integer:

\( T\Delta S \approx -1411 \, \text{kJ} \)

Conclusion:

The minimum value of \( T\Delta S \) needed to reach equilibrium is: \( 1411 \, \text{kJ}. \)