Question

Standard entropies of \(X_2\), \(Y_2\) and \(XY_5\) are 70, 50, and 110 J \(K^{-1}\) mol\(^{-1}\) respectively. The temperature in Kelvin at which the reaction \[ \frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5 \quad \Delta H = -35 \, {kJ mol}^{-1} \] will be at equilibrium is (nearest integer):

Hint:

For equilibrium calculations, use the relation \( \Delta G = \Delta H - T \Delta S \). At equilibrium, \( \Delta G = 0 \), so solving for \( T \) gives the temperature at equilibrium.

Answer 1

Step 1: Calculate the standard entropy change (\( \Delta S \)) for the reaction. 
The standard entropy change is calculated using the formula: \[ \Delta S = S_{XY_5} - \left( \frac{1}{2} S_{X_2} + \frac{5}{2} S_{Y_2} \right) \] Substitute the given entropy values: \[ \Delta S = 110 - \left( \frac{1}{2} \times 70 + \frac{5}{2} \times 50 \right) \] \[ \Delta S = 110 - (35 + 125) = 110 - 160 = -50 \, {J K}^{-1} {mol}^{-1} \] 
Step 2: Use the equation for temperature at equilibrium. At equilibrium, the change in Gibbs free energy (\( \Delta G \)) is zero, and we use the following relation: \[ \Delta G = \Delta H - T \Delta S \] Since \( \Delta G = 0 \) at equilibrium: \[ 0 = \Delta H - T \Delta S \] Rearranging to solve for \( T \): \[ T = \frac{\Delta H}{\Delta S} \] Substitute the known values for \( \Delta H \) (in J/mol) and \( \Delta S \): \[ \Delta H = -35 \, {kJ/mol} = -35000 \, {J/mol} \] \[ T = \frac{-35000 \, {J/mol}}{-50 \, {J K}^{-1} {mol}^{-1}} = 700 \, {K} \] Thus, the temperature at equilibrium is \( \boxed{700} \, {K} \).

Answer 2

Given Reaction:

The reaction under consideration is: \[ \frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5 \]

Step 1: Calculate the change in entropy (\( \Delta S^0_{\text{rxn}} \))

The change in entropy for the reaction is given by: \[ \Delta S^0_{\text{rxn}} = 110 - \left[ \left( \frac{1}{2} \times 70 \right) + \left( \frac{5}{2} \times 50 \right) \right] \] Substituting the values: \[ \Delta S^0_{\text{rxn}} = 110 - 160 = -50 \, \text{J K}^{-1} \, \text{mol}^{-1}. \]

Step 2: Determine \( \Delta G^0 \) at equilibrium

At equilibrium, the standard Gibbs free energy change (\( \Delta G^0 \)) is zero: \[ \Delta G^0 = 0 \, \text{at equilibrium}. \] The Gibbs free energy change is also related to enthalpy and entropy by the equation: \[ \Delta G^0 = \Delta H^0 - T \Delta S^0. \] Substitute the known values: \[ 0 = -35000 - T(-50). \]

Step 3: Solve for temperature \( T \)

Solving the equation for \( T \): \[ 0 = -35000 + 50T, \] \[ 50T = 35000, \] \[ T = \frac{35000}{50} = 700 \, \text{Kelvin}. \]

Final Answer:

The temperature \( T \) is \( \boxed{700} \, \text{Kelvin}. \)