Question

One mole each of \(A_2(g)\) and \(B_2(g)\) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K: \(A_{2}(g)+B_{2}(g) \rightleftharpoons 2AB(g)\). The value of x (missing enthalpy of \(B_2\) or related parameter) is ______ . (Nearest integer)}
  

Hint:

Use \(\Delta G = \Delta H - T\Delta S\) for equilibrium thermodynamics. Ensure \(\Delta S\) (usually J) is converted to kJ if \(\Delta H\) is in kJ.

Answer 1

Correct Answer: 26

   Step 1: Calculate \(\Delta G^\circ\) from K
Given \(\log K = 2.2\). \[ \Delta G^\circ = -2.303 RT \log K \] \[ \Delta G^\circ = -2.303 \times 8.3 \times 500 \times 2.2 \] \[ \Delta G^\circ \approx -21027 \text{ J/mol} \approx -21 \text{ kJ/mol} \] Step 2: Calculate \(\Delta S^\circ_{rxn}\)
\[ \Delta S^\circ_{rxn} = 2 S^\circ(AB) - [S^\circ(A_2) + S^\circ(B_2)] \] \[ \Delta S^\circ_{rxn} = 2(222) - (146 + 280) = 444 - 426 = 18 \text{ J K}^{-1} \text{ mol}^{-1} \] Step 3: Calculate \(\Delta H^\circ_{rxn}\)
Using \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\): \[ -21000 = \Delta H^\circ - 500(18) \] \[ -21000 = \Delta H^\circ - 9000 \] \[ \Delta H^\circ = -12000 \text{ J/mol} = -12 \text{ kJ/mol} \] Step 4: Solve for Missing Enthalpy (\(H_{B2}\))
\[ \Delta H^\circ_{rxn} = 2 \Delta H(AB) - [\Delta H(A_2) + H_{B2}] \] \[ -12 = 2(32) - (6 + H_{B2}) \] \[ -12 = 64 - 6 - H_{B2} \] \[ -12 = 58 - H_{B2} \] \[ H_{B2} = 58 + 12 = 70 \text{ kJ/mol} \] Final Answer: 70.