Question

Calculate standard internal energy change for

$OF_{2(g)} + H_2O_{(g)} \rightarrow 2HF_{(g)} + O_{2(g)}$ at 300 K, if

$\Delta_fH^\circ$ of $OF_{2(g)}$, $H_2O_{(g)}$ and $HF_{(g)}$ are 20, -250 and -270 kJ $mol^{-1}$ respectively.

[R = 8.314 J $K^{-1}$ $mol^{-1}$]

• A. -307.50 kJ
• B. -342.48 kJ
• C. -412.00 kJ
• D. -214.48 kJ

Answer 1

The closest answer is -307.50 kJ, but the calculated value is approximately -312.50 kJ.

Answer 2

To calculate the standard internal energy change ($\Delta U^\circ$) for the given reaction:

1. Calculate the standard enthalpy change ($\Delta H^\circ$) for the reaction:

   $\Delta H^\circ = \sum \Delta_fH^\circ(\text{products}) - \sum \Delta_fH^\circ(\text{reactants})$

   Given:

   • $\Delta_fH^\circ(OF_{2(g)}) = 20 \, \text{kJ/mol}$
   • $\Delta_fH^\circ(H_2O_{(g)}) = -250 \, \text{kJ/mol}$
   • $\Delta_fH^\circ(HF_{(g)}) = -270 \, \text{kJ/mol}$
   • $\Delta_fH^\circ(O_{2(g)}) = 0 \, \text{kJ/mol}$ (since it's an element in its standard state)

   $\Delta H^\circ = [2 \times (-270) + 0] - [20 + (-250)] = -540 - (-230) = -310 \, \text{kJ}$

2. Relate $\Delta U^\circ$ and $\Delta H^\circ$ using the formula:

   $\Delta H^\circ = \Delta U^\circ + \Delta n \, RT$ where:

   • $\Delta n =$ (moles of gaseous products) - (moles of gaseous reactants)
   • $R = 8.314 \, \text{J/mol·K} = 0.008314 \, \text{kJ/mol·K}$
   • $T = 300 \, \text{K}$

3. Calculate $\Delta n$:

   $\Delta n = (2 \, \text{mol} \, HF + 1 \, \text{mol} \, O_2) - (1 \, \text{mol} \, OF_2 + 1 \, \text{mol} \, H_2O) = 3 - 2 = 1$

4. Calculate $\Delta n \, RT$:

   $\Delta n \, RT = 1 \times 0.008314 \, \text{kJ/mol·K} \times 300 \, \text{K} = 2.4942 \, \text{kJ}$

5. Calculate $\Delta U^\circ$:

   $\Delta U^\circ = \Delta H^\circ - \Delta n \, RT = -310 \, \text{kJ} - 2.4942 \, \text{kJ} = -312.4942 \, \text{kJ} \approx -312.50 \, \text{kJ}$

Therefore, the standard internal energy change for the reaction is approximately -312.50 kJ.