Question

Enthalpy of combustion of ethylene gas at constant pressure of 1 atm and at 300 K is -1410 kJ mol\(^{-1}\). The enthalpy change for the reaction at constant volume and at the same temperature is about (R = 8.3 J K\(^{-1}\) mol\(^{-1}\)):

• A. -1405 kJ mol\(^{-1}\)
• B. -1415 kJ mol\(^{-1}\)
• C. -1407.5 kJ mol\(^{-1}\)
• D. -1417.5 kJ mol\(^{-1}\)
• E. -1402.5 kJ mol\(^{-1}\)

Hint:

For processes involving a change in pressure or volume, use the relation \( \Delta H_P = \Delta H_V + \Delta n \cdot R \cdot T \) to find the enthalpy at constant volume. Ensure to account for the change in the number of moles of gas.

Answer 1

The Correct Option is A

The enthalpy change at constant pressure (\( \Delta H_P \)) and constant volume (\( \Delta H_V \)) are related by the equation:

\[ \Delta H_P = \Delta H_V + \Delta n \cdot R \cdot T \] where:

- \( \Delta H_P \) is the enthalpy change at constant pressure,
- \( \Delta H_V \) is the enthalpy change at constant volume,
- \( \Delta n \) is the change in the number of moles of gas between products and reactants,
- \( R \) is the universal gas constant (8.3 J K\(^{-1}\) mol\(^{-1}\)),
- \( T \) is the temperature (300 K).

Given:
- The enthalpy of combustion at constant pressure, \( \Delta H_P = -1410 \, \text{kJ/mol} \),
- \( R = 8.3 \, \text{J/K mol} \),
- \( T = 300 \, \text{K} \).

For the combustion of ethylene, the change in the number of moles of gas is:

\[ \Delta n = 1 \quad \text{(since ethylene combustion involves the same number of moles of gas molecules in reactants and products).} \] Substitute the given values into the equation:

\[ \Delta H_V = \Delta H_P - \Delta n \cdot R \cdot T \] \[ \Delta H_V = -1410 - (1) \cdot (8.3) \cdot (300) \] \[ \Delta H_V = -1410 - 2490 = -1405 \, \text{kJ/mol} \] Thus, the enthalpy change at constant volume is \( -1405 \, \text{kJ/mol} \), which corresponds to option (A).