Question

The standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are –382.64 kJ mol^–1 and –145.6 JK^–1 mol^–1, respectively. Standard Gibbs energy change for the same reaction at 298 K is

• A. –339.3 kJ mol^–1
• B. – 439.3 kJ mol^–1
• C. –523.2 kJ mol^–1
• D. –221.1 kJ mol^–1

Answer 1

The Correct Option is A

To find the standard Gibbs energy change \((\Delta G^\circ)\) for the oxidation of ammonia at 298 K, we can use the Gibbs free energy equation:

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

where:

• \(\Delta H^\circ\) is the standard enthalpy change.
• \(T\) is the absolute temperature in Kelvin.
• \(\Delta S^\circ\) is the standard entropy change.

Given values:

• \(\Delta H^\circ = -382.64 \text{ kJ mol}^{-1}\)
• \(T = 298 \text{ K}\)
• \(\Delta S^\circ = -145.6 \text{ J K}^{-1} \text{ mol}^{-1}\)

Note that \(\Delta S^\circ\) must be converted to kJ K^-1 mol^-1 to match the units of \(\Delta H^\circ\). This conversion is done by dividing by 1000:

\(\Delta S^\circ = -145.6 \text{ J K}^{-1} \text{ mol}^{-1} = -0.1456 \text{ kJ K}^{-1} \text{ mol}^{-1}\)

Now substitute the values into the Gibbs free energy equation:

\(\Delta G^\circ = (-382.64 \text{ kJ mol}^{-1}) - (298 \text{ K})(-0.1456 \text{ kJ K}^{-1} \text{ mol}^{-1})\)

Calculate the temperature-entropy product:

\(T\Delta S^\circ = (298 \text{ K})(-0.1456 \text{ kJ K}^{-1} \text{ mol}^{-1}) = -43.3888 \text{ kJ mol}^{-1}\)

Substitute this back into the equation:

\(\Delta G^\circ = -382.64 \text{ kJ mol}^{-1} + 43.3888 \text{ kJ mol}^{-1}\)

Simplify the equation:

\(\Delta G^\circ = -339.2512 \text{ kJ mol}^{-1}\)

Rounding to match the precision of the given options, we obtain:

\(\Delta G^\circ \approx -339.3 \text{ kJ mol}^{-1}\)

Thus, the correct answer is -339.3 kJ mol^-1.