Question

If the standard Gibbs energy, enthalpy and entropy changes \(\Delta G^\circ\), \(\Delta H^\circ\) and \(\Delta S^\circ\) for a reaction are \(-34.44~\text{kJ mol}^{-1}\), \(-6.44~\text{kJ mol}^{-1}\), and \(80~\text{J K}^{-1} \text{mol}^{-1}\) respectively, the temperature of the reaction (in \(^\circ\text{C}\)) is

• A. 350
• B. 427
• C. 77
• D. 35

Hint:

Use \(\Delta G = \Delta H - T \Delta S\). Watch out for unit conversions!

Answer 1

The Correct Option is C

We use the Gibbs free energy relation: \[ \Delta G = \Delta H - T\Delta S \Rightarrow T = \frac{\Delta H - \Delta G}{\Delta S} \] Convert all to consistent units (J/mol):
- \(\Delta G = -34.44 \times 10^3 = -34440\) J/mol
- \(\Delta H = -6.44 \times 10^3 = -6440\) J/mol
- \(\Delta S = 80\) J/K·mol
\[ T = \frac{-6440 - (-34440)}{80} = \frac{28000}{80} = 350~\text{K} \Rightarrow 350 - 273 = \boxed{77^\circ\text{C}} \]

Answer 2

If the standard Gibbs energy, enthalpy and entropy changes \(\Delta G^\circ\), \(\Delta H^\circ\), and \(\Delta S^\circ\) for a reaction are:
\(\Delta G^\circ = -34.44~\text{kJ mol}^{-1}\)
\(\Delta H^\circ = -6.44~\text{kJ mol}^{-1}\)
\(\Delta S^\circ = 80~\text{J K}^{-1} \text{mol}^{-1}\)

Step 1: Use the relation:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
We are given all values except T. Convert entropy to kJ units:
\[ \Delta S^\circ = \frac{80}{1000} = 0.080\ \text{kJ K}^{-1} \text{mol}^{-1} \]
Substitute the values into the equation:
\[ -34.44 = -6.44 - T \times 0.080 \]

Step 2: Solve for T:
\[ -34.44 + 6.44 = -0.080T \]
\[ -28.00 = -0.080T \]
\[ T = \frac{28.00}{0.080} = 350\ \text{K} \]

Step 3: Convert temperature to Celsius:
\[ T = 350 - 273 = 77^\circ\text{C} \]

Final Answer:
\[ \boxed{77^\circ\text{C}} \]