Question

For a certain reaction $\Delta H = -225$ kJ and $\Delta S = -150$ J K$^{-1}$. Find the temperature so that $\Delta G$ is zero.

• A. 1500 K
• B. 1450 K
• C. 1340 K
• D. 1300 K

Hint:

Always ensure $\Delta H$ and $\Delta S$ are in the same units (usually Joules) before dividing.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the temperature at which a reaction is at equilibrium (\( \Delta G = 0 \)) given the enthalpy change and entropy change.
Step 2: Key Formula or Approach:
Gibbs Free Energy equation:
\[ \Delta G = \Delta H - T\Delta S \]
For \( \Delta G = 0 \), we have \( T = \frac{\Delta H}{\Delta S} \).
Step 3: Detailed Explanation:
Given:
\( \Delta H = -225 \text{ kJ} = -225 \times 10^3 \text{ J} \)
\( \Delta S = -150 \text{ J K}^{-1} \)
Substituting the values:
\[ T = \frac{-225000 \text{ J}}{-150 \text{ J K}^{-1}} \]
\[ T = \frac{22500}{15} \text{ K} \]
\[ T = 1500 \text{ K} \]
Step 4: Final Answer:
The temperature is 1500 K.