Question

For a certain reaction, $\Delta H = -210$ kJ and $\Delta S = -150$ J K$^{-1}$. Find the temperature so that $\Delta G = 0$.

• A. 1100 K
• B. 1200 K
• C. 1400 K
• D. 1300 K

Hint:

Always double-check unit consistency! Enthalpy is almost always given in kilojoules (kJ) while entropy is usually given in joules (J). Failing to convert kJ to J is the most common mistake made in these thermodynamic calculations.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are provided with the enthalpy change ($\Delta H$) and the entropy change ($\Delta S$) for a reaction.
We need to calculate the exact equilibrium temperature where the Gibbs Free Energy change ($\Delta G$) equals zero.

Step 2: Key Formula or Approach:
The fundamental thermodynamic relationship between free energy, enthalpy, and entropy is given by the Gibbs Free Energy equation:
$$\Delta G = \Delta H - T\Delta S$$
At equilibrium, $\Delta G = 0$, which simplifies the equation to:
$$T = \frac{\Delta H}{\Delta S}$$

Step 3: Detailed Explanation:
Before substituting the values, we must ensure that both $\Delta H$ and $\Delta S$ share the same energy units (usually Joules).
Given:
$\Delta H = -210 \text{ kJ} = -210,000 \text{ J}$
$\Delta S = -150 \text{ J K}^{-1}$
Substitute these values into the derived equilibrium formula:
$$T = \frac{-210,000}{-150}$$
$$T = \frac{210000}{150}$$
$$T = \frac{2100}{15}$$
$$T = 1400 \text{ K}$$

Step 4: Final Answer:
The temperature at which $\Delta G = 0$ is 1400 K, which corresponds to option (c).