Question

Calculate $\Delta S_{\text{total}$ for a certain reaction if $\Delta H=-150\text{ kJ}$ and $\Delta S=32\text{ J K}^{-1}$ at 300 K.

• A. 266.00 $\text{J K}^{-1}$
• B. 532.00 $\text{J K}^{-1}$
• C. 798.00 $\text{J K}^{-1}$
• D. 468.00 $\text{J K}^{-1}$

Hint:

Exothermic reactions ($\Delta H < 0$) release heat to the surroundings, increasing the disorder of the surroundings, making $\Delta S_{\text{surr}}$ strongly positive! Always match energy units (J vs kJ) before adding.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We must calculate the total entropy change of the universe ($\Delta S_{\text{total}}$) using the given enthalpy change of the system ($\Delta H_{\text{sys}}$) and entropy change of the system ($\Delta S_{\text{sys}}$) at a constant temperature.

Step 2: Detailed Explanation:
The total entropy change is the sum of the entropy change of the system and the surroundings:
$\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$
We are given:
$\Delta S_{\text{sys}} = 32 \text{ J K}^{-1}$
$\Delta H_{\text{sys}} = -150 \text{ kJ} = -150,000 \text{ J}$ (Notice the unit conversion to Joules!)
$T = 300 \text{ K}$
The entropy change of the surroundings ($\Delta S_{\text{surr}}$) is driven by the heat released or absorbed by the system. At constant pressure and temperature:
$\Delta S_{\text{surr}} = \frac{-\Delta H_{\text{sys}}}{T}$
Calculate $\Delta S_{\text{surr}}$:
$\Delta S_{\text{surr}} = \frac{-(-150,000 \text{ J})}{300 \text{ K}}$
$\Delta S_{\text{surr}} = \frac{150,000}{300} = +500 \text{ J K}^{-1}$
Now, sum them up to find the total entropy change:
$\Delta S_{\text{total}} = 32 \text{ J K}^{-1} + 500 \text{ J K}^{-1}$
$\Delta S_{\text{total}} = 532 \text{ J K}^{-1}$

Step 3: Final Answer:
The $\Delta S_{\text{total}}$ is 532.00 $\text{J K}^{-1}$, matching option (b).