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 Concept:
Total entropy change ($\Delta S_{total}$) is the sum of the entropy change of the system ($\Delta S_{sys}$) and the entropy change of the surroundings ($\Delta S_{surr}$).

Step 2: Formula Application:
$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$
Where $\Delta S_{surr} = \frac{-\Delta H_{sys}}{T}$

Step 3: Explanation:
Given: $\Delta H = -150 \text{ kJ} = -150000 \text{ J}$, $\Delta S_{sys} = 32 \text{ J K}^{-1}$, $T = 300 \text{ K}$. $\Delta S_{surr} = \frac{-(-150000)}{300} = 500 \text{ J K}^{-1}$. $\Delta S_{total} = 32 + 500 = 532 \text{ J K}^{-1}$.

Step 4: Final Answer:
The total entropy change is 532.00 J K$^{-1}$.