Q48 / 150 · MHT CET 2025 (26 Apr Shift 1) · Chemistry
Calculate $\Delta S_{\text {total }}$ for a certain reaction if $\Delta H=-150 \mathrm{~kJ}$ and $\Delta S=32 \mathrm{JK}^{-1}$ at 300 K .
- A$266 \times 00 \mathrm{JK}^{-1}$
- Correct answerB$532 \times 00 \mathrm{JK}^{-1}$
- C$798 \times 00 \mathrm{JK}^{-1}$
- D$468 \times 00 \mathrm{JK}^{-1}$
Solution
The total entropy change for a reaction combines system and surroundings contributions: $\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}$.
The entropy change of the surroundings follows from the enthalpy change at constant temperature: $\Delta S_{\text{surroundings}} = -\frac{\Delta H}{T}$.
Given $\Delta H = -150 \times 10^3\ \mathrm{J}$ and $T = 300\ \mathrm{K}$, substitution yields $\Delta S_{\text{surroundings}} = -\frac{-150 \times 10^3}{300} = 500\ \mathrm{JK}^{-1}$.
Adding the system entropy change $\Delta S_{\text{system}} = 32\ \mathrm{JK}^{-1}$ gives $\Delta S_{\text{total}} = 32 + 500 = 532\ \mathrm{JK}^{-1}$.
The value $532\ \mathrm{JK}^{-1}$ corresponds to option B.
Hence, the correct answer is option: 'B'.