JEE Main 2023 · 24 January · morning · Numerical answer · +4 / −0
For independent process at 300 K
The number of non-spontaneous process from the following is __________
| Process | Δ H/kJ mol⁻¹ | Δ S/J K⁻¹ |
|---|---|---|
| A | -25 | -80 |
| B | -22 | 40 |
| C | 25 | -50 |
| D | 22 | 20 |
Answer and solution
Answer 2
Solution
Chemistry ma’am · 2 steps
Do not judge by signs alone: work out Δ G = Δ H - TΔ S at 300 K for each process.
- The test. A process is spontaneous when Δ G < 0 and non-spontaneous when Δ G > 0. Convert Δ S from J to kJ first, so that it matches the units of Δ H.
Δ G = Δ H - TΔ S, T = 300 K
Process Δ H (kJ) TΔ S (kJ) Δ G (kJ) Verdict A -25 300 × (-0.080) = -24 -25 + 24 = -1 spontaneous B -22 300 × 0.040 = 12 -22 - 12 = -34 spontaneous C 25 300 × (-0.050) = -15 25 + 15 = 40 non-spontaneous D 22 300 × 0.020 = 6 22 - 6 = 16 non-spontaneous - So. Only C and D have a positive Δ G at 300 K, so the number of non-spontaneous processes is 2.
The common slip: 3. Calling A non-spontaneous only because its Δ S is negative gives 3. At 300 K its Δ G is -1 kJ mol⁻¹, so A is spontaneous.
The idea to keep. When Δ H and Δ S have the same sign, spontaneity depends on temperature. Always calculate Δ G at the given T, with Δ S converted to kJ.
Answer from the NTA final answer key. Solution by PrepWiser.