For a reaction, Δ H = -40 kJ/mol and Δ S = -100 J/mol·K at T = 400 K — JEE Chemistry
For a reaction, $\Delta H = -40$ kJ/mol and $\Delta S = -100$ J/mol·K at $T = 400$ K. Determine if the reaction is spontaneous.
1 Answer
VCVarun Choudhary
✓ Accepted
· 1mo ago
▲ 5
$$\Delta G = \Delta H - T\Delta S = -40000 - 400\times(-100)$$
$$= -40000 + 40000 = 0 J/mol$$
The system is at equilibrium at 400 K. Below 400 K, $\Delta G < 0$ (spontaneous). Above 400 K, $\Delta G > 0$ (non-spontaneous).
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Discussion (2)
N
For revision — the key formula used here comes up almost every year.
AS
Thanks a ton, I was stuck on this exact problem for an hour.