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Question
For the reaction \[\ce{Ag2O_{(s)} -> 2Ag_{(s)} + 1/2O2_{(g)}}\]: ΔH = 30.56 kJ mol−1 and ΔS = 6.66 JK−1 mol−1 (at 1 atm). Calculate the temperature at which ΔG is equal to zero. Also predict the direction of the reaction
- at this temperature and
- below this temperature.
Solution
Given, ∆H = 30.56 kJ mol−1 = 30560 J mol−1
∆S = 6.66 × 10−3 kJ K−1 mol−1
T = ?
at which ∆G = 0
∆G = ∆H – T∆S
0 = ∆H – T∆S
T = `(Δ"H")/(Δ"S")`
T = `(30.56 "kJ mol"^-1)/(6.66 xx 10^-3 "kJ mol"^-1)`
T = 4589 K
- At 4589K; ∆G = 0 the reaction is in equilibrium.
- at temperature below 4598 K; ∆H > T∆S
∆G = ∆H – T∆S > 0, the reaction in the forward direction, is non spontaneous. In other words the reaction occurs in the backward direction.
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