Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK–1 mol–1, respectively. For the reaction,X2 + Y2 → XY3, ΔH = – 30 kJ, to be at equilibrium, the temperature will be1250 K500 K750 K1000 K
Question
Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK–1 mol–1, respectively. For the reaction,X2 + Y2 → XY3, ΔH = – 30 kJ, to be at equilibrium, the temperature will be1250 K500 K750 K1000 K
Solution
The reaction given is X2 + Y2 → XY3.
The standard entropy change for the reaction, ΔS°, can be calculated using the formula:
ΔS° = ΣS°(products) - ΣS°(reactants)
For the given reaction, this becomes:
ΔS° = S°(XY3) - [S°(X2) + S°(Y2)] = 50 - (60 + 40) = -50 JK^-1 mol^-1
The standard enthalpy change for the reaction, ΔH°, is given as -30 kJ. To convert this to the same units as ΔS°, we multiply by 1000, giving -30000 J.
The equation for the temperature at equilibrium, T, is given by:
T = ΔH° / ΔS°
Substituting the given values into this equation gives:
T = -30000 / -50 = 600 K
So, none of the given options (1250 K, 500 K, 750 K, 1000 K) are correct. The temperature at equilibrium would be 600 K.
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