Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectivel...
Calculating the Equilibrium Temperature
Given:
Standard entropy of X2 = 60 J/K/mol
Standard entropy of Y2 = 40 J/K/mol
Standard entropy of XY3 = 50 J/K/mol
Enthalpy change for the reaction 1/2 X2 + 3/2Y2 → XY3 = -30 kJ/mol
To calculate the equilibrium temperature of the reaction, we can use the following formula:
ΔG° = -RT ln(K)
Where:
ΔG° = Standard free energy change
R = Gas constant (8.314 J/K/mol)
T = Temperature in Kelvin
ln(K) = Natural logarithm of the equilibrium constant (K)
At equilibrium, ΔG° = 0, and we can rearrange the formula to solve for T:
T = -ΔG° / (R ln(K))
Calculating the Equilibrium Constant
The equilibrium constant (K) can be calculated using the following formula:
ΔG° = -RT ln(K)
K = e^(-ΔG° / (RT))
To calculate ΔG°, we need to use the following formula:
ΔG° = ΔH° - TΔS°
Where:
ΔH° = Enthalpy change
ΔS° = Standard entropy change
Substituting the given values, we get:
ΔH° = -30 kJ/mol
ΔS° = (1/2)(60 J/K/mol) + (3/2)(40 J/K/mol) - (1)(50 J/K/mol) = 40 J/K/mol
ΔG° = -30,000 J/mol - T(40 J/K/mol)
At equilibrium, ΔG° = 0, so:
0 = -30,000 J/mol - T(40 J/K/mol)
T = -30,000 J/mol / (40 J/K/mol)
T = 750 K
Therefore, the equilibrium temperature of the reaction is 750 K, which is option (d).
Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectivel...
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