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Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?
a)1250k
b)500 k
c)1000k
d)750k?
Most Upvoted Answer
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).
Community Answer
Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectivel...
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Standard entropy of X2 Y2 and XY3 are 60 40 and 50 J/K/mol respectively. For the reaction 1/2 X2+3/2Y2_XY3 delta H=-30 kj to be at equilibrium the temperature will be ?a)1250kb)500 kc)1000kd)750k?
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