In: Chemistry
Answer ALL parts. (a) The reaction B + C → D just becomes spontaneous at a temperature of 78.1 °C. When the temperature is increased to 220 °C: (i) Calculate Gibbs free energy (ii) Calculate equilibrium constant (iii) State whether the products or reactants are favoured at this temperature and explain your reasoning (iv) State the assumptions associated with your calculations.
Useful data for this question:
ΔSreaction (B + C → D) = 259.4 J K–1 mol–
Given transformation,
B + C → D
Given that, at 78.1 0C reaction becomes just spontaneous that means ΔG0 attains least possible –Ve value [Because for spontaneous reaction Gibbs free energy change (ΔG0) is -ve.] Below that temperature reaction would be non-spontaneous with ΔG0 ≥ 0 (0 or +ve).
We can assume that at 78 0C ΔG0 = 0.
We know the relation between ΔG0 , ΔH0 , ΔS0 and T as,
ΔG0 = ΔH0 -TΔS0 ……….. (1)
At T = 78 0C = 78 + 273.15 = 351.15 K, ΔG0 = 0
Hence, eq. (1) takes form,
ΔH0 = TΔS0
With T = 351.15 K and ΔS0 = 259.4 J.K-1.mol-1 we get,
ΔH0 = 351.15 x 259.4
ΔH0 = +91088.31 J/mol
ΔH0 = +91.088 kJ/mol.
1)Calculation of ΔG0 at T = 220 0C,
We have, ΔH0 = +91088.31 J/mol, ΔS0 = 259.4 J.K-1.mol-1 and T = 220 0C = 273.15 + 220 = 493.15 K.
Using eq. (1) let us calculate ΔG0 for given transformation at given temperature 220 0C,
ΔG0 = ΔH0 -TΔS0
ΔG0 = (+91088.31) - 493.15 x 259.4
ΔG0 = -36835 J/mol
ΔG0 = -36.835 kJ/mol
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2)Calculating equilibrium constant K,
Equilibrium constant K and ΔG0 related by formula,
ΔG0 = -RTln(K) …………. (2)
Where, R = 8.314 J.K-1.mol-1.
T = 220 0C = 273.15 + 220 = 493.15 K
ΔG0 = -36835 J/mol
K = ?
Let us all the known things in eq. (2)
-36835 = - 8.314 x 493.15 ln(K)
36835 = 8.314 x 493.15 ln(K)
ln(K) = (36835) / (8.314 x 493.15)
ln(K) = 8.984
K = e8.984
K = 7.97 x 103
Equilibrium constant for given transformation at given temperature is 7.97 x 103
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3)
As, ΔG0 = -36835 J/mol and K = 7.97 x 103
i.e. ΔG0 = -ve and High equilibrium constant value
Hence, forward reaction will be favored i.e. Product (D) formation will be favored.
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4) Assumption is stated at the beginning only.
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