Homework Answers
1.
(a) For reaction 1,
Equilibrium expression = [NO2]^2/[N2O4]
For reaction 2,
Equilibrium expression = [H2CO3]/[H2O][CO2]
(b) For reaction 1,
dGo = 2 x 51.3 - 97.82 = 4.78 kJ
dGo = -RTlnKc
4780 = -8.314 x 298 lnKc
Kc = 0.1452
For reaction 2,
dGo = -623.1 - (-394.39 - 228.61) = -0.1 kJ
dGo = -RTlnKc
-100 = -8.314 x 298 lnKc
Kc = 1.041
(c) Direction of reaction in case of,
Reaction 1 Is towards reactant as it free energy change dGo is +ve
Reaction 2 Is towards products as it free energy change dGo is -ve
(d) The reaction is not at equilibrium in first case
The reaction is almost at equilibrium in the second case.
(e) When temperature is increased, reaction 1, equilibrium shifts towards products. More products are formed.
When temperature is increased, reaction 1, equilibrium shifts towards reactants. More reactants are formed.
(f) Increasing concentration of reactant would enhance product formation in both the cases.
(g) Decreasing colume reduces pressure and reaction 1 equilibrium shifts towards product end (higher moles).
Decreasing colume reduces pressure and reaction 1 equilibrium shifts towards reactant end (higher moles).
(h) Increasing pressure would have reverse effect to that we have seen in (g).
2. Changing volume and effect on the equilibrium state of the reaction,
(a) no effect : same number of moles on both ends
(b) left shift to higher number of moles
(c) right shift to higher number of moles
3. For the given endothermic reaction,
(a) dGo = dGoproducts - dGoreactants
= 236.65 - (-16.4 - 237.14)
= 490.19 kJ
dGo = -RTlnKc
490.19 x 1000 = -8.314 x 298 lnKc
Kc = 1.19 x 10^-86
(b) Increasing temperature would enhance product formation for this endothermic reaction
(c) Decreasing temperature would reverse the reaction towards reactant side.
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