Estimate the value of the equilibrium constant at 610 KK for each of the following reactions. ΔG∘fΔGf∘ for BrCl(g)BrCl(g) is −−1.0 kJ/molkJ/mol. The standard molar entropy, S∘S∘, for BrCl(g)BrCl(g) is 240.0 J/mol⋅KJ/mol⋅K.
Part A
2NO2(g)⇌N2O4(g).2NO2(g)⇌N2O4(g). ΔH∘fΔHf∘ for N2O4(g)N2O4(g) is 9.16 kJ/molkJ/mol.
Express your answer using three significant figures.
K= ???
Part B
Br2(g)+Cl2(g)⇌2BrCl(g).Br2(g)+Cl2(g)⇌2BrCl(g). ΔH∘fΔHf∘ for BrCl(g)BrCl(g) is 14.6 kJ/molkJ/mol.
Express your answer using three significant figures.
K= ???
PART A:
Given reaction:
2 NO2 <===> N2O4
Given that:
ΔHo = 9.16 kJ/mol
ΔSo = -175.83 J/mol.K
Therefore, ΔGo = ΔHo - TΔSo
ΔGo = (9.16 kJ/mol) - (610 x -175.83 x 10-3 kJ/mol)
ΔGo = 116.42 kJ/mol
We have, equilibrium constant :
ΔGo = -RT ln K
116.42 kJ/mol = (-8.314 x 10-3 x 610) x ln K
ln K = -0.04356
K = e-0.04356
∴ K = 0.9574 --------- (answer)
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PART B:
Given reaction:
Br2 + Cl2 <=====> 2 BrCl
Given that:
ΔHo = 14.6 kJ/mol
ΔSo = 11.471 J/mol.K
Therefore, ΔGo = ΔHo - TΔSo
ΔGo = (14.6 kJ/mol) - (610 x 11.471 x 10-3 kJ/mol)
ΔGo = 21.6 kJ/mol
We have, equilibrium constant :
ΔGo = -RT ln K
21.6 kJ/mol = (-8.314 x 10-3 x 610) x ln K
ln K = -0.2348
K = e-0.2348
∴ K = 0.7907 --------- (answer)
Estimate the value of the equilibrium constant at 610 KK for each of the following reactions....
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