Question

For the reaction 2 HBr(g) + Cl_2(g) rightarrow 2 HCl(g) + Br_2(g) a. Write the equilibrium constant expression for the reaction. b. Using the following G degree values, calculate Delta G degree for the reaction. HBr(g) = -53.22 kJ/mol HCl(g) = -95.27 kJ/mol Cl_2(g) = 0 kJ/mol Br_2(g) = 3.14 kJ/mol c. Calculate the equilibrium constant K_eq be at 298 K. d. Does this equilibrium lie more with reactants or products?

Answer 1

a)

K = [products]^p /[ reactants]^r

K = [HCl]^2[Br] /([HBr]^2[Cl2]

b)

dG = Gproducts - Greactants

dG = 2*-95.27 + 3.14 - (2*-53.22 + 0)

dG = -80.96 kJ/mol

c)

Keq --> dG = -RT*ln(Keq)

Keq = exp(-dG/(RT))

Keq = exp(80960/(8.314*298))

Keq = 1.55420*10^14

d)

this favours strongly the products