Question

Q1)Determine the equilibrium constant for the following reaction at 498 K.
2 Hg(g) + O₂(g) → 2 HgO(s) ΔH° = -304.2 kJ; ΔS° = -414.2 J/K

a)1.87 × 1010

b)2.31 × 10-22

c)4.33 × 1021

d)8.10 × 1031

e)5.34 × 10-11

Q2) Determine the equilibrium constant for the following reaction at 655 K.

HCN(g) + 2 H₂(g) → CH₃NH₂(g) ΔH° = -158 kJ; ΔS°= -219.9 J/K

a)2.51 × 10-13

b)3.99 × 1012

c)3.26 × 10-12

d)13.0

e)3.07 × 1011

Answer 1

1.2Hg + O₂ ---> 2HgO

Again we know,

and    , K_eq=equillibrium constant

so,

R = 8.314 J/K/mol

T = 498K

- 8.314 X 498 ln K_eq = (-304.2 X1000) - ( 498 X -414.2)

or, K_eq = 1.87 X 10¹⁰

so the equillibrium constant is (a) 1.87 X 10¹⁰

2. HCN + 2H₂ --> CH₃NH₂

similarly like the previous answer,

or, -8.314 X 655 ln K_eq = (-158 X 1000) - (655 X -219.9)

or, K_eq = 12.994 = 13

so the equillibrium constant is (d)13.0