Question

For the following reaction, the equilibrium constant is 3.9 X 10s at 300. K and 0.12 at 500. K. Calculate ??? and ?So for this reaction. N2(g) +3 H2(g) 2 NH3(g)

Answer 1

Given: Equilibrium constant, K at 300K is 3.9 x 10⁵ and Equilibrium constant, K at 500K is 0.12

from the given information we can calculate delta G at 300K and 500K, using the formula:

delta G = -RTlnK

delta G₃₀₀ = - RTlnK = - 8.314 J/ mol.K * 300K * ln (3.9 x 10⁵) = - 32110.1 J/mol

delta G₅₀₀ = - RTlnK = - 8.314 J/ mol.K * 500K * ln (0.12) = - 8813.9 J/mol

We also know that delta G = delta H - T delta S

implies:

- 32110.1 J/mol = delta H - 300 delta S ------1

- 8813.9 J/mol = delta H - 500 delta S -------2

Subtract 2 from 1,

- 32110.1 J/mol - (-8813.9 J/mol) = delta H - 300 delta S - {delta H - 500 delta S}

- 32110.1 + 8813.9 = -300 delta S + 500 delta S

200 delta S = -23296.19 J/mol

delta S = -116.48 J/mol

Substituting value of delta S in 1 or 2 will give us delta H.

subrtituting delta S in 1,

- 32110.1 J/mol = delta H - 300* (-116.48) J/mol

-32110.1 J/mol = delta H + 34944.3 J/mol

delta H = - 67054.5 J/mol = - 67.05 KJ/mol