Question

For the reaction, CO(g) + 3 H2(g) CH4(g) + H2O(g), Kp = 2.15 X 10*11 at 200 °C and Kp = 4.56 X 10*8 at 260 °C. Determine H° by using the van’t Hoff equation.

Answer 1

∆H° = -215.28 kJ

Explanation

From Van't Hoff equation following relation is derived

2.303log(K₁/K₂) = (∆H°/R)(1/T₂ - 1/T₁)

where,

K₁ = equilibrium constant 1, 2.15 × 10¹¹

K₂ = equilibrium constant 2 , 4.56 ×10⁸

T₁ = Tempeeature 1, 200℃ = 473.15K

T₂ = Temperature 2, 260℃ = 533.15K

R = gas constant, 8.314J/K mol

2.303log( 2.15 ×10¹¹/4.56 ×10⁸) = ( ∆H°/8.314J/K mol)(1/ 533.15K - 1/473.15K)

2.303log( 471.49) = ( ∆H° / 8.314 J/K mol)(-0.0002378)

6.157 = ∆H° × -2.860 ×10^-5

∆H^° =- 215280J

∆H° = -215.28kJ