Question

Thermodynamics Free energy and chemical equilibrium Quiz: Te equilibrium constant for the reaction N29) + O22) + 2 NO 262) Decreases from 1.5 * 109 at 430 °C to 23 at 1000°C. From these data: a-calculate enthalpy change of this reaction; b- equilibrium constant at 1500 °C.

Answer 1

a. From Vanthoff equation

ln (K2/k1) = $\Delta$ H /R*[1/T1 - 1/T2]
k1 = 1.5*10^5 , T1 = 430+273 = 703 k

k2 = 23, T2 = 1000+273 = 1273 k

$\Delta$H = enthalpy change in kj/mol ,     R = 8.314*10^-3 kj.k-1.mol-1

ln((23)/(1.5*10^5)) = ($\Delta$H)/(8.314*10^-3))((1/703)-(1/1273)

$\Delta$H = -114.64 kj/mol

The enthalpy change of the reaction = -114.64 kj/mol

b. Again,

ln (K2/k1) = $\Delta$ H /R*[1/T1 - 1/T2]
k1 = 1.5*10^5 , T1 = 430+273 = 703 k

k2 = ?, T2 = 1500+273 = 1773 k

$\Delta$H = enthalpy change = -114.64 kj/mol ,     R = 8.314*10^-3 kj.k-1.mol-1

ln((k2)/(1.5*10^5)) = (-114.64)/(8.314*10^-3))((1/703)-(1/1773)

the equilibrium constant at 1500 oC = 1.084