Question

The free energy of formation of nitric oxide, NO, at 1000 K (roughly the temperature in an automobile engine during ignition) is about 78 kJ/mol. Calculate the equilibrium constant Kp for the reaction N2(g) + O2(g) 2NO(g) at this temperature.

Answer 1

Answer: The value of $K_p$ for the chemical equation is $8.341\times 10^{-5}$

Explanation:

For the given chemical equation:

$N_2(g)+O_2(g)\rightarrow 2NO(g)$

To calculate the $K_p$ for given value of Gibbs free energy, we use the relation:

$\Delta G=-RT\ln K_p$

where,

$\Delta G$ = Gibbs free energy = 78 kJ/mol = 78000 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = 1000 K

$K_p$ = equilibrium constant in terms of partial pressure = ?

Putting values in above equation, we get:

$78000J/mol=-(8.314J/Kmol)\times 1000K\times \ln K_p\\\\Kp=8.341\times 10^{-5}$

Hence, the value of $K_p$ for the chemical equation is $8.341\times 10^{-5}$