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* 10^(-5)`

Step-by-step 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)* 1000K* \ln K_p\\\\Kp=8.341* 10^(-5)`

Hence, the value of
`K_p` for the chemical equation is
`8.341* 10^(-5)`