Answer: The value of [tex]K_p[/tex] for the chemical equation is [tex]8.341\times 10^{-5}[/tex]
Explanation:
For the given chemical equation:
[tex]N_2(g)+O_2(g)\rightarrow 2NO(g)[/tex]
To calculate the [tex]K_p[/tex] for given value of Gibbs free energy, we use the relation:
[tex]\Delta G=-RT\ln K_p[/tex]
where,
[tex]\Delta G[/tex] = Gibbs free energy = 78 kJ/mol = 78000 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = 1000 K
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = ?
Putting values in above equation, we get:
[tex]78000J/mol=-(8.314J/Kmol)\times 1000K\times \ln K_p\\\\Kp=8.341\times 10^{-5}[/tex]
Hence, the value of [tex]K_p[/tex] for the chemical equation is [tex]8.341\times 10^{-5}[/tex]
