Answer :
Answer: The [tex]K_c[/tex] for the given reaction is [tex]2.67\times 10^{-2}[/tex]
Explanation:
For the given chemical equation:
[tex]2NO(g)+Cl_2\rightleftharpoons 2NOCl(g)[/tex]
Relation of [tex]K_p\text{ with }K_c[/tex] is given by the formula:
[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]
where,
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = [tex]6.5\times 10^{-4}[/tex]
[tex]K_c[/tex] = equilibrium constant in terms of concentration = ?
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = 308 K
[tex]\Delta n_g[/tex] = change in number of moles of gas particles [tex]n_{products}-n_{reactants}=2-3=-1[/tex]
Putting values in above equation, we get:
[tex]6.5\times 10^{-4}=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{6.5\times 10^{-4}}{(0.0821\times 500)^{-1})}=2.67\times 10^{-2}[/tex]
Hence, the [tex]K_c[/tex] for the given reaction is [tex]2.67\times 10^{-2}[/tex]