Question

give an example for gaseous reversible reaction for which kp is equal to KC

Answer 1

For kp to be equal to kc, ∆n should be zero
where,
∆n=no. of gaseous moles in product side -
no. of gaseous moles in the reactant side
From the formula:
Kp=Kc x [RT]^∆n
In the following reaction,
N2(g) + O2(g) ---------> 2NO(g)
∆n=2-2=0
Therefore, Kp=Kc

Answer 2

Answer:

gaseous reversible reaction for which $\mathrm{K}_{\mathrm{p}}$ is equal to $\bold{\mathrm{K}_{\mathrm{c}}}$ is $\bold{\mathrm{H}_{2}+\mathrm{I}_{2} \rightarrow 2 \mathrm{HI}}$

Explanation:

$\bold{\mathrm{K}_{\mathrm{p}} \text { and } \mathrm{K}_{\mathrm{c}}}$ are the equilibrium constants for gaseous reactions and reactions in solution state respectively. They are ratio of product and reactant pressure or concentrations at equilibrium stage.  

The relation between $\bold{\mathrm{K}_{\mathrm{p}} \text { and } \mathrm{K}_{\mathrm{c}}}$ is:

$\bold{\mathrm{K}_{\mathrm{n}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}}}$

Here R = gas constant, T  = temperature, ∆n = number of product molecules – number of reactant molecules.

The reaction in which ∆n is zero, it will show $\bold{\mathrm{Kp}=\mathrm{K}_{\mathrm{c}}}$ for example:

$\bold{\mathrm{H}_{2}+\mathrm{I}_{2} \rightarrow 2 \mathrm{HI}}$