Question

The ratio of kp/ kc for a reaction 2NO+Cl2=2NOCl is

Answer 1

Explanation:

2NO (g) + Cl2 (g) ⇄ 2NOCl (g)

• $K_{p}$ = $K_{c}$ $(RT)^{n}$

         $K_{p}$ : equilibrium constant used when equilibrium concentrations are expressed in atmospheric pressure

          $K_{c}$ : equilibrium constant used when equilibrium concentrations are expressed in molarity.

• Δn=moles of gaseous products - moles of gaseous reactants  

            = 2 - (2+1) =2-3 = -1

• $K_{p}$ = $K_{c}$ $(RT)^{-1}$

Answer:

$\frac{K_{p} }{K_{c} }$=$(RT)^{-1}$

Answer 2

To find:

The ratio of kp/ kc for a reaction 2NO+Cl2=2NOCl?

Calculation:

$2NO(g)+Cl_{2}(g)\rightarrow 2NOCl(g)$

Now, we know the relationship between $K_{p}$ and $K_{c}$ is as follows:

$\boxed{ \bf \: K_{p} = K_{c} \times RT^{(\Delta n_{g})}}$

• Note that $\Delta n_{g}$ refers only to gas molecules/atoms in the equilibrium.

From the Equation:

$\rm \implies\: K_{p} = K_{c} \times RT^{(\Delta n_{g})}$

$\rm \implies\: \dfrac{K_{p}}{K_{c}} = RT^{(\Delta n_{g})}$

$\rm \implies\: \dfrac{K_{p}}{K_{c}} = RT^{(2 - 1 - 2)}$

$\rm \implies\: \dfrac{K_{p}}{K_{c}} = RT^{( - 1)}$

So, the required ratio is RT^(-1).