# Question #8697a

Apr 29, 2016

${K}_{c} = 6.55 \times {10}^{14} R$

#### Explanation:

$2 N {O}_{\left(g\right)} + {O}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 N {O}_{2 \left(g\right)}$

${K}_{p} = \frac{{p}_{N {O}_{2}}^{2}}{{p}_{N O}^{2} \times {p}_{{O}_{2}}}$

$p$ represents the partial pressure of the gas.

The relationship between ${K}_{p}$ and ${K}_{c}$ is given by:

${K}_{p} = {K}_{c} {\left(R T\right)}^{\Delta n}$

$R$ is the gas constant

$T$ is the absolute temperature

$\Delta n$ is the difference between the no. moles of product molecules - the number of moles of reactant molecules.

In this case you can see that $\Delta n = 2 - 3 = - 1$

$\therefore {K}_{p} = {K}_{c} {\left(R T\right)}^{- 1}$

${K}_{p} = {K}_{c} / \left(R T\right)$

$\therefore {K}_{c} = {K}_{p} \times R T$

${K}_{c} = 2.2 \times {10}^{12} \times R \times 298$

${K}_{c} = 6.55 \times {10}^{14} R$

You can't evaluate this numerically since no units are given for ${K}_{p}$ so you can't chose a value for $R$.