K_p for the given reaction is (36 × 10^–2 atm^–1). Find out K_c (M^–1) (nearest integer). (2NO₂ ⇌ N₂O₄) (R = 0.0821 atm.L/mol.K) (T = 300 K)

The correct answer is 9.

Step 1: Understand the given data
We are given the equilibrium constant in terms of partial pressures ($K_p$):
$K_p=36\times {10}^{-2}{\text{atm}}^{-1}$

Universal gas constant ($R$) = $0.0821\text{atm·L/mol·K}$
Temperature ($T$) = $300\text{K}$

Step 2: Write the chemical equation and calculate $\Delta n_g$
The given chemical equilibrium reaction is:
$2{\text{NO}}_2\rightleftharpoons {\text{N}}_2{\text{O}}_4$

The change in the number of moles of gaseous components ($\Delta n_g$) is given by:
$\Delta n_g=n_{\text{products (g)}}-n_{\text{reactants (g)}}$

$\Delta n_g=1-2=-1$

Step 3: Apply the relationship between $K_p$ and $K_c$
The formula linking $K_p$ and $K_c$ is:
$K_p=K_c{\left(RT\right)}^{\Delta n_g}$

Substituting $\Delta n_g=-1$ into the equation:
$K_p=K_c{\left(RT\right)}^{-1}=\frac{K_c}{RT}$

Rearranging the equation to solve for $K_c$:
$K_c=K_p\times \left(RT\right)$

Step 4: Perform the calculation
Substitute the given numerical values:
$K_c=(36\times {10}^{-2})\times (0.0821\times 300)$

$K_c=0.36\times 24.63$

$K_c=8.8668$

Rounding to the nearest integer gives:
$K_c\approx 9$