Reaction and given condition

The equilibrium is $Cl_2(g) \rightleftharpoons 2Cl(g)$

We are told that at equilibrium, there are equal numbers of $Cl_2$ molecules and $Cl$ atoms.

Hence, the mole numbers are equal: $n_{Cl_2} = n_{Cl}$

Mole fraction at equilibrium

Let $n_{Cl_2} = n_{Cl} = n$

Then total moles: $n_{\text{total}} = n + n = 2n$

Therefore mole fractions are $y_{Cl_2} = \frac{n}{2n} = \frac{1}{2}, \qquad y_{Cl} = \frac{n}{2n} = \frac{1}{2}$

Partial pressures

Total pressure is $1\,\text{atm}$, so $P_{Cl_2} = y_{Cl_2} P_{\text{total}} = \frac{1}{2}\times 1 = 0.5\,\text{atm}$ $P_{Cl} = y_{Cl} P_{\text{total}} = \frac{1}{2}\times 1 = 0.5\,\text{atm}$

Expression for $K_p$

For $Cl_2(g) \rightleftharpoons 2Cl(g)$

$K_p = \frac{(P_{Cl})^2}{P_{Cl_2}}$

Substitute the values: $K_p = \frac{(0.5)^2}{0.5} = \frac{0.25}{0.5} = 0.5$

Match with the form given

We need $K_p = x \times 10^{-1}$

Since $0.5 = 5 \times 10^{-1}$

therefore, $x = 5$

Comparison with stored answer

Derived answer: $5$

Stored correct answer: $5$

They match.