At 1990 K and 1 atm pressure there are equal number of $C l_{2}$ molecules and Cl atoms in the reaction mixture. The value of $K_{p}$ for the reaction $C l_{2 (g)} ⇌ 2 C l_{(g)} u n d e r t h e a b o v e c o n d i t i o n s i s$ x $\times 10^{- 1}$ . The value of x is ________. (Rounded off to the nearest integer).

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answer is 5.

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Detailed Solution

It is given to find out the value of x.

$K_{p} = \frac{[P_{C l}]^{2}}{P_{C l_{2}}}$

moles of both of Cl₂ and Cl molecule be x.

Partial pressure of Cl is, $P_{C l}$ = $\frac{x}{2 x} \times 1$ = $\frac{1}{2}$

Partial pressure of $C l_{2}$ is, $P_{C l_{2}}$ = $\frac{x}{2 x} \times 1$ = $\frac{1}{2}$

$K_{p} = \frac{[P_{C l}]^{2}}{P_{C l_{2}}}$

= $\frac{[\frac{1}{2}]^{2}}{[\frac{1}{2}]}$

= $\frac{1}{2}$ = 0.5 OR $5 \times 10^{- 1}$

Hence, the answer is 5.

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