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Standard XII

Chemistry

Introduction to Alcohols

At 1990 K and...

Question

At $1990 K$ and $1 a t m$ pressure, there are equal number of $C l_{2}$ molecules and $C l$ atoms in the reaction mixture. The value of $K_{p}$ for the reaction $C l_{2 (g)} ⇌ 2 C l_{(g)}$ under the above condition is $x \times 10^{- 1}$ . The value of $x$ is ______.
(rounded off to the nearest integer)

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Solution

$C l_{2 (g)} ⇌ 2 C l_{(g)}$
$∵ No. of atoms of C l = No. of molecules of C l_{2}$
i.e. $n_{C l (g)} = n_{C l_{2} (g)}$
'n' represent number of moles
$P_{τ} = 1 a t m$
Using Dalton's law of partial pressure:
$P_{C l (g)} = 0.5 a t m = P_{C l_{2} (g)}$
$∴ K_{P} = \frac{(0.5)^{2}}{0.5} = 5 \times 10^{- 1}$
Hence, $x = 5$

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At 1990 K and 1 atm pressure, there are equal number of Cl2 molecules and Cl atoms in the reaction mixture. The value of Kp for the reaction Cl2(g)⇌2Cl(g) under the above condition is x×10−1. The value of x is ______. (rounded off to the nearest integer)

Cl2(g)⇌2Cl(g) ∵ No. of atoms of Cl=No. of molecules of Cl2 i.e. nCl(g)=nCl2(g) 'n' represent number of moles Pτ=1 atm Using Dalton's law of partial pressure: PCl(g)=0.5 atm=PCl2(g) ∴KP=(0.5)20.5=5×10−1 Hence, x=5