At 1990 K and 1 atm pressure, there are equal number


At $1990 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure, there are equal number of $\mathrm{Cl}_{2}$ molecules and $\mathrm{Cl}$ atoms in the reaction mixture. The value $\mathrm{K}_{\mathrm{P}}$ for the reaction $\mathrm{Cl}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{Cl}_{(\mathrm{g})}$ under the above conditions is $x \times 10^{-1}$. The value of $x$ is___________

(Rounded of to the nearest integer)


$\mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{Cl}$

Let mol of both of $\mathrm{Cl}_{2}$ and $\mathrm{Cl}$ is $x$

$\mathrm{P}_{\mathrm{Cl}}=\frac{\mathrm{X}}{2 \mathrm{x}} \times 1=\frac{1}{2}$

$\mathrm{P}_{\mathrm{Cl}_{2}}=\frac{\mathrm{x}}{2 \mathrm{x}} \times 1=\frac{1}{2}$

$K_{p}=\frac{\left(\frac{1}{2}\right)^{2}}{\frac{1}{2}}=\frac{1}{2}=0.5 \Rightarrow 5 \times 10^{-1}$

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