JEE MAIN - Chemistry (2024 - 6th April Morning Shift - No. 4)
At $$-20^{\circ} \mathrm{C}$$ and $$1 \mathrm{~atm}$$ pressure, a cylinder is filled with equal number of $$\mathrm{H}_2, \mathrm{I}_2$$ and $$\mathrm{HI}$$ molecules for the reaction $$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$$, the $$\mathrm{K}_{\mathrm{p}}$$ for the process is $$x \times 10^{-1}$$.
$$\mathrm{x}=$$ __________.
[Given : $$\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$]
Explanation
At $$-20^{\circ} \mathrm{C}$$ and a pressure of $$1 \mathrm{~atm}$$, a cylinder contains equal quantities of $$\mathrm{H}_2, \mathrm{I}_2,$$ and $$\mathrm{HI}$$ molecules. The equilibrium constant for the reaction
$$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$$
denoted as $$\mathrm{K}_{\mathrm{p}}$$, is given by the expression $$x \times 10^{-1}$$.
To solve for $$x$$, use the provided equilibrium expression:
$$\mathrm{K_{P} = \mathrm{K_C}} =\frac{\left[\mathrm{HI}\right]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=1 = 10 \times 10^{-1}$$
Therefore, $$x = 10$$.
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