At -20^\circ C and 1 ~atm pressure, a cylinder is filled with equal number of H_2, I_2 and HI molecules for the reaction H_2(~g)+I_2(~g) leftharpoons 2 HI(g) , the K_p for the process is x \times 10^-1 . x= \underline\hspace2cm . [Given : R=0.082 ~L ~atm ~K^-1 ~mol^-1 ]

Chemical Equilibrium PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q4 with Solution

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Correct answer: C

For the reaction

\mathrm{H_2(g)}+\mathrm{I_2(g)}\rightleftharpoons 2\mathrm{HI(g)}

we have

K_p=\frac{(P_{\mathrm{HI}})^2}{P_{\mathrm{H_2}}\,P_{\mathrm{I_2}}}

The cylinder contains equal number of molecules of $\mathrm{H_2}$ , $\mathrm{I_2}$ and $\mathrm{HI}$ .

Hence their mole numbers are equal, so their mole fractions are equal:

y_{\mathrm{H_2}}=y_{\mathrm{I_2}}=y_{\mathrm{HI}}=\frac13

At total pressure $P=1\,\text{atm}$ , the partial pressures are

P_{\mathrm{H_2}}=P_{\mathrm{I_2}}=P_{\mathrm{HI}}=\frac13\,\text{atm}

K_p=\frac{\left(\frac13\right)^2}{\left(\frac13\right)\left(\frac13\right)}=1

So,

K_p=1=10\times 10^{-1}

Therefore,

x=10

Since $K_p=x\times 10^{-1}=1$ , we get $x=10$ .

So the correct option is C.