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QJEEM23Ch0324: JEE Main 2023 Chemistry - Class 11 JEEM23Ch0324 Solution: (i) (ii) If the degree of dissociation and initial…

(i) X(g)⇌Y(g)+Z(g)Kp1=3\mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 1}=3

(ii) A(g)⇌2 B(g)Kp2=1\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 2}=1

If the degree of dissociation and initial concentration of both the reactants X(g)\mathrm{X}(\mathrm{g}) and A(g)\mathrm{A}(\mathrm{g}) are equal, then the ratio of the total pressure at equilibrium (p1p2)\left(\frac{p_{1}}{p_{2}}\right) is equal to x:1\mathrm{x}: 1 . The value of x\mathrm{x} is _____________ (Nearest integer)

Step-by-Step Verified Solution

JEE Main 2023 (Online) 1st February Morning Shift Chemistry - Chemical Equilibrium Question 28 English Explanation 1

amp;kp1=(α1+α×p1)21−α1+αp1amp;3=α2×p11−α2\begin{aligned}& \mathrm{k}_{\mathrm{p}_1}=\frac{\left(\frac{\alpha}{1+\alpha} \times \mathrm{p}_1\right)^2}{\frac{1-\alpha}{1+\alpha} \mathrm{p}_1} \\\\& 3=\frac{\alpha^2 \times \mathrm{p}_1}{1-\alpha^2}\end{aligned}

JEE Main 2023 (Online) 1st February Morning Shift Chemistry - Chemical Equilibrium Question 28 English Explanation 2

amp;kp2=(2α1+α×p2)21−α1+α×p2amp;1=4α2×p21−α2amp;kp1kp2=p14p2amp;31=p14p2amp;p1:p2=12:1amp;∴x=12\begin{aligned}& \mathrm{k}_{\mathrm{p}_2}=\frac{\left(\frac{2 \alpha}{1+\alpha} \times \mathrm{p}_2\right)^2}{\frac{1-\alpha}{1+\alpha} \times \mathrm{p}_2} \\\\& 1=\frac{4 \alpha^2 \times \mathrm{p}_2}{1-\alpha^2} \\\\& \frac{\mathrm{k}_{\mathrm{p}_1}}{\mathrm{k}_{\mathrm{p}_2}}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\& \frac{3}{1}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\& {\mathrm{p}_1} : {\mathrm{p}_2} = 12 : 1 \\\\& \therefore \mathrm{x}=12\end{aligned}

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