Verified Solution • 2025–26 Session moderate

QJEEM22Chc642: JEE Main 2022 Chemistry - Class 11 JEEM22Chc642 Solution: The equilibrium constant for the reversible reaction 2A(g)…

The equilibrium constant for the reversible reaction

2A(g) ⇌\rightleftharpoons 2B(g) + C(g) is K1

32{3 \over 2} A(g) ⇌\rightleftharpoons 32{3 \over 2} B(g) + 34{3 \over 4} C(g) is K2.

K1 and K2 are related as :

Multiple Choice Options

  • A K1=K2{K_1} = \sqrt {{K_2}}
  • B K2=K1{K_2} = \sqrt {{K_1}}
  • C K2=K13/4{K_2} = K_1^{3/4}
  • D K1=K23/4{K_1} = K_2^{3/4}

Step-by-Step Verified Solution

2A(g)=2B( g)+C(g)K1...(i)2 \mathrm{A}(\mathrm{g})=2 \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \quad K_{1} \quad\quad ...(i)

32A(g)=32B(g)+34C(g)K2...(ii)\frac{3}{2} \mathrm{A}(\mathrm{g})=\frac{3}{2} \mathrm{B}(\mathrm{g})+\frac{3}{4} \mathrm{C}(\mathrm{g}) \quad K_{2}\quad\quad...(ii)

eq. (ii) is 34\frac{3}{4} times of eq. (i), hence,

K2=(K1)34K_{2}=\left(K_{1}\right)^{\frac{3}{4}}

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