Verified Solution • 2025–26 Session moderate

QJEEM24Ch6018: JEE Main 2024 Chemistry - Class 11 JEEM24Ch6018 Solution: At and pressure, a cylinder is filled with equal number of and…

At −20∘C-20^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm} pressure, a cylinder is filled with equal number of H2,I2\mathrm{H}_2, \mathrm{I}_2 and HI\mathrm{HI} molecules for the reaction H2( g)+I2( g)⇌2HI(g)\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) , the Kp\mathrm{K}_{\mathrm{p}} for the process is x×10−1x \times 10^{-1} .

x=\mathrm{x}= __________.

[Given : R=0.082 L atm K−1 mol−1\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ]

Multiple Choice Options

  • A 2
  • B 1
  • C 10
  • D 0.01

Step-by-Step Verified Solution

At −20∘C-20^{\circ} \mathrm{C} and a pressure of 1 atm1 \mathrm{~atm} , a cylinder contains equal quantities of H2,I2,\mathrm{H}_2, \mathrm{I}_2, and HI\mathrm{HI} molecules. The equilibrium constant for the reaction

H2( g)+I2( g)⇌2HI(g)\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})

denoted as Kp\mathrm{K}_{\mathrm{p}} , is given by the expression x×10−1x \times 10^{-1} .

To solve for xx , use the provided equilibrium expression:

KP=KC=[HI]2[H2][I2]=1=10×10−1\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=10x = 10 .

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