Q15 / 54 · JEE Advanced 2016 (Paper 2) · Chemistry

Paragraph: Thermal decomposition of gaseous $\mathrm{X}_{2}$ to gaseous $\mathrm{X}$ at $298 \mathrm{~K}$ takes place according to the following equation: $\mathrm{X}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{X}(\mathrm{g})$ The standard reaction Gibbs energy, $\Delta_{r} G^{\circ}$, of this reaction is positive. At the start of the reaction, there is one mole of $\mathrm{X}_{2}$ and no $\mathrm{X}$. As the reaction proceeds, the number of moles of $\mathrm{X}$ formed is given by $\beta$. Thus, $\beta_{\text {equilibrium }}$ is the number of moles of $\mathrm{X}$ formed at equilibrium. The reaction is carried out at a constant total pressure of $2$ bar. Consider the gases to behave ideally. (Given : $R=0.083 \mathrm{~L}$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )
Question: The equilibrium constant $K_{P}$ for this reaction at $298 \mathrm{~K},$ in terms of $\beta_{equilibrium}$ is

  • A
    8βequilibrium22-βequilibrium
  • Correct answer
    B
    8βequilibrium24-βequilibrium2
  • C
    4βequilibrium22-βequilibrium
  • D
    4βequilibrium24-βequilibrium2

Solution

X 2 ( g ) at  equilibrium ⇌ 2X( g ) β equilibrium
n T = 1− β equilibrium 2 + β equilibrium
=1+ β equilibrium 2
1 - β equilibrium 2
KP=Px2Px2=βequilibrium1+βequilibrium2PT21-βequilibrium21+βequilibrium2PT
KP=βequilibrium21-βequilibrium24PT=2βequilibrium21-βequilibrium4
=8βequilibrium24-βequilibrium2

Hence, the correct answer is option: 'B'.

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