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JEE Advanced · Chemistry · 7. Chemical Equilibrium

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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

  1. A 8βequilibrium22-βequilibrium
  2. B 8βequilibrium24-βequilibrium2
  3. C 4βequilibrium22-βequilibrium
  4. D 4βequilibrium24-βequilibrium2
Verified Solution

Answer & Solution

Correct Answer

(B) 8βequilibrium24-βequilibrium2

Step-by-step Solution

Detailed explanation

X 2 ( g ) atequilibrium 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
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