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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 INCORRECT statement among the following, for this reaction, is

  1. A Decrease in the total pressure will result in formation of more moles of gaseous X
  2. B At the start of the reaction, dissociation of gaseous X2 takes place spontaneously
  3. C βequilibrium=0.7
  4. D KC<1
Verified Solution

Answer & Solution

Correct Answer

(C) βequilibrium=0.7

Step-by-step Solution

Detailed explanation

A] On decreasing PTQ=nx2 PTnx2 nT Q will be less than Kp reaction will move in forward direction
B] At the start of the reaction G=Go+RT ln Q
t=0, Q=0rxnG= ve (spontaneous)
C] If βequilibrium=0.7
Kp=8×0.494-0.49=3.923.51
Kp>1
Since it is given that
Go>0Kp<1
This is incorrect
D] Kp=KC×RTng
KC=KpR×2981
KC<1
From JEE Advanced
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