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JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)

At \(320\,K,\) a gas \(A_2\) is \(20\%\)  dissociated to \(A(g).\) The standard free energy change at  \(320\,K\)  and \(1\,atm\)   in \(J\,mo1^{-1}\)  is approximately \((R= 8.314\,\,JK^{-1}\,\,mol^{-1};\,\,ln\,2 = 0.693;\,\,ln\,3 = 1.098)\)

  1. A \(1844\)
  2. B \(2068\)
  3. C \(4281\)
  4. D \(4763\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4281\)

Step-by-step Solution

Detailed explanation

In the reaction \({A_2} \leftrightarrow 2A\) Initially, Let \([{A_2}] = 1\,M\) and \([A] = 0\,M\) After \(20\%\) dissociation , \(80\%\) of \(A_2\) remains. \([{A_2}] = 1 \times \frac{{80}}{{100}} = 0.8\,M\) \(20\%\) of \(1\,M\) is…
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