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 \((R= 8.314\,\,JK^{-1}\,\,mol^{-1};\,\,ln\,2 = 0.693;\,\,ln\,3 = 1.098)\)
- A \(1844\)
- B \(2068\)
- C \(4281\)
- D \(4763\)
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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