JEE MAIN - Chemistry (2018 - 16th April Morning Slot - No. 23)
At 320 K, a gas A2 is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in J mol-1 is approximately : (R = 8.314 JK-1 mol-1; ln2 = 0.693; ln 3 = 1.098)
4763
2068
1844
4281
Explanation
A2 (g) $$\rightleftharpoons$$ 2 A (g)
Assume initially concentration of A2 = [A2] = 1 m
at equilibrium [A2] = 1 $$ \times $$ $${{80} \over {100}}$$ = 0.8 M
and 20% of [A2] = 1 $$ \times $$ $${{20} \over {100}}$$ = 0.2 M
$$\therefore\,\,\,\,$$ [A] = 2 $$ \times $$ 0.2 = 0.4 M
Equilibrium constant
K = $${{{{[A]}^2}} \over {[{A_2}]}}$$ = $${{{{\left[ {0.4} \right]}^2}} \over {\left[ {0.8} \right]}}$$ = 0.2
$$\Delta $$Go = $$-$$ RT $$\ell $$nK
= $$-$$ 8.314 $$ \times $$ 320 $$ \times $$ $$\ell $$n(0.2)
= 4281 J/mol.
Assume initially concentration of A2 = [A2] = 1 m
at equilibrium [A2] = 1 $$ \times $$ $${{80} \over {100}}$$ = 0.8 M
and 20% of [A2] = 1 $$ \times $$ $${{20} \over {100}}$$ = 0.2 M
$$\therefore\,\,\,\,$$ [A] = 2 $$ \times $$ 0.2 = 0.4 M
Equilibrium constant
K = $${{{{[A]}^2}} \over {[{A_2}]}}$$ = $${{{{\left[ {0.4} \right]}^2}} \over {\left[ {0.8} \right]}}$$ = 0.2
$$\Delta $$Go = $$-$$ RT $$\ell $$nK
= $$-$$ 8.314 $$ \times $$ 320 $$ \times $$ $$\ell $$n(0.2)
= 4281 J/mol.
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