JEE MAIN - Chemistry (2021 - 16th March Morning Shift - No. 17)
For the reaction $$A(g) \rightleftharpoons B(g)$$ at 495 K, $$\Delta$$rG$$^\circ$$ = $$-$$9.478 kJ mol$$-$$1.
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol$$-$$1 K$$-$$1; ln 10 = 2.303]
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol$$-$$1 K$$-$$1; ln 10 = 2.303]
Answer
20
Explanation
$$\Delta$$Go = $$-$$RT ln Keq
$$-$$9.478 $$\times$$ 103 = $$-$$495 $$\times$$ 8.314 ln Keq
ln Keq = 2.303 = ln 10
So, Keq = 10
Now, A(g) $$\rightleftharpoons$$ B(g)
$$\matrix{ {t = 0} & {22} & 0 \cr {t = t} & {22 - x} & x \cr } $$
$$Keq = {{[B]} \over {[A]}} = {x \over {(22 - x)}} = 10$$
x = 20
So, millimoles of B = 20
$$-$$9.478 $$\times$$ 103 = $$-$$495 $$\times$$ 8.314 ln Keq
ln Keq = 2.303 = ln 10
So, Keq = 10
Now, A(g) $$\rightleftharpoons$$ B(g)
$$\matrix{ {t = 0} & {22} & 0 \cr {t = t} & {22 - x} & x \cr } $$
$$Keq = {{[B]} \over {[A]}} = {x \over {(22 - x)}} = 10$$
x = 20
So, millimoles of B = 20
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