JEE MAIN - Chemistry (2021 - 18th March Evening Shift - No. 19)
The gas phase reaction $$2A(g) \rightleftharpoons {A_2}(g)$$ at 400 K has $$\Delta$$Go = + 25.2 kJ mol-1.
The equilibrium constant KC for this reaction is ________ $$\times$$ 10$$-$$2. (Round off to the Nearest Integer).
[Use : R = 8.3 J mol$$-$$1 K$$-$$1, ln 10 = 2.3 log10 2 = 0.30, 1 atm = 1 bar]
[antilog ($$-$$0.3) = 0.501]
The equilibrium constant KC for this reaction is ________ $$\times$$ 10$$-$$2. (Round off to the Nearest Integer).
[Use : R = 8.3 J mol$$-$$1 K$$-$$1, ln 10 = 2.3 log10 2 = 0.30, 1 atm = 1 bar]
[antilog ($$-$$0.3) = 0.501]
Answer
2
Explanation
Using formula,
$$\Delta$$G$$^\circ$$ = $$-$$ RTln(Kp)
$$ \Rightarrow $$ 25.2 $$\times$$ 103 = $$-$$8.3 $$\times$$ 400 $$\times$$ 2.3 log (Kp)
$$ \Rightarrow $$ Kp = 10$$-$$3.3
= 10$$-$$3 $$\times$$ 0.501
= 5.01 $$\times$$ 10$$-$$4 Bar$$-$$1
Also,
$${{{K_p}} \over {{K_c}}} = {(RT)^{\Delta {n_g}}}$$
$$ \Rightarrow {{{K_p}} \over {{K_c}}} = {(RT)^{ - 1}}$$
$$ \Rightarrow {K_c} = {K_p}(RT)$$
$$ = 5.01 \times {10^{ - 4}} \times 8.3 \times 400$$
$$ = 1.66 \times {10^{ - 5}}$$ m3/mole
$$ = 1.66 \times {10^{ - 2}}$$ L/mol
$$\Delta$$G$$^\circ$$ = $$-$$ RTln(Kp)
$$ \Rightarrow $$ 25.2 $$\times$$ 103 = $$-$$8.3 $$\times$$ 400 $$\times$$ 2.3 log (Kp)
$$ \Rightarrow $$ Kp = 10$$-$$3.3
= 10$$-$$3 $$\times$$ 0.501
= 5.01 $$\times$$ 10$$-$$4 Bar$$-$$1
Also,
$${{{K_p}} \over {{K_c}}} = {(RT)^{\Delta {n_g}}}$$
$$ \Rightarrow {{{K_p}} \over {{K_c}}} = {(RT)^{ - 1}}$$
$$ \Rightarrow {K_c} = {K_p}(RT)$$
$$ = 5.01 \times {10^{ - 4}} \times 8.3 \times 400$$
$$ = 1.66 \times {10^{ - 5}}$$ m3/mole
$$ = 1.66 \times {10^{ - 2}}$$ L/mol
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