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Law of Mass Action

For the equil...

Question

For the equilibrium reaction. $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); Δ G^{o} = - 30 k J$
In which of the following case, the reaction will move (spontaneous) in forward direction to achieve equilibrium?
(Given: $2.303 R T = 5, 705$ kJ)

P_{N_{2}} = 1

atm,

P_{H_{2}} = 1

atm and

P_{N H_{3}} = 1

atm at 298 K.

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P_{N_{2}} = 10

atm,

P_{H_{2}} = 10

atm and

P_{N H_{3}} = 0.01

atm at 298 K.

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P_{N_{2}} = 1

atm,

P_{H_{2}} = 1

atm and

P_{N H_{3}} = 0.001

atm at 298 K.

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P_{N_{2}} = 0.01

atm,

P_{H_{2}} = 0.001

atm and

P_{N H_{3}} = 0.01

atm at 298 K.

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Solution

The correct option is A $P_{N_{2}} = 1$ atm, $P_{H_{2}} = 1$ atm and $P_{N H_{3}} = 1$ atm at 298 K.
Reaction constant $(Q_{C}) = \frac{P_{N H_{3}}^{2}}{P_{N H_{2}} P_{H_{2}}^{3}}$
For forward reaction, $Q_{C} < K_{C}$
$Δ G^{o} = - R T l n K_{C}$
i.e. $\frac{Δ G^{o}}{- 2.303 R T} = log K_{C}$
$∴ log K_{C} = 5.26 \times 10^{- 3}$
$∴ K_{C} = 1.012$
For A:- $Q_{C} = \frac{1^{2}}{1 \times 1^{3}} = 1$
Therefore, $Q_{C} < K_{C}$ for (A)
Therefore, option (A) is the answer.

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For the equilibrium reaction. N2(g)+3H2(g)⇌2NH3(g);ΔGo=−30kJIn which of the following case, the reaction will move (spontaneous) in forward direction to achieve equilibrium?(Given: 2.303RT=5,705 kJ)

The correct option is A PN2=1 atm, PH2=1 atm and PNH3=1 atm at 298 K.Reaction constant (QC)=P2NH3PNH2P3H2For forward reaction, QC<KCΔGo=−RTlnKCi.e. ΔGo−2.303RT=logKC∴logKC=5.26×10−3∴KC=1.012For A:- QC=121×13=1Therefore, QC<KC for (A)Therefore, option (A) is the answer.

For the equilibrium reaction. $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); Δ G^{o} = - 30 k J$
In which of the following case, the reaction will move (spontaneous) in forward direction to achieve equilibrium?
(Given: $2.303 R T = 5, 705$ kJ)