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Haber Bosch Process

In the follow...

Question

In the following Haber synthesis of $N H_{3}$ , the equilibrium constant for $N H_{3}$ formation, on increase in temperature, is:

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 H N_{3} (g)$ + heat energy

Decreased

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Increased

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First decreased and then increased

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Not altered

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Solution

The correct option is A Decreased
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) + h e a t e n e r g y$ .

The given reaction is exothermic and equilibrium constant for an exothermic reaction decreases with rise in temperature.

Hence, option $A$ is correct.

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Similar questions

N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} : Δ H = - 92 k J

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N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} : Δ H = - 92 k J

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N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} : Δ H = - 92 k J

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(K c)

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

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1.2 \times 10^{- 2}, 3.0 \times 10^{- 2}

1.5 \times 10^{- 2} M

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[N_{2}] = 1.5 \times 10^{- 2} M

[H_{2}] = 3.0 \times 10^{- 2} M

[N H_{3}] = 1.2 \times 10^{- 2} M

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