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Le Chatelier's Principle for Del N Equals to 0

In the manufa...

Question

In the manufacture of ammonia by Haber's process, $N_{2} (g) + 3 H_{2} (g) ⇋ 2 N H_{3} (g) + 92.3$ kJ. Which of the following conditions is unfavourable?

Reducing the temperature

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Removing ammonia as it is formed

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Increasing the temperature

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Increasing the pressure

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Solution

The correct option is C Increasing the temperature
The process of manufacturing of ammonia is an exothermic process,
On increasing temperature, the reaction will try to move in backward direction to neutralize its effect given by Le chatelier principle.

Hence, increasing the temperature becomes unfavourable condition for manufacturing ammonia.

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Similar questions

N_{2} (g) + 3 H_{2} ⇌ 2 N H_{3} (g) + 92.3 k J

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g); Δ H^{o} = - 22.4 k J / m o l

200

500^{o} C

F e

500^{o} C

N_{2} (g) + 3 H_{2} (g) ⟶ 2 {N H}_{3} (g)

Δ H_{25^{o} C}^{o} = - 92.2 k J

Molecules	$N_{2} (g)$	$H_{2} (g)$	${N H}_{3} (g)$
$C_{P} J K^{- 1} {m o l}^{- 1}$	$29.1$	$28.8$	$35.1$

C_{P}

100^{o} C

25^{o} C

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g) + 22.4 k c a l

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g) : H_{25^{\circ} C}^{0} = - 92.2 k J

Molecule	$N_{2} (g)$	$H_{2} (g)$	$N H_{3} (g)$
$C_{p} (J K^{- 1} m o l e^{- 1})$	29.1	28.8	35.1

C_{p}

100^{\circ} C

25^{\circ} C

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