Oxidation of ammonia is exothermic:
$4 N H_{3} (g) + 5 O_{2} (g) ⇌ 4 N O (g) + 6 H_{2} O (g) + 905.6 k J$
Equilibrium system suggests that:

on increasing temperature, equilibrium mixture will have less

[N O]

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on increasing temperature, equilibrium constant remains uncharged

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on increasing pressure, equilibrium mixture will have more

[N H_{3}]

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addition of

H e

at constant pressure promote the oxidation of

N H_{3}

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Solution

The correct option is B on increasing temperature, equilibrium mixture will have less $[N O]$
In an exothermic reaction, increasing the temp. will lead the reaction in a backward direction, so on increasing the temperature, equilibrium will have less $N O$ .

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Q. Assertion :

In the first step of Ostwald's process for synthesis of nitric acid, ammonium is oxidised to

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4 N H_{3} (g) + 5 O_{2} (g) ⇌ 4 N O (g) + 6 H_{2} O (g)

Δ H^{\circ} = - 905.6 k J

The equilibrium amount of

N O

will be less with increase in temperature. Reason: Heat is absorbed by products on increasing the temperature and the equilibrium shifts towards left.

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Oxidation of ammonia is exothermic:4NH3(g)+5O2(g)⇌4NO(g)+6H2O(g)+905.6kJEquilibrium system suggests that:

The correct option is B on increasing temperature, equilibrium mixture will have less [NO]In an exothermic reaction, increasing the temp. will lead the reaction in a backward direction, so on increasing the temperature, equilibrium will have less NO.

Oxidation of ammonia is exothermic:
$4 N H_{3} (g) + 5 O_{2} (g) ⇌ 4 N O (g) + 6 H_{2} O (g) + 905.6 k J$
Equilibrium system suggests that:

The correct option is B on increasing temperature, equilibrium mixture will have less $[N O]$
In an exothermic reaction, increasing the temp. will lead the reaction in a backward direction, so on increasing the temperature, equilibrium will have less $N O$ .