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Le Chatelier's Principle for Del N Greater Than Zero

In a flask co...

Question

In a flask colourless $N_{2} O_{4}$ is in equilibrium with brown coloured ${N O}_{2}$ . At equilibrium when the flask is heated at $100^{o} C$ , the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, $Δ H$ for formation of ${N O}_{2}$ is:

Negative

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Positive

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Zero

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Undefined

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Solution

The correct option is B Positive
Equillibrium is represented as:
$N_{2} O_{4} c o l o r l e s s ⇌ 2 {N O}_{2} b r o w n$
Upon heating equilibrium goes forward, and revert back on cooling
Therefore energy is absorbed by the system to produce $N O_{2}$ , hence it is an endothermic process and thus the sign of enthalpy of formation of $N O_{2}$ is positive.
Option B is correct.

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Similar questions

Q. In a flask colourless

N_{2} O_{4}

is in equilibrium with brown coloured

{N O}_{2}

. At equilibrium when the flask is heated at

100^{0}

, the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy,

△ H

for the formation of

{N O}_{2}

is:

Q. In a flask colourless

N_{2} O_{4}

is in equilibrium with brown coloured

{N O}_{2}

. At equilibrium, when the flask is heated to

100^{o} C

the brown colour deepens and on cooling, the brown colour became less coloured. The change in enthalpy

Δ H

of the system is:

Q. In a flask colourless

N_{2} O_{4}

is in equilibrium with brown coloured

N O_{2}

. At equilibrium when the flask is heated at

100^{o}

, the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy,

Δ

H for formation of

N O_{2}

is?

Q. One mole of

N_{2} O_{4}

is heated in a flask with a volume of

0.1 {d m}^{3}

. At equilibrium,

1.708

moles of

{N O}_{2}

and

0.146

mole of

N_{2} O_{4}

were found

134^{o} C

. The equilibrium constant will be:

Q. Assertion :

N O_{2}

is reddish brown and its dimer

N_{2} O_{4}

is colourless Reason: There is a dative bond formation between two

N O_{2}

molecules

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In a flask colourless N2O4 is in equilibrium with brown coloured NO2. At equilibrium when the flask is heated at 100oC, the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, ΔH for formation of NO2 is:

In a flask colourless $N_{2} O_{4}$ is in equilibrium with brown coloured ${N O}_{2}$ . At equilibrium when the flask is heated at $100^{o} C$ , the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, $Δ H$ for formation of ${N O}_{2}$ is: