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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

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 $100^{0}$ , the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, $△ H$ for the formation of ${N O}_{2}$ is:

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
$N_{2} O_{4} ⇌ 2 N O_{2}$
Since, the heat is absorbed in the reaction, therefore $Δ H_{f}^{N O_{2}}$ must be greater than zero as reaction is endothermic.

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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^{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:

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. 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. 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 1000, the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, △H for the formation of NO2 is:

The correct option is B positiveN2O4⇌2NO2Since, the heat is absorbed in the reaction, therefore ΔHNO2f must be greater than zero as reaction is endothermic.

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:

The correct option is B positive
$N_{2} O_{4} ⇌ 2 N O_{2}$
Since, the heat is absorbed in the reaction, therefore $Δ H_{f}^{N O_{2}}$ must be greater than zero as reaction is endothermic.