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Standard XII

Chemistry

Characteristics of Equilibrium Constant

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 $100^{o}$ , 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
Here, heat favours formation of product ( NO $_{2}$ ). It means the reaction is endothermic so after getting energy in the form of heat, forward reaction is favoured producing more NO $_{2}$ . This, causes the colour to deepen.
For endothermic reactions ∆H is positive.
So, the correct option is 'Positive'.

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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 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 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. Assertion :On cooling, the brown colour of

N O_{2}

disappears. Reason: On cooling

N O_{2}

forms dimer

(N_{2} O_{4})

resulting in the pairing of the odd electrons in

N O_{2}

Q. Of the following statements which are true for

N O_{2}

i)

Nitrogen dioxide brown liquid turns pale on cooling and becomes colourless solid
ii)

N O_{2}

undergoes dimerisation to form

N_{2} O_{4}

iii)

N O_{2}

is paramagnetic,

N_{2} O_{4}

is diamagnetic

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In a flask colourless N2O4 is in equilibrium with brown coloured NO2. At equilibrium when the flask is heated at 100o, 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 $100^{o}$ , the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, $Δ$ H for formation of $N O_{2}$ is?