The brown gas prepared by the action of concentrated nitric acid on copper is an equilibrium mixture of dinitrogen tetraoxide and nitrogen dioxide :

$N_{2} O_{4} (g) E n d o t h e r m i c ⇌ 2 N O_{2} (g); Δ H = + v e$
Which one of the following changes would result in a darkening of the colour?

Increase in pressure

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Increase in temperature

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Addition of a catalyst

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Removal of dinitrogen tetraoxide by liquefaction

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Solution

The correct option is B Increase in temperature
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); Δ H = + v e$
Brown colour is formed due to $N O_{2}$ gas. Darkening of colour is possible when more $N O_{2}$ is formed. When temperature increases the equilibrium shifted to the right side and more $N O_{2}$ is formed since the reaction is endothermic.

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Similar questions

Q. The degree of dissociation of dinitrogen tetraoxide

N_{2} O_{4} (g) ⟶ {2 N O}_{2} (g)

at temperature

T

and total pressure

p

α

.Which one of the following is the correct expression for the equilibrium constant

(K_{p})

at this temperature ?

Q. In the commercial preparation of dinitrogen by the liquefaction and fractional distillation,dinitrogen is obtained in:

Which of the following reaction is not true for Nitric acid ( $H N O_{3}$ )?

Hydrogen may be prepared in the laboratory by the action of a metal on an acid.

(a) Which of the metals copper, zinc , magnesium or sodium would be the most suitable ?

(b) Which of the acids dilute sulphuric, concentrated sulphuric, dilute nitric acid and concentrated nitric would you choose ? Explain why you would not use the acids you reject.

Q. Does Le Chatelier's principle predict a change of equilibrium concentration for the following reaction if the gas mixture is compressed?

N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)

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The brown gas prepared by the action of concentrated nitric acid on copper is an equilibrium mixture of dinitrogen tetraoxide and nitrogen dioxide : N2O4(g)Endothermic⇌2NO2(g);ΔH=+veWhich one of the following changes would result in a darkening of the colour?

The correct option is B Increase in temperature N2O4(g)⇌2NO2(g);ΔH=+veBrown colour is formed due to NO2 gas. Darkening of colour is possible when more NO2 is formed. When temperature increases the equilibrium shifted to the right side and more NO2 is formed since the reaction is endothermic.

The brown gas prepared by the action of concentrated nitric acid on copper is an equilibrium mixture of dinitrogen tetraoxide and nitrogen dioxide :

$N_{2} O_{4} (g) E n d o t h e r m i c ⇌ 2 N O_{2} (g); Δ H = + v e$
Which one of the following changes would result in a darkening of the colour?