At a certain temperature the equilibrium constant of the reaction N2 - O2- 2NO is 8-10-4- Assuming air to be mixture of four volumes of N2 and one volume of O2- the

#160; #160; N_2 #160; #160; #160; #160; #160; #160; O_2 #160; #160; #160; #160; #160; #160; NO Initial vol ...

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Byju's Answer

Standard XII

Chemistry

Introduction to Alcohols

At a certain ...

Question

At a certain temperature the equilibrium constant of the reaction $N_{2} + O_{2} ⇌ 2 N O$ is $8 \times 10^{- 4}$ . Assuming air to be mixture of four volumes of $N_{2}$ and one volume of $O_{2}$ , the % by volume of $N O$ formed if air is allowed to reach this equilibrium is $X$ . What is the value of $X$ to the nearest integer?

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Solution

$N_{2}$ $O_{2}$ $N O$
Initial volumes
4
1
0
Equilibrium volumes
$4 - x$ $1 - x$ $2 x$
The volumes are directly proportional to the number of moles and the molar concentrations.
$K = \frac{[N O]}{[N_{2}] [O_{2}]}$
$8 \times 10^{- 4} = \frac{(2 x)^{2}}{(4 - x) (1 - x)}$
Since, the value of K is very small, $4 - x ≃ x$ and $1 - x ≃ 1$
$x^{2} = 8 \times 10^{- 4}$
$x = 0.02828$
$[N O] = 2 x = 0.05657$
The percentage by volume of NO is $= \frac{0.05657}{5} \times 100 = 1.11 = X$

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Q. At a certain temperature the equilibrium constant of the reaction :

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	$N_{2}$	$O_{2}$	$N O$
Initial volumes	4	1	0
Equilibrium volumes	$4 - x$	$1 - x$	$2 x$

At a certain temperature the equilibrium constant of the reaction N2+O2⇌2NO is 8×10−4. Assuming air to be mixture of four volumes of N2 and one volume of O2, the % by volume of NO formed if air is allowed to reach this equilibrium is X. What is the value of X to the nearest integer?