At a certain temperature the equilibrium constant of the reaction - N2 - O2- 2NO- is 8-10-4- Assuming air to be a mixture of 4 volumes of N2 and 1 volume of O2- The

N_2 O_2 NO Initial volume 410Equilibrium volume 4 x 1 x 2x K=(2x)^2/(4 x)(1 x)=8 × 10^ 4 4x^2/4=8 × 10^ 4 x=0.0283 ...

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Standard XII

Chemistry

Reaction Quotient

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 a mixture of 4 volumes of $N_{2}$ and 1 volume of $O_{2}$ . The % by volume of $N O$ formed if air is allowed to reach this equilibrium is $X$ . $100 X$ is _________.

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Solution

$N_{2}$
$O_{2}$ $N O$
Initial volume
4
1
0
Equilibrium volume
$4 - x$ $1 - x$ $2 x$
$K = \frac{(2 x)^{2}}{(4 - x) (1 - x)} = 8 \times 10^{- 4}$
$\frac{4 x^{2}}{4} = 8 \times 10^{- 4}$
$x = 0.0283$

$N_{2}$
$O_{2}$ $N O$
Equilibrium volume
3.9717
0.9717
0.0566
The percentage by volume of nitrogen formed $\frac{0.0566}{5} \times 100 = 1.11 = X$
$100 X = 111$

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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 volume	4	1	0
Equilibrium volume	$4 - x$	$1 - x$	$2 x$

At a certain temperature the equilibrium constant of the reaction : N2+O2⇌2NO, is 8×10−4. Assuming air to be a mixture of 4 volumes of N2 and 1 volume of O2. The % by volume of NO formed if air is allowed to reach this equilibrium is X. 100X is _________.

N2O2NOInitial volume 410Equilibrium volume4−x1−x2xK=(2x)2(4−x)(1−x)=8×10−44x24=8×10−4x=0.0283N2O2NOEquilibrium volume 3.97170.97170.0566The percentage by volume of nitrogen formed 0.05665×100=1.11=X100X=111