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Reaction Quotient

In the reacti...

Question

In the reaction, $N_{2} + O_{2} ⇌ 2 N O$ , the moles/litre of $N_{2}$ , $O_{2}$ and NO respectively $0.25$ , $0.05$ and $1.0$ at equilibrium, the initial concentration of $N_{2}$ and $O_{2}$ will be respectively.

0.75

mol/litre,

0.55

mole/litre

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0.50

mole/litre,

0.75

mole/litre

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0.25

mole/litre,

0.50

mole/litre

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0.25

mole/litre,

1.0

mole/litre

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Solution

The correct option is A $0.75$ mol/litre, $0.55$ mole/litre
Let the initial concentration of $N_{2}$ be $x$ and $O_{2}$ be $y$ .
$N_{2} + O_{2} ⇋ 2 N O$
$t = 0$ $x$ $y$ $0$
$t = e q u i l i b r i u m$ $x - α$ $y - α$ $2 α$

$x - α = 0.25$
$y - α = 0.05$
$2 α = 1$
$α = 0.5$
so,
$x = 0.25 + 0.5 = 0.75$
$y = 0.05 + 0.5 = 0.55$
Hence, option(A) will be correct with initial concentration will be:
$N_{2} = 0.75$
$O_{2} = 0.55$

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In the reaction, N2+O2⇌2NO, the moles/litre of N2, O2 and NO respectively 0.25, 0.05 and 1.0 at equilibrium, the initial concentration of N2 and O2 will be respectively.

In the reaction, $N_{2} + O_{2} ⇌ 2 N O$ , the moles/litre of $N_{2}$ , $O_{2}$ and NO respectively $0.25$ , $0.05$ and $1.0$ at equilibrium, the initial concentration of $N_{2}$ and $O_{2}$ will be respectively.