# 40% of a mixture of 0.2 mole of $N_{2}$ and 0.6 mole of $H_{2}$ react to give $NH_{3}$ according to the equation, $N_{2} (g)+3H_{2}(g)\leftrightarrow 2NH_{3}(g)$ at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is Option 1) 4:5 Option 2) 5:4 Option 3) 7:10 Option 4) 8:5

As learnt in

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

$aA+bB\rightleftharpoons cC+dD$

$K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}$

$[A],\:[B],\:[C]\:[D]$

are equilibrium concentration

The reaction is as follows:

N2        +        3H2        $\rightleftharpoons$        2NH3

at t=0     0.2                0.6                        0

at t1         0.2 - x          0.6 - 3x                    2x

We know that, 0.4 x 0.8 = x + 3x  (given)

$\Rightarrow$ x = 0.08

$\therefore$ N2 = 0.12

H2 = 0.36

NH3 = 0.16

Gases initial = 0.8

Gases final = 0.64

$\Rightarrow \frac{final}{initial} = \frac{4}{5}$

Option 1)

4:5

This option is correct

Option 2)

5:4

This option is incorrect

Option 3)

7:10

This option is incorrect

Option 4)

8:5

This option is incorrect

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