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

 

Answers (1)

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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