Question

9.2 grams of $N_2{O}_4$ is taken in a closed one litre vessel and heated till the following equilibrium is reached $N_{2}O-4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ At equilibrium, $50\%N_2{O}_4\left(g\right)$ is dissocated.What is the equilibrium constant (in mol $litr{e}^{-1}$)? (molecular mass of $N_2{O}_4=92$)

• A. 0.1
• B. 0.4
• C. 0.2
• D. 2

Answer 1

The correct option is C 0.2

The equilibrium reaction given to us

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

Use the molar mass of dinitrogen tetroxide to find the number of moles present in the sample

$9.2g\times \frac{1moleof{N}_2{O}_4}{92.011g}=0.100molesof{N}_2{O}_4$

This means that the initial concentration of dinitrogen tetroxide is equal to

$0.100mol{L}^{-1}$

Now the problem tells us that at equilibrium $50\%$ of the added dinitrogen tetroxide is dissociated.

So, if half of the initial concentration gets converted to nitrogen dioxide, we can say that te equilibrium concentration of dinitrogen tetroxide will be

$[N_2{O}_4{]}_{eq}=\frac{1}{2}\times 0.100mol{L}^{{}^{-}1}=0.0500mol{L}^{{}^{-}1}$

Notice that for every molecule of dinitrogen tetroxide that dissociates, $2$ molecules of nitrogen dioxide are produced.

$[N{O}_2{]}_{eq}=2\times 0.0500mol{L}^{{}^{-}1}$

By defination , the equilibrium constant for this reation will be

$K_c=\frac{{\left(0.100mol{L}^{-1}\right)}^2}{0.0500mol{L}^{-1}}$

$K_c=0.2mol{L}^{-1}$

Hence, option c is correct.