Question

At particular temperature, Kp = 70.9 for the following reactionN2O4(g) == 2NO2(g)1. A certain pressure of N2O4 is initially added to an otherwise evacuated.2. At equilibrium, 25.8% of N2O4 remains.3. What is the partial pressure of NO2 at equilibrium?

Answer 1

Answer : The partial pressure of $NO_2$ is, 12.34  atm

Explanation :

For the given chemical reaction:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{(P_{NO_2})^2}{P_{N_2O_4}}$         ........(1)

The equilibrium reaction is:

                     $N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial             x                    0

At eqm       (x-y)                 2y

Putting values in expression 1, we get:

$70.9=\frac{(2y)^2}{(x-y)}$           ..............(2)

As we are given that, 25.8 % of $N_2O_4$ remains at equilibrium. That means,

$25.8\% \times x=(x-y)$

$\frac{25.8}{100}\times x=(x-y)$

$0.258\times x=(x-y)$

$0.742x=y$       ..............(3)

Now put equation 3 in 2, we get the value of 'x'.

$70.9=\frac{(2\times 0.742x)^2}{(x-0.742x)}$

$x=8.31$

Now put the value of 'x' in equation 3, we get:

$0.742x=y$

$0.742\times 8.31=y$

$y=6.17$

Now we have to calculate the new partial pressure of $NO_2$ at equilibrium.

Partial pressure of $NO_2$ = (2y) = (2\times 6.17) = 12.34  atm

Hence, the partial pressure of $NO_2$ is, 12.34  atm