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pogonyaev
3 years ago
7

At particular temperature, Kp = 70.9 for the following reactionN2O4(g) == 2NO2(g)1. A certain pressure of N2O4 is initially adde

d to an otherwise evacuated.2. At equilibrium, 25.8% of N2O4 remains.3. What is the partial pressure of NO2 at equilibrium?
Chemistry
1 answer:
elixir [45]3 years ago
8 0

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

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