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

To determine the partial pressure of NO2 at equilibrium, use the equilibrium expression and substitute the given values into the equation.

Step-by-step explanation:

To determine the partial pressure of NO2 at equilibrium, we will use the equilibrium expression for the reaction:

Kp = (P(NO2))^2 / P(N2O4)

Given that the equilibrium constant Kp is 70.9 and the equilibrium concentration of N2O4 is 25.8% of its initial concentration, we can calculate the partial pressure of NO2 at equilibrium:

P(NO2) = sqrt(Kp * P(N2O4))

Substituting the values, we get:

P(NO2) = sqrt(70.9 * (1 - 0.258))

P(NO2) = sqrt(70.9 * 0.742)

Answer 2

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=((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=((2y)^2)/((x-y))` ..............(2)

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

`25.8\% * x=(x-y)`

`(25.8)/(100)* x=(x-y)`

`0.258* x=(x-y)`

`0.742x=y` ..............(3)

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

`70.9=((2* 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* 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