Respuesta :
Answer: The partial pressure of nitrogen dioxide after it is reached the second time is 1.454 atm
Explanation:
We are given:
Initial partial pressure of nitrogen dioxide gas = 1.9 atm
First equilibrium partial pressure of nitrogen dioxide gas = 1.1 atm
The chemical equation for the reaction of nitrogen dioxide to dinitrogen tetroxide follows:
[tex]2NO_2\rightleftharpoons N_2O_4[/tex]
Initial: 1.9
At eqllm: 1.9-2x x
Evaluating the value of 'x'
[tex]\Rightarrow (1.9-2x)=1.1\\\\x=0.4[/tex]
So, equilibrium partial pressure of dinitrogen tetroxide = x = 0.4 atm
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}[/tex] ..........(1)
Putting values in above expression, we get:
[tex]K_p=\frac{0.4}{(1.1)^2}\\\\K_p=0.3306[/tex]
Now, when more nitrogen dioxide is added, equilibrium is re-established:
Partial pressure of nitrogen dioxide added = 0.95 atm
[tex]2NO_2\rightleftharpoons N_2O_4[/tex]
Initial: 2.05 0.4
At eqllm: 2.05-2x 0.4+x
Putting values in expression 1, we get:
[tex]0.3306=\frac{(0.4+x)}{(2.05-2x)^2}\\\\x=2.51,0.298[/tex]
Neglecting the value of x = 2.51 because the equilibrium partial pressure of nitrogen dioxide will become negative, which is not possible.
So, the equilibrium partial pressure of nitrogen dioxide after it is reached the second time = (2.05 - 2x) = [2.05 - 2(0.298)] = 1.454 atm
Hence, the partial pressure of nitrogen dioxide after it is reached the second time is 1.454 atm