Respuesta :
Answer: The equilibrium constant, [tex]K_p[/tex] for the given reaction is 6.653
Explanation:
We are given:
Initial partial pressure of nitrogen dioxide = 1.00 atm
Initial partial pressure of dinitrogen tetraoxide = 1.500 atm
Equilibrium partial pressure of nitrogen dioxide = 0.512 atm
For the given chemical equation:
[tex]2NO_2(g)\rightleftharpoons N_2O_4(g)[/tex]
Initial: 1.00 1.500
At eqllm: 1.00-2x 1.500+x
Evaluating the value of 'x'
[tex]\Rightarrow (1.00-2x)=0.512\\\\x=0.244[/tex]
So, equilibrium partial pressure of dinitrogen tetraoxide = (1.500 + x) = [1.500 + 0.244] = 1.744 M
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{1.744}{(0.512)^2}\\\\K_p=6.653[/tex]
Hence, the equilibrium constant, [tex]K_p[/tex] for the given reaction is 6.653
