Respuesta :
Answer: The value of equilibrium constant is 0.997
Explanation:
We are given:
Percent degree of dissociation = 39 %
Degree of dissociation, [tex]\alpha[/tex] = 0.39
Concentration of [tex]N_2O_4[/tex], c = [tex]\frac{1mol}{1L}=1M[/tex]
The given chemical equation follows:
[tex]N_2O_4\rightleftharpoons 2NO_2[/tex]
Initial: c -
At Eqllm: [tex]c-c\alpha[/tex] [tex]2c\alpha[/tex]
So, equilibrium concentration of [tex]N_2O_4=c-c\alpha =[1-(1\times 0.39)]=0.61M[/tex]
Equilibrium concentration of [tex]NO_2=2c\alpha =[2\times 1\times 0.39]=0.78M[/tex]
The expression of [tex]K_{c}[/tex] for above equation follows:
[tex]K_{c}=\frac{[NO_2]^2}{[N_2O_4]}[/tex]
Putting values in above equation, we get:
[tex]K_{c}=\frac{(0.78)^2}{0.61}\\\\K_{c}=0.997[/tex]
Hence, the value of equilibrium constant is 0.997
