Answer: The value of equilibrium constant at -40°C is [tex]1.26\times 10^{-5}[/tex]
Explanation:
We are given:
Percent degree of dissociation = 0.456 %
Degree of dissociation, [tex]\alpha[/tex] = 0.00456
Concentration of [tex]N_2O_4[/tex], c = 0.15 M
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 =[0.15-(0.15\times 0.00456)]=0.1493M[/tex]
Equilibrium concentration of [tex]NO_2=2c\alpha =[2\times 0.15\times 0.00456]=0.00137M[/tex]
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=\frac{[NO_2]^2}{[N_2O_4]}[/tex]
Putting values in above equation, we get:
[tex]K_{eq}=\frac{(0.00137)^2}{0.1493}\\\\K_{eq}=1.26\times 10^{-5}[/tex]
Hence, the value of equilibrium constant at -40°C is [tex]1.26\times 10^{-5}[/tex]
