Answer :
Answer: The value of [tex]K_c[/tex] for the reaction at 550.3 K is 247.83
Explanation:
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{c}[/tex]
For a general chemical reaction:
[tex]aA+bB\rightarrow cC+dD[/tex]
The expression for [tex]K_{c}[/tex] is written as:
[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]
The chemical equation for the production of methanol follows:
[tex]CO+2H_2\rightleftharpoons CH_3OH[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[CH_3OH]}{[CO][H_2]^2}[/tex]
We are given:
[tex][CH_3OH]=0.0401mol/L[/tex]
[tex][CO]=0.02722mol/L[/tex]
[tex][H_2]=0.07710mol/L[/tex]
Putting values in above equation, we get:
[tex]K_c=\frac{0.0401}{0.02722\times (0.07710)^2}\\\\K_c=247.83[/tex]
Hence, the value of [tex]K_c[/tex] for the reaction at 550.3 K is 247.83