We have an equation (1) in equilibrium and we are given the value of its equilibrium constant Kc1 = 4.4x10^8.
The second reaction corresponds to the same initial reaction but now the coefficients are multiplied by two. Let us determine the Kc equations for each reaction.
Reaction 1:
[tex]Kc_1=\frac{\lbrack COCl_2\rbrack\lbrack Cl_2\rbrack}{\lbrack CCl_4\rbrack\lbrack O_2\rbrack^{1/2}}=4.4\times10^8[/tex]
Reaction 2:
[tex]Kc_2=\frac{\lbrack COCl_2\rbrack^2\lbrack Cl_2\rbrack^2}{\lbrack CCl_4\rbrack^2\lbrack O_2\rbrack}[/tex]
Now, if we compare both equations we can notice that the equation of the second reaction is equivalent to the first reaction squared, that is:
[tex]\left(Kc_1\right)^2=\frac{\lbrack COCl_2\rbrack^2\lbrack Cl_2\rbrack^2}{\lbrack CCl_4\rbrack^2\lbrack O_2\rbrack}=Kc_2[/tex]
Therefore, the value of Kc for the second reaction will be:
[tex]\begin{gathered} Kc_2=Kc_1^2=(4.4\times10^8)^2 \\ Kc_2=1.9\times10^{17} \end{gathered}[/tex]
Answer: Kc for the reaction 2CCl4 + O2 <--->2COCl2 + 2Cl2 is 1.9 x 10^17
