Question

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.(see image)

Answer 1

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:

`Kc_1=(\lbrack COCl_2\rbrack\lbrack Cl_2\rbrack)/(\lbrack CCl_4\rbrack\lbrack O_2\rbrack^(1/2))=4.4*10^8`

Reaction 2:

`Kc_2=(\lbrack COCl_2\rbrack^2\lbrack Cl_2\rbrack^2)/(\lbrack CCl_4\rbrack^2\lbrack O_2\rbrack)`

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:

`\left(Kc_1\right)^2=(\lbrack COCl_2\rbrack^2\lbrack Cl_2\rbrack^2)/(\lbrack CCl_4\rbrack^2\lbrack O_2\rbrack)=Kc_2`

Therefore, the value of Kc for the second reaction will be:

`\begin{gathered} Kc_2=Kc_1^2=(4.4*10^8)^2 \\ Kc_2=1.9*10^(17) \end{gathered}`

Answer: Kc for the reaction 2CCl4 + O2 <--->2COCl2 + 2Cl2 is 1.9 x 10^17