Question

At 700 K, CCl4 decomposes to carbon and chlorine. The Kp for the decomposition is 0.76. Find the starting pressure of CCl4 at this temperature that will produce a total pressure of 1.2 atm at equilibrium.

P=?

Answer 1

`p_(CCl_4)^0=1.523atm`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`CCl_4(g)\rightarrow C(s)+2Cl_2(g)`

Therefore, the law of mass action is:

`Kp=(p_(Cl_2)^2)/(p_(CCl_4))`

Pressure of carbon at equilibrium is not considered since it is solid. In such a way, based on the law of mass action, at equilibrium we have:

`P_T=1.2atm=p_(CCl_4)+p_(Cl_2)=(p_(CCl_4)^0-x)+2x`

And in the law of mass action:

`0.75=((2x)^2)/(p_(CCl_4)^0-x)`

Now, from the total pressure, we have that:

`(p_(CCl_4)^0-x)+2x=1.2atm\\\\p_(CCl_4)^0=1.2atm-2x+x\\\\p_(CCl_4)^0=1.2atm-x`

And we combine:

`0.75=((2x)^2)/(1.2atm-x-x)\\\\0.75=((2x)^2)/(1.2atm-2x)`

Thus, solving for
`x` we have to roots:

`x_1=-0.698atm\\\\x_2=0.323atm`

Clearly, the solution must be 0.323 atm, in such a way, the initial pressure turns out:

`p_(CCl_4)^0=1.2atm+x=1.2atm+0.323atm\\\\p_(CCl_4)^0=1.523atm`

Regards.