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

Answer:

$p_{CCl_4}^0=1.523atm$

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:

$Kp=\frac{p_{Cl_2}^2}{p_{CCl_4}}$

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:

$0.75=\frac{(2x)^2}{p_{CCl_4}^0-x}$

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:

$0.75=\frac{(2x)^2}{1.2atm-x-x}\\\\0.75=\frac{(2x)^2}{1.2atm-2x}$

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.