Answer:
[tex]p_{CCl_4}^0=1.523atm[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]CCl_4(g)\rightarrow C(s)+2Cl_2(g)[/tex]
Therefore, the law of mass action is:
[tex]Kp=\frac{p_{Cl_2}^2}{p_{CCl_4}}[/tex]
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:
[tex]P_T=1.2atm=p_{CCl_4}+p_{Cl_2}=(p_{CCl_4}^0-x)+2x[/tex]
And in the law of mass action:
[tex]0.75=\frac{(2x)^2}{p_{CCl_4}^0-x}[/tex]
Now, from the total pressure, we have that:
[tex](p_{CCl_4}^0-x)+2x=1.2atm\\\\p_{CCl_4}^0=1.2atm-2x+x\\\\p_{CCl_4}^0=1.2atm-x[/tex]
And we combine:
[tex]0.75=\frac{(2x)^2}{1.2atm-x-x}\\\\0.75=\frac{(2x)^2}{1.2atm-2x}[/tex]
Thus, solving for [tex]x[/tex] we have to roots:
[tex]x_1=-0.698atm\\\\x_2=0.323atm[/tex]
Clearly, the solution must be 0.323 atm, in such a way, the initial pressure turns out:
[tex]p_{CCl_4}^0=1.2atm+x=1.2atm+0.323atm\\\\p_{CCl_4}^0=1.523atm[/tex]
Regards.
