Question

At elevated temperature, carbon tetrachloride decomposes to its elements: CCl4(g) C(s) 2Cl2(g). At 700 K, if the initial pressure of CCl4 is 1. 00 atm and at equilibrium the total pressure is 1. 35 atm, then calculate K

Answer 1

Answer:

$K=0.0131M$

Explanation:

Hello,

In this case, the undergoing chemical reaction turns out:

$CCl_4(g) \rightleftharpoons C(s)+ 2Cl_2(g)$

In such a way, by means of the mass of law action for such reaction, which is given below:

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

And in terms of the change $x$ due to reaction extent:

$p_{CCl_4}=p_{CCl_4}^0-x\\p_{Cl_2}=2x\\p_T=p_{CCl_4}+p_{Cl_2}=1.00-x+2x\\p_T=1.00+x$

$x$ results:

$x=1.35 atm -1.00atm=0.35atm$

In such a way, Kp:

$Kp=\frac{(2x)^2}{1.00-x} =\frac{(2*0.35atm)^2}{1.00atm-0.35atm}=0.754atm$

Nonetheless, K is asked instead of Kp, thus:

$K=\frac{Kp}{(RT)^{\Delta \nu _{gas}}}$

Whereas:

$\Delta \nu _{gas}=2-1=1$

Which is the change in the moles of gaseous species chlorine and carbon tetrachloride. Hence, we finally obtain:

$K=\frac{0.754atm}{(0.082\frac{atm*L}{mol*K}*700K)^{1}}\\\\K=0.0131mol/L=0.0131M$

Best regards.