Question

PCl5 dissociates according to the reaction: PCl5(g) = PCl3(g) + Cl2(g) One mole of PCl5 was placed in one liter of solution. When equilibrium was established, 0.2 mole of PCl5 remained in the mixture. What is the equilibrium constant for this reaction? (Hint: remember the ICE procedure? initial, change, and equilibrium)

Answer 1

Answer:

$K=3.2$

Explanation:

Hello,

In this case, for the given equilibrium, we write the law of mass action:

$K=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Next, in terms of the change $x$ due to the reaction extent (ICE procedure):

$K=\frac{x*x}{1M-x}$

Clearly, the initial concentration phosphorous pentachloride is 1 M (one mole per litre), therefore, since the equilibrium concentration is 0.2 M (same volume) we can compute $x$:

$x=1M-0.2M=0.8M$

Thus, we compute the equilibrium constant:

$K=\frac{0.8*0.8}{0.2} \\\\K=3.2$

Regards.