Christmas Chemistry

Chemistry Level 2

Given the reaction below and the information provided, determine the K p K_p of the reaction.

P C l 5 ( g ) ⇌ C l 2 ( g ) + P C l 3 ( g ) \mathrm{PCl}_5 (g)\rightleftharpoons \mathrm{Cl}_2 (g)+\mathrm{PCl}_3(g)

Initially, 2.00 a t m 2.00 \ \mathrm{atm} of P C l 5 \mathrm{PCl}_5 is found in a container. After establishing equilibrium, 1.00 a t m 1.00 \ \mathrm{atm} of C l 2 \mathrm{Cl}_2 and 1.00 a t m 1.00 \ \mathrm{atm} of P C l 3 \mathrm{PCl}_3 is found.

1
<1 Can not be determined 1

Hamza A by Hamza A , 19, USA

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1 solution

Hamza A
Dec 25, 2018

Using a RICE table we have the following

R P C l 5 ( g ) ⇌ C l 2 ( g ) + P C l 3 ( g ) R \ \mathrm{PCl}_5 (g)\rightleftharpoons \mathrm{Cl}_2 (g)+\mathrm{PCl}_3(g)

I 1.00 a t m 0.00 a t m 0.00 a t m I \ 1.00 \mathrm{atm} \ \ \ \ \ 0.00\mathrm{atm} \ \ \ \ \ \ 0.00 \mathrm{atm}

C − x + x + x C \ -x \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ +x

E ? 1.00 a t m 1.00 a t m E \ \ \ ? \ \ \ \ \ \ \ \ \ \ \ \ 1.00\mathrm{atm} \ \ \ \ \ \ \ \ \ 1.00 \mathrm{atm}

Therefore,

x = 1.00 a t m ⟹ [ P C l 5 ] E = 2.00 a t m − x = 1.00 a t m x=1.00 \mathrm{atm} \implies [\mathrm{PCl}_5]_{E}=2.00\mathrm{atm}-x=1.00 \mathrm{atm}

Hence, K p = P C l 2 P P C l 3 P P C l 5 = 1.00 ⋅ 1.00 1.00 = 1.00 K_p=\frac{P_{\mathrm{Cl}_2} P_{\mathrm{PCl}_3}}{P_{\mathrm{PCl}_5}}=\frac{1.00 \cdot 1.00}{1.00}=1.00

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