Question

An evacuated reaction vessel is filled with NOBr to an initial pressure of 5.00 atm. When the system reaches equilibrium according to the reaction below, there are 0.750 atm of Br₂. What is Kp for this reaction?

2 NOBr (g) ⇌ 2 NO (g) + Br₂ (g)

Answer 1

Consider this reaction process as shown below:

Reaction : $2 NOBr (g) ⇌ 2 NO (g) + Br₂ (g)$

At \ t=0\ \ \ \ \ \ \ 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \

At $\ t=T$$\ \ \ \ \ 5-2\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha$

And, at equilibrium time ,pressure of $Br_2$ is$\ \alpha =0.750\ atm$

So $,\ p_{NO}=2\alpha=2\times 0.750=1.5\ atm$

And,$p_{NOBr}=5-2\times 0.750=5-1.5=3.5\ atm$

$K_p=\frac{[p_{NO}]^2[p_{Br_2}]}{[p_{NOBr}]^2}$ $=\frac{1.5^2\times 0.750}{3.5^2}\ atm=0.1377\ atm\approx0.14\ atm$