Answer to Question #173187 in General Chemistry for Justin

Question #173187

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)

Expert's answer

Consider this reaction process as shown below:

Reaction : "2 NOBr (g) \u21cc 2 NO (g) + Br\u2082 (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"