Question

Consider the following reaction (assume an ideal gas mixture): 2NOBr(g) 2NO(g) + Br2(g)A 1.0-liter vessel was initially filled with pure NOBr, at a pressure of 4.0 atm, at 300 K. After equilibrium was established, the partial pressure of NOBr was 3.1 atm. What is Kp for the reaction?

Answer 1

Answer:kp= 0.27

Explanation:

The equation of reaction is given as;

2 NOBr(g) ⇋ 2 NO(g) + Br₂(g)

The partial pressure in equilibrium are related as;

Kp = (P_NO)²∙(P_Br₂ )/ (P_NOBr)².

From the question, the parameters given are; P= 4.0atm, V= 1.0L, T=300k.

PV= nRT--------(ideal gas law equation). Where P= pressure, V= volume, n= number of moles, T= temperature and R= constant.

Making number of moles,n to be the subject of the formula, we have;

n(NOBr) = pV/RT = 4.0 x1.0/(0.082 x 300) = 0.163 mol.

Kp = P(Br2) x P(NO)^2 / P (NOBr)^2.

Partial pressure,P1 = X1 x P(total), whereX1= mole fraction = n1 / Sn

X(NOBr) = P(NOBr) / P(total) = 2.5/4.0 = 0.625

X(Br2) = (1 - 0.625)/3 = 0.1875

X(NO) = 2 x X(Br2) = 2 x 0.1875 = 0.375

P(Br2) = X(Br2) x P(total) = 0.1875 x 4.0 =0.75 atm

P(NO) = X(NO) x P(total) = 0.375 x 4.0 = 1.5atm

Kp = 0.75 x 1.5^2 / 2.5^2 = 0.27

Kp= 0.27