Question

A sample of NOCl decomposes according to the following equation:

2 NOCl(g) 2 NO(g) + 1 Cl2(g)

An equilibrium mixture in a 3-L vessel at 165 oC, contains 0.0199 g of NOCl, 0.00456 g of NO, and 0.0272 g of Cl2.

(a) Calculate KP for this reaction at this temperature. KP = ?

(b) What is the total pressure exerted by the equilibrium mixture of gases? Ptotal = ___atm.

Answer 1

2NOCl(g) -------> 2NO(g) + Cl₂(g)

Molar mass of Cl₂ = 71 g/mole

molar mass of NO = 30 g/mole

Molar mass of NOCl = 65.5 g/mole

Thus, moles of NOCl in 0.0199 g of it = mass/molar mass = 0.0199/65.5 = 3.04*10^-4

moles of NO in 0.00456 g of it = mass/molar mass = 0.00456/30 = 0.000152

moles of Cl₂ in 0.0272 g of it = mass/molar mass = 0.0272/71 = 3.831*10^-4

Now, at eqb., [NOCl] = mass/volume of solution in litres = 1.013*10^-4 M

[NO] = 5.067*10^-5 m ; [Cl₂] = 0.000128 M

Thus, K_c = {[NO]²*[Cl₂]}/[NOCl]² = 3.2*10^-5

Now, K_p = K_c*(R*T)ⁿ ; where n = moles of gaseous products - moles of gaseous reactants = 1

Thus, K_p = 0.00115

Now, total moles of gases in the equilibrium mixture = 0.0008391

Applying Ideal Gas Equation, we get

total pressure, P = n*R*T/V = 0.0008391*0.0821*438/3 = 0.0101 atm