Answer to Question #207930 in General Chemistry for Earl

Question

Answer the following problem

A. 1.00 Mole sample of NOCI was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mole of CI2. Calculate K at this temperature. The equation is as follows:

2NOCI(g) <--> 2NO(g) + CI2(g)

Answer 1

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

2mol NOCl is consumed when 2mol NO and 1mol Cl2 produced .

Therefore, when 0.056mol Cl2 is there , then 2× 0.056 = 0.112mol of NO(g) must be there and moles of NOCl must be decreased by 0.112mol ,then available mol of NOCl = 1mol - 0.112mol = 0.888mol NOCl .

Thus equilibrium concentration of NO(g) = 0.112mol , Cl2(g) = 0.056mol and that of NOCl(g) = 0.888mol

Equilibrium constant (K) = equilibrium concentration of products with raise to power of stoichiometric coefficient ÷ equilibrium concentration of reactants with raise to power stoichiometric constant.

K for given reaction = [NO]^2[Cl2]/[NOCl]^2

K = (0.112)^2(0.056)/(0.888)^2 = 0.000702464/0.78854 = 8.91×10^-4

Thus equilibrium constant = 8.91×10^-4