Answer to Question #269372 in General Chemistry for Aaa

Question

A 1.86 mol sample of NOBr is placed in a 5.0 liter evacuated flask at 25oC. Analysis of the mixture at equilibrium shows that 0.164 mol of NO is present.What is the value of the equilibrium constant, Kc for this reaction? 2NOBr⇌2NO+Br2

Answer 1

Reaction is, 

2NOBR$\iff$ 2NO + Br₂

Mole ratio for reaction is:

2 :2 : 1 respectively for the equation.

Initial moles of NOBr is 1.86moles

At equilibrium, the moles of NO present are 0.164 moles

From Mole ratio, mole of Br_{2 }will be 0.082 moles.

Remaining moles of NOBr is (1.86-0.164)=0.704 moles.

Equilibrium constant Kc = [C]^c [D]^d /[A] ^a[B]^b

Molarity of NOBr =1.696÷5=0.3392

Molarity of NO= 0.164 ÷5=0.0328

Molarity of Br = 0.082 ÷5= 0.0164

Kc=[0.0328]²[0.0164]/[0.3392]²

^{Answer is} 0.0001533486