Answer to Question #118033 in General Chemistry for Crizzia Paula Lobrido

Question

1. Consider the following exchange of reactions :

2BrCl(g) ⇄ Cl2(g) + Br2(g) ; Kp =0.169
2IBr(g) ⇄ Br2(g) + I2(g); Kp = 0.0149

a. Determine the value of Kp on the following reaction:

BrCl(g) + ½ I2(g) ⇄ IBr(g) + ½ Cl2(g); Kp=?

b. Determine Kc at 25°C

Answer 1

subtract reaction 2 from reaction 1 we will get

2BrCl(g)+I₂(g)<=>Cl₂(g)+2IBr(g)

so Kp for this reaction will Kp1/Kp2

now divide the obtained reaction by 2 we will get

BrCl(g)+1/2I₂(g)<=>1/2Cl₂(g)+IBr(g)

Kp(new) will be √(Kp1/Kp2)

hence √(0.169/0.0149)=3.36

for part B)Kp=Kc(RT)^(∆n)

here ∆n is change in no of moles of gases

in above reaction ∆n is zero

so Kp=Kc

Kc = 3.36