Problem 15.52 For the equilibrium Br2( g )+Cl2( g )2BrCl( g ) at 400 K, K c = 7.
ID: 938247 • Letter: P
Question
Problem 15.52
For the equilibrium
Br2(g)+Cl2(g)2BrCl(g)
at 400 K, Kc = 7.0.
Part A
If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Br2?
Express your answer to two significant figures and include the appropriate units.
SubmitMy AnswersGive Up
Part B
If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Cl2?
Express your answer to two significant figures and include the appropriate units.
SubmitMy AnswersGive Up
Part C
If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of BrCl?
Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
Br2 + Cl2 <> 2BrCl
K = [BrCl]^2 / [Br2][Cl2]
K = 7
a)
If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Br2?
initiallly:
[Br2] = 0.25/3 = 0.0833
[Cl2] = 0.55/3 = 0.183
[BrCl2] = 0
then; in equilibrium
[Br2] = 0.0833-x
[Cl2] = 0.183-x
[BrCl2] = 0 +2x
substitute in K
K = [BrCl]^2 / [Br2][Cl2]
7 = (2x)^2/ (0.0833-x)(0.183-x)
(0.0833*0.183 -(0.0833+0.183)x + x^2) = 4/7x^2
0.0152439 - 0.2663x + x^2= 0.5714285x^2
0.42857x^2 - 0.2663x + 0.0152439 = 0
x = 0.0637
substitute
[Br2] = 0.0833-x = 0.0833-0.0637 = 0.0196 M
[Br2] = 0.0196 M
[Cl2] = 0.183-x = 0.183-0.0637
[Cl2] = 0.1193 M
[BrCl2] = 0 +2x = 2*0.0637 = 0.1274
[BrCl2] =0.1274 M