Question

Calculate the partial pressure (in atm) of C2H5Br at equilibrium when 1.23 atm of C2H6 and 6.21 atm of Br2 react at 2500 K according to the following chemical equation:

C2H6 (g) + Br2 (g) ⇌ C2H5Br (g) + HBr (g) Kp = 6.87

Report your answer to three significant figures in scientific notation.

Answer 1

C2H6 (g) + Br2 (g) ⇌ C2H5Br (g) + HBr (g) Kp = 6.87

Kp = P-C2H5Br * P-HBr / P-C2H6 * P-Br2

P-x = partial pressure of x

P-C2H5Br= 0

P-HBr = 0

P-C2H6= 1.23

P-Br2 = 6.21

in equilibrium

P- C2H5Br= 0 + x

P-HBr = 0 + x

P- C2H6 = 1.23 - x

P-Br2 = 6.21 - x

Solve

Kp = x*x/(1.23 - x)(6.21 - x)

6.87 = x^2/(1.23 - x)(6.21 - x)

(1.23 - x)(6.21 - x) = x^2 / 6.87

1.23*6.21 -(1.23+6.21)x + x^2 = (1/6.87) x^2

(1/6.87 - 1) x^2 - (1.23+6.21)x + 1.23*6.21 = 0

-0.85443x^2 - 7.44x +7.6383 = 0

x = 0.927795

substitute

P - C2H5Br = 0 + 0.927795 = 0.927795 atm