Question

For the reaction: H2(g) + Br2(g) ⇌ 2 HBr(g), Kc = 7.5 × 102 at a...

For the reaction: H2(g) + Br2(g) ⇌ 2 HBr(g), Kc = 7.5 × 102 at a certain temperature. If 2 mole each of H2 and Br2 are placed in 2-L flask, what is the concentration of H2 at equilibrium?

A) 0.96 B) 0.93 C) 1.86 D)0.04 E) 0.07

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Homework Answers

Answer #1

Initial concentration of H2 = mol of H2 / volume in L

= 2 mol / 2 L

= 1.00 M

Initial concentration of Br2 = mol of Br2 / volume in L

= 2 mol / 2 L

= 1.00 M

ICE Table:

Equilibrium constant expression is

Kc = [HBr]^2/[H2]*[Br2]

750.0 = (2*x)^2/(1-1*x)^2

sqrt(750.0) = (2*x)/(1-1*x)

27.386 = (2*x)/(1-1*x)

27.39-27.39*x = 2*x

27.39-29.39*x = 0

x = 0.932

At equilibrium:

[H2] = 1.0 - x = 1 - 0.932 = 0.068 M

Answer: E

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