# chemistry?

The equilibrium constant Kc for the following reaction is 2.18 106 at 730°C.

H2(g) + Br2(g) ⇌2 HBr(g)

Starting with 3.40 moles of HBr in a 12.0 L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium.

Update:

2.18*10^6

### 1 Answer

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- hcbiochemLv 72 months ago
Set up an ICE table:

.................[H2]......[Br2]........[HBr]

Initial...........0...........0.............0.283

change.......+x.........+x.............-2x

Equilib.........x............x........ 0.283-2x

Kc = [HBr]^2 / [H2][Br2] = 2.18X10^6

Kc = (0.283-2x)^2 / x^2 = 2.18X10^6

As the first step, take the square root of both sides to give:

0.283-2x/x = 1476

0.283 = 1478 x

x = 1.92X10^-4 M = [H2] = [Br2]

[HBr] = 0.283-2x = 0.283 M

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