Question

A sealed 1.5-L container initially holds 0.00623 mol H2, 0.00414 mol Br2, and 0.0244 mol of HBr at 550 K. when equilibrium is established, [H2]= 0.00467M

H2(g)+Br2(g) <-> 2HBr(g)

A. what are [HBr}eq and [Br2}eq?

B. what are Kc and Kp at 550 K?

C.A 0.00209 mol sample of Br2 is added to the equilibrium mixture of gases. What are the partial pressures of all species once equilibrium is reestablished?

Answer 1

A) the calculations are given below

We would create an ICE table to calculate the change in concentration and euilibrium concentrations of all the species

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Reaction :         H2    +    Br2 -------------------> 2HBr

Initial               0.00623   0.00414                    0.0244

change            -0.0015    -0.0015                    +0.0030

Equilibrium      0.00467    0.00276                    0.0274

----------------------------------------------------------------------------

Thus, the equilibrium concentrations of the species are,

Molarity of H2 = 0.00467/1.5 = 0.00311 M

molarity of Br2 = 0.00276/1.5 = 0.00176 M

molarity of HBr = 0.0274/1.5 = 0.018 M

B.

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

     = (0.018)^2/(0.00311)(0.00176)

     = 59.20

pH2 = 0.00311 x 0.0821 x 550 = 0.14

pBr2 = 0.0794

pHBr = 0.812

Kp = (0.812)^2/(0.14)(0.0794)

     = 59.31

C. When 0.00209 mol of Br2 was added at equilibrium,

new concentration of Br2 = 0.00176 + 0.00209 / 1.5 = 0.00315

pBr = 0.142

Kp = 33.17