12) A container holds a mixture of H2(g) and Br2(g) at 1000 K. HBr(g) The two ga

Question

12) A container holds a mixture of H2(g) and Br2(g) at 1000 K. HBr(g) The two gases react to form H2(g) + Br(g) -2 HBr(g) The following data are collected in an experiment: Initial pressure of H2(g)-20.0 atm Initial pressure of Br2(g)- 10.0 atm Equilibrium pressure of Br2(g) = 0.00105 atm a) Calculate Kp for this reaction at 1000 K. b) Calculate the equilibrium pressure of Br2 that will be observed when pure HBr at a pressure of 30.0 atm is placed into a container and allowed to come to equilibrium with H2 and Br2 at 1000 K

Explanation / Answer

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

Kp = (p(HBr))^2/p(H2) p(Br2)

at equillibrium

p(H2) = 20 - x

p(Br2) = 10-x

p(HBr) = 2x

10 -x = 0.00105atm

x = 10 - 0.00105atm

= 9.99895atm

p(H2) = 20 - 9.99895atm

= 10.00105atm

p(HBr) = 2× 9.99895atm = 19.9979atm

Kp = (19.9979atm)^2/(10.00105atm)(0.00105atm)

= 3.8×10^4

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

Kp = 1/3.8×10^4 = 2.63 ×10^-5

at equillibrium ,

p(HBr) = 30 - 2x   

p(H2) = x

p(Br2) = x

x^2/(30 -2x)^2= 2.63×10^-5

x / 30 - 2x = 5.13×10^-3

x = 0.1539 - 0.01026x

x = 0.1539/ 1.01026

= 0.1523

Therefore, at equillibrium

p(H2) = 0.1523atm

p(Br2) = 0.1523atm

p(HBr ) = 30 - 2(0.1523) =29.69 atm

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