1. For the reaction in the previous problem, that is, 2HI(g) ? H 2 (g) + I 2 (g)
ID: 846408 • Letter: 1
Question
1. For the reaction in the previous problem, that is,
2HI(g) ? H2(g) + I2(g) Keq = 0.016
Initially a container contains 0.32 M HI and no product. What is the equilibrium concentration of H2?
2.For the reaction:
2HBr(g) ? H2(g) + Br2(g)
Initially a container contains 0.71 M HBr and no product. What is the equilibrium concentration of HBr if the equilibrium concentration of H2 is 0.19 M?
3
2HI(g) ? H2(g) + I2(g) Keq = 0.016
Initially a container contains 0.60 M HI, 0.038 M H2, and 0.15 M I2 at equilibrium. What is the new equilibrium concentration of H2, if the H2 concentration is increased by 0.276 M?
Explanation / Answer
2HBr(g) <--> H2(g) + Br2(g)
[HBr] = 0.86 - 2X
[H2] = 0.00 + X
[Br2] = 0.00 + X
Keq = 0.016 = [H2][Br2] / [HBr]^2
0.016 = X^2 / (0.86-X)^2
Take the square root of both sides
0.1264911 = X/(0.86-X)
0.1087824 - 0.1264911X = X
0.1087824 = 1.1264911X
X = 0.0966
[H2] = 0.0966 M (0.097 M; 2 s.f.)
Same problem with new numbers-
[HI] = 0.81 + 2X
[H2] = 0.038 + 0.280 - X = 0.318 - X
[I2] = 0.15 - X
Keq = 0.016 = [H2][I2] / [HI]^2
Substitute the values for yhe unknowns, then
expand and solve the quadratic equation for X
[H2] = 0.318 - X
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