For the reaction in the previous problem, that is, 1. 2HI(g) ? H 2 (g) + I 2 (g)

Question

For the reaction in the previous problem, that is,

1. 2HI(g) ? H2(g) + I2(g) Keq = 0.016

Initially a container contains 0.46 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 0.64 M HBr and no product. What is the equilibrium concentration of HBr if the equilibrium concentration of H2 is 0.16 M? (Hint provided in feedback.)

3.

For the reaction:

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 0.124 M?

Explanation / Answer

here also , we first write ICE table

                        2HI(g) --------------H2(g) +    I2(g)               Keq = 0.016
Initial---           0.6 M                    0.038               0.15
New                 0.6          0.038+0.124 M             0.15

change           +2x                       -x                          -x

Final             0.6+2x                0.162-x                    0.15-x

addition of H2 will take the reaction in backward direction . so concentration of HI must increase and that if I2 must decrease

Set up the equilibrium expression and solve for x
Keq = [H2] [I2] / [HI]^2
0.016 = (0.162-x) (0.15-x) / (0.60+2x)^2
x = 0.08874....

Plug in 0.08874 for x in the expressions for E in your ICE table:
[HI] = 0.77748 M
[H2] = 0.07326 M
[I2] = 0.06126 M

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