Thanks in advance. Please answer 1 and 2 1. The equilibrium constant K c for the

Question

Thanks in advance. Please answer 1 and 2

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

Starting with 3.30 moles of HBr in a 12.0 L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium.

2. Consider the following equilibrium process at 686°C.

The equilibrium concentrations of the reacting species are [CO] = 0.050 M, [H2] = 0.045 M, [CO2] = 0.086 M, and [H2O] = 0.040 M.

(a) Calculate Kc for the reaction at 686°C.

(b) If we add CO2 to increase its concentration to 0.40 mol/L, what will the concentrations of all the gases be when equilibrium is reestablished?

Explanation / Answer

Answer – Q 1) We are given, Kc = 2.18 106 , moles of HBr = 3.30 moles , volume = 12.0 L

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

[HBr] = 3.30 moles / 12.0 L

            = 0.275 M

So, reaction gets reversed, when we are given HBr in starting

2 HBr(g) <----> H2(g) + Br2(g) Kc = 1/ 2.18*106 = 4.59*10-7

Now we need to put ICE chart

   2 HBr(g) <----> H2(g) + Br2(g)

I     0.275                0           0

C      -2x                  +x         +x

E   0.275-2x              +x        +x

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

4.59*10-7 = x*x/(0.275-2x)2

4.59*10-7*(0.275-2x)2 = x2

We can neglect 2x from the (0.275-2x)2, since the Kc value is too small

So, 4.59*10-7*(0.275)2 = x2

x2 = 3.469*10-8

So, x = 1.86*10-4 M

We know, at equilibrium

[H2] = x = 1.86*10-4 M

[Br2] = x = 1.86*10-4 M

[HBr] = 0.275-2x

           = 0.275-2*1.86*10-4 M

           = 0.274 M

Q 2) a) We are given , reaction - CO2(g) + H2(g) <-----> CO(g) + H2O(g)

At equilibrium concentrations

[CO] = 0.050 M,

[H2] = 0.045 M,

[CO2] = 0.086 M, and [H2O] = 0.040 M.

Kc = [CO][H2O] /[CO2] [H2]

      = 0.050 M * 0.040 M / 0.086 M * 0.045 M

       = 0.517

b) When we increase the [CO2] = 0.40M

   CO2(g) + H2(g) <-----> CO(g) + H2O(g)

I 0.40        0.045               0.050      0.040

C -x             -x                     +x           +x

E 0.40-x   0.045-x             0.050+x   0.040+x

We know,

Kc = [CO][H2O] /[CO2] [H2]

0.517 = (0.050+x)(0.040+x) / (0.40-x) (0.045-x)

0.517[(0.40-x) (0.045-x)] = (0.050+x)(0.040+x)

0.517x2-0.230x+0.00931 = x2 +0.09x+0.002

x2-0.517x2 +0.09x+0.230x + 0.002 -0.00931

0.483x2 +0.32x -0.00731 = 0

So, x = 0.0221

So at equilibrium,

[CO2] = 0.40 -0.0221 = 0.378 M

[H2] = 0.045 -0.0221 = 0.0229 M

[CO] = 0.050+0.0221 = 0.0721 M

[H2O] = 0.040+0.0221 = 0.0621 M

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