The equilibrium constant K_c for the reaction I_2(g) 2I(g) is 3.8 times 10^-5 at

Question

The equilibrium constant K_c for the reaction I_2(g) 2I(g) is 3.8 times 10^-5 at 727degreer C. Calculate K_c and K_p for the equilibrium 2I(g) I_2(g) 2. The equilibrium constant K_c for the reaction O_2(g) 2/3O_3(g) is 3.34 times 10^-9. Calculate the equilibrium constant (K_c) for 3O_2(g) 2O_3(g)3. The equilibrium constant K_c for the reaction below is 92.8 at a certain temperature. H_2(g) + I_2(g) 2HI(g) If you start with 0.245 M hydrogen iodide, what will the concentration of H_2, I_2, and HI be at equilibrium?

Explanation / Answer

Answer - 1) In this one we are given reaction.

I2(g) <---> 2I(g) , Kc = 3.8*10-5 , T = 727oC +273 = 100.15 K

We need to calculate the Kc and Kp for the following reaction

2I(g) <---> I2(g)

We know this reaction is reverse it the original and when the reaction reverse then the equilibrium constant gets inverse.

2I(g) <---> I2(g)   Kc’ = 1/Kc

                                   = 2.63*104

We know , Kp = Kc *(RT)n

n = 1 moles – 2 moles = -1 moles

                       Kp = 2.63*104 * (0.0821 *1000.15)-1

                            = 320.5

2) ) In this one we are given reaction.

O2(g) <---> 2/3 O3(g) , Kc = 3.34*10-9

We need to calculate the Kc for the following reaction

3O2(g) <---> 2 O3(g)

We need to multiply by the 3 by the given original, so when we multiply the reaction by 3 then equilibrium constant gets cube root.   

3O2(g) <---> 2 O3(g) Kc’ = (3.34*10-9)3

                                       = 3.72*10-26

3) Give, reaction H2(g) + I2(g) <---->2HI(g) , Kc = 92.8

[HI] = 0.245 M

Reaction gets reversed, since are given first product concentration and we know when reaction reversed its Kc gets inversed means

Kc for reverse reaction = 1/92.8 = 0.0108

So reverse reaction with Kc is -

2 HI (g) <----->  H2 (g) + I2 (g) Kc= 0.0108

We need to put ICE table

       2 HI (g) <----->  H2 (g) + I2 (g)

I    0.245                     0           0

C    -2x                      +x         +x

E   0.245-2x               +x        +x

Kc = [H2 (g)] [I2 (g)] / [HI(g)]2

0.0108= x *x / (0.245-2x)

0.0108 (0.245-2x) = x2

0.00265 -0.0216x = x2

x2 + 0.0216x – 0.00265 = 0

a= 1, b = - 0.0216 , c = 0.00265

we know the quadratic formula

x = -b+/-b2-4ac / 2a

by placing the value in it and calculate x value

x = 0.0418

so at equilibrium concentrations

[HI] = 0.245-2x

        = 0.245* 2*0.0418 M

        = 0.161 M

x = [H2] = 0.0418 M

x = [I2] = 0.0418 M

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