Question 1 The equilibrium constant is given for two of the reactions below. Det

Question

Question 1

The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant.

            2A(g) + B(g) A2B(g)         Kc = ?

            A2B(g) + B(g) A2B2(g)    Kc = 16.4

            2A(g) + 2B(g) A2B2(g)    Kc = 28.2

Question 2

In which of the following reactions will Kc = Kp?

Question 3

Consider the following reaction:

                        2 H2O(g) + 2 SO2(g) 2 H2S(g) + 3 O2(g)

A reaction mixture initially contains 2.8 M H2O and 2.6 M SO2. Determine the equilibrium concentration of H2S if Kc for the reaction at this temperature is 1.3 × 10-6.

Question 4

. Consider the following reaction at equilibrium. What effect will adding 1 mole of Ar to the reaction mixture have on the system?

                        2 H2S(g) + 3 O2(g) 2 H2O(g) + 2 SO2(g)

Question 5

Consider the following reaction and its equilibrium constant:

            4 CuO(s) + CH4(g) CO2(g) + 4 Cu(s) + 2 H2O(g)          Kc = 1.10

A reaction mixture contains 0.22 M CH4, 0.67 M CO2 and 1.3 M H2O. Which of the following statements is TRUE concerning this system?

Explanation / Answer

Given :

1.).   2A(g) + B(g) A2B(g)         Kc = ?

            2).     A2B(g) + B(g) A2B2(g)    Kc = 16.4

            3).     2A(g) + 2B(g) A2B2(g)    Kc = 28.2

We are interested to get Kc of reaction 1

Lets reverse reaction 2 and then add to reaction three

Reverse of reaction 2

    A2B2(g) A2B(g) + B(g)   Kc (reverse) = 1/ 16.4

     2A(g) + 2B(g) A2B2(g)    Kc = 28.2

-------------------------------------------------

   2A(g) + B(g) A2B(g)          Kc = (1/16.4 ) x 28.2 = 1.72

Kc for first reaction = 1.72

So answer is : 1.72

Question 2

Kc = kp/ (RT)delta n

We use above formula to answer this question.

Delta n is number of electrons product –number of electrons of reactant

We need delta n = 0

Then kc = kp

If we look at second reaction :

SO3(g) + NO(g) SO2(g) + NO2(g)

In this reaction Number of moles product = 2 , Number of moles of product = 2

Delta n = 2 – 2 = 0

So Kc = kp / (RT)0 = kp

So answer is

SO3(g) + NO(g) SO2(g) + NO2(g)

Question 3 :

2 H2O(g) + 2 SO2(g) 2 H2S(g) + 3 O2(g)

Lets set up ICE

            2 H2O(g) + 2 SO2(g) 2 H2S(g) + 3 O2(g)

I           2.8                   2.6                   0                      0

C         -2x                   -2x                   +2x                  +3x

E          (2.8-2x)         (2.6-2x)             2x                    3x

Kc = [ 2x ]2 [3x]3 /[ (2.8-2x)]2 [ (2.6-2x)]2

1.3E-6 =     [ 2x ]2 [3x]3 /[ (2.8-2x)]2 [ (2.6-2x)]2

Since value of Kc is small so we neglect x in the denominator.

1.3E-6 =     [ 2x ]2 [3x]3 /[ (2.8]2 [ (2.6)]2

1.3E-6 = 4x2 X 27x3 /(52.998)

6.89E-5 = 108 x5

x = 0.0577

[H2S]= 2x = 2 x 0.0577 = 0.1153 M

And so answer is : 0.12 M

Q. 4)

2 H2S(g) + 3 O2(g) 2 H2O(g) + 2 SO2(g)

After addition of extra mole of Ar, There is no any effect on the reaction.

Since it does not involved in the reaction .

Q. 5

4 CuO(s) + CH4(g) CO2(g) + 4 Cu(s) + 2 H2O(g)          Kc = 1.10

By using all the concentrations we find Qc

Qc = [ CO2] [ H2O]2 / [CH4]

= 0.67 *1.32 / 0.22

= 5.15

Qc = 5.14 which is more than Kc

If this happens then there is reversal of reaction at equilibrium.

And reaction goes to left side.

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