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6 months ago

Class 11th Equilibrium

7.15. At 700 K, equilibrium constant for the reaction:

H2?(g) + I2 (g) ?2HI (g)

is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2?(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?  
0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

H2 (g) + I2 (g)? 2HI (g); K=64, T=700K

2HI? H2 + I2 K=1/64700K

  a         0      0

a (1−α)   aα/2   aα/2

0.5         x        x

x2 / (0.5)2  = 1 / 54.8

x2 =   0.25 / 54.8

x =  = 0.068 M

At equilibrium, [H2] = [I2] = 0.068 M

New answer posted

6 months ago

Class 11th Equilibrium

7.14. One mole of H2O and one mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation.

H2O (g) + CO (g) ? H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.     
0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Number of moles of water originally present = 1 mol
Percentage of water reacted= 40%
Number of moles of water reacted= 1 x 40/100 = 0.4 mol
Number of moles of water left= (1 – 0.4) = 0.6 mole

According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
Thus, the molar conc., per litre of the reactants and products before the reaction and at the equilibrium point are as follows:

 

H2O

CO

H2

CO2

Initial moles / litre

1/10

1/10

0

0

At Equilibrium

(1 – 0.4) / 10 = 0.6/10

(1 – 0.4) / 10 = 0.6/10

0.4/10

0.4/10

 

Equilibrium constant, Kc= [H2] [CO2] / [H2O] [CO]

= [ (0.4/10) x (0.4/10)] / [ (0.6/10) x (0.6/10)]

= 0.16 / 0.36 = 0.44

New answer posted

6 months ago

Class 11th Equilibrium

7.13. The equilibrium constant expression for a gas reaction is,        
Write the balanced chemical equation corresponding to this expression.       
0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Balanced chemical equation for the reaction is

4 NO (g) + 6 H2O (g)? 4 NH3 (g) + 5 O2 (g)

New answer posted

6 months ago

Class 11th Equilibrium

7.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant Kc for the reaction

N2?(g) + 3H2? (g) ? 2NH3? (g) is 1.7*102
Is this reaction at equilibrium? If not, what is the direction of net reaction?           
0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

According to the given equation, concentration quotient,

Qc = [NH3]2 / [N2] [ H2]3

= (8.13/20 mol L-1)2 / (1.57 / 20 mol L-1) x (1.92 / 20 mol L-1)3

= 2.38 x 103

The equilibrium constant (Kc) for the reaction = 1.7 x 10-2

As Qc ≠ Kc, the reaction is not in a state of equilibrium. 

New answer posted

6 months ago

Class 11th Equilibrium

7.11. A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?

2HI (g) ? H2 (g) + I2 (g)
0 Follower 1 View

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Vishal Baghel

Contributor-Level 10

pHI = 0.04 atm, pH2 = 0.08 atm, pI2 = 0.08 atm

Kp= (pH2 x pI2) / (pHI)2 = (0.08 x 0.08) / (0.04 x 0.04)

= 4.0

New answer posted

6 months ago

Class 11th Equilibrium

7.10. At 450 K, Kp = 2.0 * 1010/bar for the given reaction at equilibrium.

2SO2(g) + O2(g) ?? 2SO3 (g)

What is Kc at this temperature?
0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Kp = Kc (RT)? ng

=> Kc = Kp (RT)-? ng

Putting the values of Kp= 2.0x1010 bar-1, R= 0.083 L bar K-1 mol-1, T = 450 K, and? ng = 2-3= -1

=>  Kc = (2.0 x 1010 bar-1) x [ (0.083 L bar K-1 mol-1) x (450 K)]- (-1)

= 7.47 x 1011 mol-1 L

= 7.47 x 1011 M-1

New answer posted

6 months ago

Class 11th Equilibrium

7.9. Nitric oxide reacts with Br2and gives nitrosyl bromide as per reaction given below:

                                                2NO (g) + Br2 (g) ? 2NOBr (g)
When 0.087 mole of NO and 0.0437 mole of Br2 are mixed in a closed container at constant temperature, 0.0518 mole of NOBr is obtained at equilibrium. Determine the compositions of the equilibrium mixture.
0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

According to the equation, 2 moles of NO (g) react with 1 mole of Br2  (g) to form 2 moles of NOBr  (g). The composition of the equilibrium mixture can be calculated as follows:
No. of moles of NOBr  (g) formed at equilibrium = 0.0518 mol
No. of moles of NO (g) taking part in reaction = 0.0518 mol
No. of moles of NO (g) left at equilibrium = 0.087 – 0.0518 = 0.0352 mol
No. of moles of Br2  (g) taking part in reaction = 1/2 x 0.0518 = 0.0259 mol
No. of moles of Br2  (g) left at equilibrium = 0.0437 – 0.0259 = 0.0178 mol
The initial molar concentration and equilibrium molar concentration of different spe

...more

New answer posted

6 months ago

Class 11th Equilibrium

7.8. Reaction between nitrogen and oxygen takes place as follows:
                                                            2N2?(g) + O2?(g) ? 2N2?O(g)
If a mixture of 0.482 mol of N2 and 0.933 mol of O2 is placed in a reaction vessel of volume 10 L and allowed to form N2O at a temperature for which Kc = 2.0 x 10-37, determine the

...more
0 Follower 3 Views

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Vishal Baghel

Contributor-Level 10

Let x moles of N2 (g) take part in the reaction. According to the equation, x/2 moles of O2  (g) will react to form x moles of N2O (g). The molar concentration per litre of different species before the reaction and at the equilibrium point is:

 

[N2? ]

[O2]

[N2? O]

Initial concentration

0.482? / 10

0.933/ 10

0

Conc. at equilibrium

(0.482 – x)? / 10

(0.933 – x/2) / 10

x/ 10

The value of equilibrium constant (2.0 x 10-37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, value of x is extremely small and can be omitted as far as the reactants are concerned.

Applying Law of chemical equilibrium, Kc = [N2O]2 / [N2]2 [O2]

= Kc = 2.0 x 10-37

= (x/10)2 / [ (0.482/ 10)2 x (0.933 / 10)2]

= 0.01 x2 / (2.1676 x 10-4)

x2= 4

...more

New answer posted

6 months ago

Class 11th Equilibrium

7.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
0 Follower 3 Views

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Vishal Baghel

Contributor-Level 10

This is because molar concentration of a pure solid or liquid is independent of the amount present.
Since density as well as molar mass of pure liquid or solid is fixed; their molar concentrations are constant.

The concentration of solid or liquid is =   =   =
At constant temperature, the density and molar mass of pure solid and liquid are constant.
Due to this, their molar concentrations are constant and are not included in the equilibrium constant.

New answer posted

6 months ago

Class 11th Equilibrium

7.6. For the following equilibrium, K =6.3 x 1014 at 1000 K. 

NO (g) + O3 —–> NO2 (g) + O2(g). 

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?  
0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

For reverse reaction, KC (reverse) = 1/ KC = 1 / 6.3 x 1014 = 1.59 x 10-15

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