Fulfillment of the experiment

Сu²⁺ ions increase the rate of decomposition of thiosulphuric acid.

Pour into three test-tubes from burett by 1ml of Na₂S₂O₃ solution, then add to he first tube 1ml of catalyst solution (CuSO₄) and 3ml of water; to the second – 2ml of catasyst solution, and 2ml of water, to the third – 3ml of catalyst solution and 1ml of water. Thus, there will be 5ml of the solution with the same concentration of Na₂S₂O₃, but with different concentrations of CuSO₄in each tube.

Calculate the concentrations of Na₂S₂O₃ and Cu²⁺-ions in prepared solutions.

Pour into three other tubes by 5ml of H₂SO₄ solution. Then mix prepared solutions of Na₂S₂O₃ and H₂SO₄ by pares. Measure the time from the mixing moment till the mud appearance with help of secondmeasurer. Write down received results as:

 

№ of tube	Concentration of Cu^2+ ions, mol/l	Time of mud appearance	υrelative = 1/τ, c^-1
1
2
3

Treatment of the experiment results.

 

Calculate the relative rate of the reaction for different Cu²⁺ concentrations. Make up a plot of reaction rate versus catalyst concentration. Make a conclusion about Cu²⁺ influence on the rate of thiosulphuric acid decomposition using received dependence.

    

Class 9

Topic: chemical equilibrium

Chemical equilibrium is established in such systems, where reversible chemical reactions proceed. To characterize it quantitatively the equilibrium constant is applied (K). For any homogeneous reaction: 

                 aA + bB ↔ cC + dD

                 K = [C]^c.[D]^d / [A]^a.[B]^b, where

[A], [B], [C], [D] – equilibrium molar concentrations of substances A, B, C, D.

As for heterogeneous reaction such as:

       1) A_(g) + 2B_(solid)  ↔ AB_2(liquid)       K_c = [AB₂] / [A]

       2) A_(g) + 2B_(liq) ↔ AB_2(liq)       K_{c =} [AB₂] / [A]

       3) A_(liq) + 2B_(sol) ↔ AB_2(liq)      K_c = [AB₂] / [A]

So, concentration of solid (always), liquid (for reactions with gases)does not participate in equilibrium constant expression, similar to kinetic equation of chemical reaction.

Equilibrium constant, equal to the ratio of rate constant of direct (k₁) and reverse (k₂) reaction, depends on temperature and nature of reactants, but not on their concentrations:                                    

                                   K = k₁ / k₂

    Concentration of reactants and products, temperature, and pressure – for gases influence on the state of chemical equilibrium. If any of these parameters changes, the equilibrium is disturbed and all concentrations are changing till the new equilibrium will be established with other values of equilibrium concentrations. This transfer is called as a shift of chemical equilibrium.

Le-Shatelie principle point to direction of equilibrium shift so that the rate of reaction (direct or reverse), which weakens external effect, increases. Really, increasing of [C] must lead to decreasing of [D], and as a result – to increasing of [A] and [B], therefore it leads to shift of the equilibrium to left side (initial substances).

 

Tasks and their solution

 

1. During equilibrium state of the system N_2(g) + 3H_2(g) ↔ 2NH_3(g); ∆H = -92,4 kJ, equilibrium concentrations are equal to: [N₂] = 3 mol/l; [H₂] = 9 mol/l; [NH₃] = 4 mol/l. Solve for: a) equilibrium constant; b) initial concentrations of N₂ and H₂; c) in which direction does the equilibrium shift if the temperature increases; if the volume of reaction vessel decreases?

Solution:

 

a) K_c = [NH₃]² / [N₂ ]^.[H₂]³ = 4² / 3^.9³  = 0,0073.

K_c << 1, consiquently equilibrium is shifted to left, reverse reaction is predominating.

b)  Initial concentration is equal to equilibrium plus reacted concentrations:

C(N₂) = C*(N₂) + [N₂]

(H₂) = C*(H₂) + [H₂]

According to reaction equation: to form 2 moles of NH₃ it is necessary to consume 1 mole of N₂  and 3 moles of H₂. If at the equilibrium moment [NH₃] = 4 mol/l, C*(N₂) = 2 mol/l; C*(H₂) = 6 mol/l.

   C(N₂) = 3 + 2 = 5 (mol/l),

   C(H₂) = 9 + 6 = 15 (mol/l).

c) If the temperature increases the equilibrium will shift to left, towards endothermic reaction; if the volume of reaction vessel decreases the pressure will increase and the equilibrium will shift to right, towards decrease of number of moles of gaseous mixter.

The answer: K_c = 0,0073; 5 mol/l N₂, 15 mol/l H₂.

2. Equilibrium constant of the reaction                                                                                         FeO_(solid) + CO_g) ↔ Fe_(solid) + CO_2(g)  at the some temperature is equal to 0,5. Find the equilibrium concentrations of CO and CO₂, if the initial concentrations was: C(CO) = 0,05 mol/l; C(CO₂) = 0,01 mol/l.

Solution: 

K_c = [CO₂] / [CO] = 0,5.

Assume that x moles of CO have reacted, then,using equation, from x moles of CO x moles of CO₂ have been produced.

                [CO] = C(CO) – C*(CO) = (0,05 – x) mol/l

                [CO₂]= C(CO₂) – C*(CO₂) = (0,01+x) mol/l

                      K_c = (0,01 + x) / (0,05 – x)

(0,01+x) = 0,5(0,05-x); 0,01+x = 0,025-0,5x; 1,5x = 0,015; x = 0,015/1,5 = 0,01

                     [CO] = 0,05 – 0,01 = 0,04 (mol/l)

                     [CO₂] = 0,01 + 0,01 = 0,02 (mol/l)

The answer: [CO] = 0,04 M, [CO₂] = 0,02 M.

 

3. At poisonings with arsenates there are HAsO₄^2- ions in the blood, which are capable to substitute HPO₄^2- ions in metabolism process. Find the equilibrium concentration of AsO₄^3- ions in aqueous solution, if [HAsO₄^2-] = 10^-4mol/l, and equilibrium constant for reaction           

 HAsO₄ ^2- ↔ H⁺ + AsO₄^3- at 25⁰ is equal to 10^-12.

Solution:

By the action mass law: K_c = [H⁺]^.[AsO₄^3-] / [HAsO₄^2-] = 10^-12.

In accordance with reaction equation: [AsO4^3-] = [H⁺], assume that it is equal to x mol/l, then K_c = x² / [HAsO₄^2-] = 10^-12.

x = 10^-8 mol/l.

The answer: 10^-8 mol/l.

 

4. Solve for mass fraction of hydrogen and iodine, which convert into hydrogen iodide, if their initial quantities are the same, equal to 1 mole and equilibrium constant at given temperature is equal to 4.

 

Solution:

 

 Make up the table:                                  H₂ + I₂ ↔ 2HI

  Are taken, mol/l                                1  1

Have reacted to equilibrium moment       x  x

Have remainded                                     1-x 1-x

Have formed                                                                2x

 

If equilibrium concentrations of hydrogen and iodine are 1-x, and hydrogen iodide 2x mol/l, then

           K_c = [HI]² / [H₂]^.[I₂] = 2x² / (1-x)² 4, x = 0,5 (mol/l).

So, mass fractions of H₂ and I₂, converted into HI, are 50%.

The answer: 50%.

 

5. For reaction 2CO₂ ↔ 2CO + O₂ at 2000⁰C the composition of equilibrium mixter is expressed by volume fractions: 85,2% CO₂, 9,9% CO, 4,9% O₂ and total pressures in the system is 101,3 kPa. Solve for equilibrium constant at given temperature expressed through:

a) partial pressure of reactants (K_p)

b)  their molarityes (K_c)

 

Solution:

Partial pressure is equal to total pressure, multiplied by volume fraction of gas in mixter:

a) p(CO₂) = 101,3^.0,852 = 86,3 (kPa)

b) p(CO) = 101,3^.0,099 = 10,0 (kPa)

c) p(O₂) = 101,3^.0,049 = 4,9 (kPa)

 

           K_p = p²(CO)^.p(O₂) / p(CO₂) = 0,067

 

for this reaction Δn = 3-2 = 1. Then K_c = K_p / (RT)^Δn = 0,067 / 8,31^.2273 = 3,6

                                                      The answer: K_p = 0,067; K_c = 3,6.

 

_6. How can you shift the equilibrium in the system 2SO₂ + O₂↔ 2SO₃

(-∆H), to increase the production of SO₃  if the concentration of SO₂ is given?

 

Solution:

According to Le-Shatelie principal it can be achieved with help of:

a) increase of O₂ concentration

b) decrease of SO₃ concentration (removement from reaction mixture)

c) increase of the pressure

d) decrease of the temperature till the value, which can faciliate to the relatively fast achievement of equilibrium.