Qualitative Analysis of IV Group Cations

Group IV  cations are calcium (II) Ca²⁺, strontium (II) Sr²⁺ and barium (II) Ba²⁺. Its group reagent is 1M solution of ammonium carbonate (NH₄)₂CO₃  in neutral or alkaline medium. The medium needs to be neutral or alkaline because it is readily decomposed even by weak acids like acetic acid.

CO₃^2-+ CH₃COOH ⇌ CO₂ ↑ + H₂O + CH₃COO^-

This reagent is easily hydrolyzed but in alkaline medium, the equilibrium shifts backwards and reagent remains un-hydrolysed.

CO₃^2-+ H₂O ⇌ HCO₃^- + OH^-

Cations of group IV precipitate in the form of carbonate. For the test of IV group cations, take the filtrate ofIII(B) group and follow these steps:

1.  Transfer the filtrate of III(B) to a porcelain dish and acidify it with dilute acetic acid. Evaporate it to a pasty mass, and allow to cool.
2. Add 3-4ml concentrated nitric acid HNO₃, and heat until the mixture is dry. Then heat more strongly until all ammonium salts are volatilised and allow it to cool.
3. Add 3ml 2M hydrochloric acid HCl and 10ml water, heat till warm while stirring to dissolve the salt.
4. Add 0.25g solid ammonium chloride NH₄Cl and add concentrated ammonia NH₃ solution to make it alkaline and add group reagent 1M ammonium carbonate (NH₄)₂CO₃in slight excess. Place it in water bath at 50-60°C for 5minutes and keep stirring. Boiling the reagent is necessary for the removal of ammonium hydrogen carbonate (NH₄)HCO₃and ammonium carbamate NH₄O(NH₂)CO₃. These salts are present as impurities in commercial (NH₄)₂CO₃and they can interfere in analysis, as they are soluble in water.
5. Filter and wash with little hot water. Precipitate may contain Barium carbonate BaCO₃, Strontium carbonate SrCO₃ and Calcium carbonate CaCO₃.

          Ba²⁺+ CO₃^2- ⟶ BaCO₃↓

          Ca²⁺+ CO₃^2- ⟶ CaCO₃↓

          Sr²⁺+ CO₃^2- ⟶ SrCO₃↓

Separation of Ba²⁺ from Sr²⁺ and Ca²⁺

Carbonates of all the three cations are insoluble in water but soluble in acetic acid and in dilute mineral acid. So, we can dissolve the precipitate of IV group cations in dilute acetic acid.

BaCO₃↓ + H⁺ ⟶ Ba²⁺ + H₂O + CO₂

CaCO₃↓ + H⁺ ⟶ Ca²⁺ + H₂O + CO₂

SrCO₃↓ + H⁺ ⟶ Sr²⁺ + H₂O + CO₂

This way we get thesolution of cations of IV group, let's call it solution (A). These cations form chromate salt on reacting with potassium chromate solution. These chromate salts have different solubilities; barium chromate is insoluble in dilute acetic acid while the other two are soluble. By using this difference in solubility, we can separate Ba²⁺ from others. 

Test a small amount of solution with potassium chromate K₂CrO₄ solution, a yellow precipitate indicates Ba²⁺. Solubility product of SrCrO₄ and CaCrO₄ is much higher than BaCrO₄, so they need much higher concentration of CrO₄^2- ion to precipitate them. That’s why they can’t get precipitated from dilute solution.

Ba²⁺ + CrO₄^2- ⟶ BaCrO₄ ↓

If Barium is absent then proceed without adding potassium chromate to the whole solution. If barium is present then heat the rest of the solution (A) till boiling and add slight excess of 0.1M potassium chromate K₂CrO₄ solution until the solution becomes yellow coloured and precipitation is complete. Filter and wash the precipitate with little hot water. The precipitate contains BaCrO₄. Keep the filtrate (B) and washings for the test of Sr²⁺ and Ca²⁺.

Confirmatory Test for Ba²⁺

Wash the precipitate with hot water. Dissolve it in concentrated hydrochloric HCl acid. HCl coverts barium chromate into soluble dichromate. Divide the solution in two parts.

BaCrO₄ ↓ + H⁺  ⟶ BaCr₂O₇+ H₂O

Part 1: Dilute it with water and add dilute sulphuric H₂SO₄acid. A white precipitate of barium sulphate is formed which is insoluble in dilute acids but soluble in concentrated sulphuric H₂SO₄ acid. This confirms Ba²⁺.

BaCr₂O₇ + H₂SO₄⟶ Ba SO₄ + H₂Cr₂O₇

Part 2: Evaporate the solution to dryness and apply flame test. Green or yellowish green flame confirms Ba²⁺.

Test for Sr²⁺ and Ca²⁺

If Ba²⁺ is absent: Boil the solution (a) for 1 min to expel excess CO₂  and test for Sr²⁺ and Ca²⁺.

If Ba²⁺ is present: Take the filtrate (b), which we saved after removal of Ba²⁺ as chromate, and neutralised it with 2M ammonia NH₃ solution and add excess of ammonium carbonate (NH₄)CO₃ solution or you can add a little solid sodium carbonate Na₂CO₃. A white precipitate indicates SrCO₃or CaCO₃ or both. Wash the precipitate with hot water and dissolve it in 4ml acetic acid and boil to remove excess CO₂.

Add 2ml saturated solution of ammonium sulphate (NH₄)₂SO₄ , followed by 0.2g sodium thiosulphate Na₂S₂O₃ and heat in a water bath for 5min and allow to stand for few minutes then filter. Sulphates of Sr²⁺ and Ca²⁺ are formed. Strontium sulphate is insoluble in ammonium sulphate solution, so the  white precipitate we get  here is strontium sulphate and calcium sulphate goes in the filtrate.

Ca²⁺+ SO₄^2- ⟶ CaSO₄

Sr²⁺+ SO₄^2-  ⟶ SrSO₄↓

Confirmatory Test for Sr²⁺

Transfer the white precipitate along with filter paper to a small crucible, heat until precipitate has charred. Moisten the ash with concentrated HCl and apply flame test. Crimson flame confirms Sr²⁺.

Confirmatory Test for Ca²⁺

Take the filtrate and add a little 0.1M ammonium oxalate (NH₄)₂C₂O₄solution and 2ml of 2M acetic acid CH₃COOH, warm the solution on a water bath. White precipitate of calcium oxalate is formed which is insoluble in water as well as in acetic acid. This white precipitate of calcium oxalate CaC₂O₄ confirms Ca²⁺.

Ca²⁺ + (COO)₂^2- ⟶ Ca(COO)₂ ↓

We have successfully separated the cations of group IV. In the next post of analytical chemistry we will analyse the group V cations in the filtrate of group IV.

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Qualitative analysis of Group III(B) cations

In the last post of analytical chemistry we have separated III(B) cations from III(A) cations. On boiling with ammonium chloride NH₄Cl₃ and ammonia NH₃solution III(A) cations get precipitated as hydroxide, while III(B) cations remain in solution as sulphides. In the filtrate of III(A) we will perform tests for III(B) cations. First we have to re precipitate cations of III(B).

Step 1: Precipitation of III(B) cations as sulphide

To the filtrate, add 2-3ml of 2M ammonia NH₃ solution and heat, now pass hydrogen sulphide H₂S gas under pressure for 0.5 to 1 min. You will get the precipitate of III(B) cations which may contain CoS, NiS, MnS and ZnS. H₂S gas should be passed in hot solution for complete precipitation because ZnS and MnS form fine suspension if the solution is not hot enough.

           M²⁺ + S^2-⟶ MS

Wash the precipitate with the mixture of 1% ammonium chloride NH₄Cl solution and small amount of ammonium sulphide (NH₄)₂S solution and reject all the washings. You can get an idea of the cation by the colour of precipitate, if it is black it may contain NiS and CoS and it if white or dirty white it may contain ZnS and MnS. These cations have different solubility in HCl.ZnS and MnS are soluble in very dilute HCl while NiS and CoS require moderately dilute HCl. NiS and CoS are insoluble in cold dilute HCl.

Step 2 separation of Zn(II) and Mn(II) from Co(II) and Ni(II)

Transfer the precipitate to a small beaker. Add 5ml water and 5ml of dilute HCl, stir well and allow to stand for few minutes then filter. You will get ZnS and MnS in the form of filtrate and NiS and CoS will remain as residue.

MnS↓ + 2H⁺  ⟶ Mn²⁺ + H₂S↑

ZnS↓ + 2H⁺  ⟶ Zn²⁺ + H₂S↑

Step 3 Test for Nickel Ni(II) and Cobalt Co(II)

Dissolve the residue in aqua regia (3part conc. HCl: 1part conc. HNO₃). Solution may look white due to white sulphur, but on longer heating, sulphur is oxidised to sulphate and solution becomes clear.

CoS↓ + HNO₃ + 3HCl ⟶ Co²⁺ + S↓+ NOCl↑ + 2Cl^- + 2H₂O

NiS↓ + HNO₃ + 3HCl ⟶ Ni²⁺ + S↓+ NOCl↑ + 2Cl^- + 2H₂O

Divide the solution in two parts for the confirmatory tests of Ni(II) and Co(II).

Part 1 confirmatory test for Ni²⁺

Add 2ml ammonium chloride NH₄Cl 1M solution and 2M ammonia NH₃ solution until alkaline, and then excess of dimethylglyoxime reagent. Red precipitate of nickel dimethylglyoximeconfirms the presence of nickel.

You can also use spot test technique for the same test, place a drop of test solution and a drop of dimethylglyoxime reagent on a spot plate and add a drop of dilute ammonia solution. A red spot of nickel dimethylglyoxime is produced.

Ni²⁺  + 2C₄H₈O₂N₂⟶ Ni[C₄H₇O₂N₂]₂+ 2H⁺

Part 2 confirmatory test for Co²⁺

Add 1ml amyl alcohol, 2g solid ammonium thiocynate. Amyl alcohol layer becomes blue due to formation of tetrathiocobaltate(II) ions. This is known as Vogel reaction.

Co²⁺  + 4SCN^-⟶ [Co(SCN)₄]^2-

Step 4 Test for zinc Zn(II) and manganese Mn(II)

Filtrate may contain Mn²⁺ and Zn²⁺ and traces of Ni²⁺ and Co²⁺. Zinc and manganese both form hydroxide with NaOH but zinc hydroxide is soluble in excess of reagent. We can use this difference for the separation of Zn from Mn.

Boil the filtrate until H₂S is removed (evolution of H₂S gas can be identified with its characteristic odour. For confirmation you can  test it with lead acetate paper which turns black due to formation of lead sulphide H₂S + Pb²⁺ ⟶PbS↓ + 2H⁺). Cool the solution and add excess of 2M sodium hydroxide NaOH solution, followed by 1ml 3% hydrogen peroxide H₂O₂solution. Boil for 2-3 min then filter. You will get Mn²⁺ as precipitate and Zn²⁺ in the filtrate.

Mn²⁺ + 2OH^- ⟶ Mn(OH)₂↓

Hydrogen peroxide converts manganese (II) hydroxide into hydrated manganese dioxide.

Mn(OH)₂ ↓ + H₂O₂ ⟶ MnO(OH)₂↓ + H₂O

Zn²⁺ + 2OH^- ⇌ Zn(OH)₂↓

Zinc (II) hydroxide is soluble in excess of reagent by forming tetrahydroxozincate ion.

Zn(OH)₂ ↓ + 2OH^- ⇌ [Zn(OH)₄]^2-

Test for Zn²⁺in the filtrate

Divide the filtrate in to two parts.

Part 1: acidify with 2M acetic acid CH₃COOH and pass H₂S gas. White precipitate of zinc sulphide is formed.

Zn²⁺ + H_2­S + 2CH₃COO^- ⟶ZnS ↓ + 2CH₃COOH

Part 2: acidify with 1M sulphuric acid H₂SO₄, add 0.5ml of 0.1M cobalt acetate Co(CH₃COO)₂ solution and 0.5ml of ammonium tetratiocyanato-mercurate(II) regent and stir. Violet precipitate is obtained.

Zn²⁺ + [Hg(SCN)₄]^2-⟶ Zn[Hg(SCN)₄] ↓

Test for Mn²⁺in the precipitate

Precipitate largely contain MnO(OH)₂ and traces of Ni(OH)₂and Co(OH)₃. Dissolve the precipitate in 5ml of 8M nitric acid HNO₃, if it doesn’t dissolve completely then add few drops of 3% H₂O₂ solution. Boil to decompose excess of H₂O₂ and cool.

Add 0.05g sodium bismuthate NaBiO₃, stir and allow to settle. Purple solution of MnO₄^- is formed.

2Mn²⁺ + 5NaBiO₃ +14H⁺⟶ 2 (MnO₄)^- + 5Bi³⁺ + 5Na⁺+ 7H₂O

Now we have successfully analysed the III(B) group cations. In the filtrate of III(B) we will test for group IV cations. 

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First Order Reaction and its Half Life

In first order reaction, the rate of reactiondepends on 1^st power of one reactant. First order reactions are very important reactions as they are very helpful in medical field and they also help us to solve the mystery of our history. You may think why are we talking about history in the post of chemistry? You must have heard about carbon dating technique used to determine the age of fossils by assessing the decay of ¹⁴C in them. This decay of ¹⁴C follows first order reaction and by using the half life equation, archaeologist determine the age of fossils. Let’s derive the equation for the rate law of first order reaction and then we will compute the equation for half life of first order reaction.

Consider an equation where Reactant ‘R’ gets converted to product ‘P’

R⟶P

Rate = -dR/dt = k [R]¹               .........(1)

-d[R]/dt = k [R]

d[R]/ [R] = -k dt                        .........(2)

Integrating both sides

ln[R] = -k t + I                          .........(3)

Where I is the constant of integration

At t=0, R= [R]₀ the initial concentration

ln [R]₀ = -k × 0 + I

ln[R]₀ = I

On putting the value of I in equation 3 we get,

ln[R] = -k t + ln[R]₀                  .........(4)

-k t = ln[R] - ln[R]₀

k t = ln[R]₀ - ln[R]

k = 1  ln[R]₀                              ...........(5)

       t      [R] 

or,

k = 2.303 log [R]₀                      ...........(6)

          t           [R] 

On putting the units in this equation you can find the units of k. It is sec^-1for first order reaction. You can calculate the k by using the equation 5 or 6. If you didn’t know the initial concentration but the reactant concentration is given for two different times, then what will you do? Let’s see,

At time t₁, [R] = [R]₁

By using equation 4:

ln[R]₁ = -k t₁ + ln[R]₀                ...........(7)

At time t₂, [R] = [R]₂

ln[R]₂ = -k t₂ + ln[R]₀                ...........(8)

on subtracting eq 8 from eq 7:

ln[R]₁- ln[R]₂ = {-k t₁ + ln[R]₀}-{-k t₂+ ln[R]₀}

ln[R]₁- ln[R]₂ = -k t₁ + ln[R]₀ + k t₂- ln[R]₀

ln[R]₁- ln[R]₂ = k (t₂ – t₁)

ln [R]₁ = k (t₂ – t₁)                   ...........(9)

    [R]₂

Graphs of 1st order reaction

Half life of first order reaction

At t = t _½ , [R] = ½ [R]₀

On putting the values in equation 6

k  =  2.303 log [R]₀                   

          t _½       ½ [R]₀

t _½  =  2.303log [R]₀                 

             k          ½ [R]₀

t _½  =  2.303log 2[R]₀               

             k             [R]₀

t _½  =  2.303× 0.301                 

             k            

t _½  =  0.693                  ...........(10)      

             k            

Half life of first order reaction is independent of reactant concentration. It means that even if we reduce the reactant concentration, the half life of reaction remains constant. Let’s practice some problems to gain a better understanding:

Q. A reacts according to first order kinetics, with k = 5×10^-5s^-1. If initial concentration is 1M find the initial rate of reaction. What will be the rate after 30min?

Rate law of first order reaction is

Rate = -dR/dt = k [R]¹              

Rate = 5×10^-5 sec^-1 [1M]

Rate = 5×10^-5 M s^-1

= Initial rate is 5×10^-5 M s^-1.

To calculate the rate after 30min (= 1800s) we need to find the reactant concentration [R]₁

ln [R]₀ = k t                

    [R]

Or,

ln [R]   = -k t + ln[R]₀              

   

On raising both side of equation to the exponent e

e^ln[R]  = e^{-kt +ln[R]0}                       

[R] = [R]₀ e^-kt                ........(11)

[R] = 1M e^{-(5×10^-5)(1800)}

[R] = 0.91M

So the rate will be:

Rate = 5×10^-5 s^-1[0.91M]

Rate = 4.5 ×10^-5 M s^-1

The rate of the reaction after ½ h will be 4.5 ×10^-5 M s^-1.

Q. Sucrose decomposes to form glucose and fructose according to first order rate law, with half life 3.33h at 25^°C. What fraction of a sample of sucrose remains after 9h?

By using equation of half life we can find out the value of k. And then putting the value of k and t in equation 11 we can find the fraction of sucrose remaining [R]/[R]₀. 

t _½  =  0.693                  ...........(10)      

             k            

k = 0.693 / 3.33

k = 0.208 h^-1

[R] = [R]₀ e^-kt                ........(11)

[R]/[R]₀= e^-(0.208)9

[R]/[R]₀= e^-1.87

[R]/[R]₀= 0.154

Fraction remaining is 0.154 or 15.4% sucrose remains after 9h.

In the next post of chemical kinetics we will study second order reaction and its half life in detail.

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Zero order reaction and its half life

For a zero order reaction, the rate of reaction is independent of reactant concentration. It means that the rate remains constant throughout the reaction. How is it possible that the reactant concentration doesn’t affect the rate of reaction? Let’s try to find out the rate equation for zero order reaction:

R⟶ P

Rate = -dR/dt = k [R]⁰

-d[R]/dt = k × 1

d[R] = -k dt                  .........(1)

Integrating both sides

[R] = -k t + I                 .........(2)

Where I is the constant of integration

At t=0, R= [R]₀ the initial concentration

[R]₀ = -k × 0 + I

[R]₀ = I

On putting the value of I in equation 1 we get,

[R] = -k t + [R]₀                        .........(3)

-k t = [R] - [R]0

k t = [R]0 - [R]

k = [R]0 - [R]                ...........(4)

            t

If you fill units in equation 4 you will find the unit of rate constant for zero order reaction, which is mol L^-1sec^-1. When we plot a graph between reactant concentration and time, we will get a straight line with negative slope (-k). That means after a certain time reactant concentration will become negative. But concentration can never be negative. It means zero order is not an actual order of reaction but it seems to be for a limited period of time. You will understand it better by its example.

Graphs of Zero Order Reaction

Decomposition of ammonia NH₃on a hot (1130K) platinum surface (Pt) is an example of zero order reaction. 

2NH₃(g) ⟶ N₂(g) + 3H₂(g)

Platinum metal acts as a catalyst for this reaction. Ammonia molecules get attached to the surface of hot platinum metal and get decomposed into N₂ and H₂. Platinum metal has a limited surface area so the other molecules have to wait for their turn. The rate of reaction depends only on the ammonia molecules attached to the surface of Pt metal. When all the sites of Pt get saturated with NH₃ molecules, it doesn’t matter how much ammonia is present in the vessel. If the same reaction is carried out in a different condition, it may have different order. But in this metal catalysed reaction limited availability of metal surface makes rate independent of reactant concentration.

Decomposition of HI on hot gold surface and decomposition of nitrous oxide N₂O with catalyst Pt are other examples of zero order reactions. Enzyme catalysed reactions in organisms also have zero order because just like metal surface enzymes also have limited active sites to bind with the reactant. That’s why the rate of reaction doesn’t depend on the reactant concentration.

By using equation 3 you can calculate reactant concentration at any time of the reaction. The time taken for the completion of half of the reaction is known as half life of the reaction. Let’s try to derive equation for the half life of zero order reaction:

At t = t ½ , [R] = ½ [R]₀

On putting the values in equation 3

½ [R]₀ = -k t ½ + [R]₀    

-k t ½ = ½ [R]₀ - [R]₀

 k t ½ = [R]₀ - ½ [R]₀

t ½ = [R]₀ /2k                ......(5)

Rate of zero order reaction is independent of reactant concentration but its half life depends on initial concentration of reactant.

Let’s try to solve a few problems based on zero order reaction.

Q. For a zero order reaction R⟶P. If the initial concentration of R is 1.5 mol L^-1 and after 120 sec the concentration is reduced to 0.75 mol L^-1. Determine the rate constant.

[R]₀ = 1.5 mol L^-1, [R] = 0.75 mol L^-1

if you look at the values of [R]₀and [R] in this problem, you can see

[R] = ½ [R]₀

So, t = t ½ =120sec

Equation for half life of zero order reaction is

t ½ = [R]₀ /2k

On putting values in it,

120 sec = 1.5 mol L^-1 / 2k

k = 1.5 mol L^-1 / 240sec

k = 0.00625 mol L^-1 sec^-1

rate constant will be 0.00625 mol L^-1 sec^-1 or 0.00625 M sec^-1.

Q. If the initial concentration is reduced to 1 mol L^-1. Does this affect the half life of above reaction?

From the equation for half life of zero order reaction

t ½ = [R]₀ /2k

t ½ ∝ [R]₀

t₂ / t₁ = [R₀]₂/ [R₀]₁

t₂ / 120 = 1 / 1.5

t₂ = 80 sec

New half life will be reduced to 80 sec.

In the next post of chemical kinetics we will study first order reaction and its half life in detail.

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Chemical Kinetics

Chemical kinetics is like a speedometer of any reaction. It tells us at which speed the reaction takes place. Thermodynamics tells us the feasibility of reaction ie. whether it is spontaneous or non spontaneous. Equilibrium gives us an idea about the extent of a reaction, but none of them tells the speed or rate at which the reaction takes place. Kinetics helps you to determine the speed of a reaction. Let’s take an example:

                      

                          A         ⟶          B

Time = 0          a moles             0

Time = t           (a-x) moles       x moles

Here A gets converted to B. What is the rate at which this reaction takes place? At time zero ‘a’ moles of ‘A’ are present and zero moleof ‘B’. After some time ‘t’ x moles of ‘A’ get converted to ‘B’ so there will be (a-x) moles of ‘A’ left.

You can define the average rate either in terms of disappearance of A with time or appearance of B with time.

Rate_average = - Δ [Reactant] / Δtime

Rate_average = Δ [Product] / Δtime

Rate of reaction is a positive quantity, minus sign shows only decrease in concentration of reactant. Its unit is mol L^-1 sec^-1.

Let’s take an example:

Hg(l) + Cl₂(g) ⟶ HgCl₂(s)

You can define the rate in terms of reactants (Hg or Cl₂) or in terms of product HgCl₂.

Rate = - Δ [Hg] / Δtime = - Δ [Cl₂] / Δtime = Δ [HgCl₂] / Δtime

Let’s take another example:

2HI(g) ⟶ H₂(g) + I₂(g)

Rate = Δ [H₂] / Δtime = Δ [Cl₂] / Δtime ≠ Δ [HI] / Δtime

In this example, the rate of appearance of H₂is equal to the rate of appearance of I₂ but it cannot be equal to the disappearance of HI. Because when 2moles of HI are decomposed, 1 mole each of H₂ and I₂ is produced.

- ½ Δ [HI] / Δtime = Δ [H₂] / Δtime = Δ [Cl₂] / Δtime

To get the correct rate of reaction from balance equation, we divide the rate by stoichiometric coefficient. For example:

2N₂O₅(g) ⟶ 4NO₂(g)  + O₂(g)

Rate_average = ½ {-[Δ N₂O₅]/Δt} = ¼ {[Δ NO₂]/Δt} = ½ {[ΔO₂]/Δt}

Let's try to do a simple problem. If initially the concentration of N₂O₅ is 2.33mol L^-1 and after 184 minuts it is reduced to 2.08 mol L^-1. Calculate the rate of reaction.

Initial concentration [N₂O₅]_i= 2.33mol L^-1

Final concentration [N₂O₅]_f= 2.08mol L^-1

Δt = 184min is given. And,

Rate = ½ {-[Δ N₂O₅]/Δt}   

            = ½ {-[2.33mol L^-1 -  2.08mol L^-1] / 184min

            = 6.7 ×10-4mol L^-1 min^-1

Or,

            = ½ {-[2.33mol L^-1 -  2.08mol L^-1] / (184) 60 sec

            =1.13×10-5 mol L^-1 sec^-1

Rate of reaction can only be determined experimentally, you cannot predict it by balance chemical equation. It is totally an experimental quantity. It may depend on the concentrations of reactants or products. When it is represented by reactant concentration, it becomes the rate law. Let’s try to understand it by taking an example:

aA + bB ⟶ cC + dD

a, b, c and d are stoichiometric coefficients of balance equation. Rate of the reaction depends on the concentrations of A and B.

Rate ∝ [A]^x [B]^y

Rate of reaction depends on the concentration of A to the power x and B to the power y, where x and y has to be found out experimentally. They may be equal to a and b by coincidence but have no relation with stoichiometric coefficients of balance equation.

Rate = k [A]^x [B]^y

-d[R]/ dt = k [A]^x [B]^y

‘k’ is the proportionality constant known as rate constant. It only depends on temperature of the reaction and it has specific value for a particular reaction.

By adding the x and y you will get the order of the reaction.

Order of the reaction = (x +y)

Let’s see what it means. If the rate law for a reaction is given as,

Rate law = k [A]¹ [B]¹

Then the order of the reaction will become (1+1) = 2, which means it is a 2^nd order reaction.

Order of reaction is the extent of dependency of the rate of reaction on the reactant concentration. It is based on experimental data, so it can be fraction, integer or zero. For example:

CHCl₃ + Cl₂ ⟶ CCl₄ + HCl

Rate law for this reaction is = k [CHCl₃]¹[Cl₂]^1/2

Hence the order of reaction = 1+ ½ = 3/2 order

2NH₃ ⟶ N₂+ H₂

Rate =  k [NH₃]⁰

Order of reaction= 0 order

It is quite normal for us to assume that rate of reaction depends the reactant concentration. Order of reaction tells us how the rate is related to the reactant concentration. But what does it mean if its order is zero? It means that rate is independent of reactant concentration. Rate remains constant throughout the reaction. In the coming post we will discuss it in detail.

Both rate of reaction and order of reaction are experimental terms and we cannot predict them by balance chemical equation. So what information can we draw from balance chemical equation? Balance chemical equation gives us clues about how a particular reaction takes place. For example:

N₂ + H₂ ⟶ 2NH₃  

One molecule of N₂ and one molecule of H₂participate in the formation of NH₃. When these two molecules collide with each other, the reaction takes place and NH₃ is formed. In the language of chemistry you can say that the molecularity of this reaction is two. Let’s take another example:

2HI(g) ⟶ H₂(g) + I₂(g)

Here, two molecules of HI react with each other. So the molecularity of the reaction is 2.

Molecularity tells us how many molecules/ ions/ atoms are needed to collide at once for a reaction to happen. Since it is the number of molecules/ ions/ atoms, it can only be an integer. It can never be a fraction or zero. Molecularity can be 1, 2, 3, 4..... but how is it possible for 5 or 6 or more than 6 molecules to collide at the same time? It may happen but the probability of its happening is very rare. Then how do such reactions take place? These reactions are called as complex reactions. Actually, they don’t happen in a single step, they take place in a number of small reactions or elementary reactions. To calculate the Molecularity of such complex reactions, you have to sum up the Molecularies of all the elementary reactions. Molecularity of a complex reaction doesn’t make sense. Molecularity is applicable to elementary reactions only.

In the coming post we will discuss zero order and first order reactions in detail.

   

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