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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What is Gibbs Free Energy?

In the previous post of thermodynamics we have seen how the change in entropy decides the spontaneity of any reaction. But entropy alone cannot decide spontaneity, enthalpy of the reaction and temperature at which the reaction is being carried out also plays an important role. To sum up all these factors we have another thermodynamic property known as Gibbs free energy “G”.

G = H – TS

‘G’ is an extensive property. It is a state function as it doesn’t depend on path. Units of Gibbs energy is J.

Let’s find change in Gibbs free energy of the system at constant temperature:

ΔG_sys = ΔH_sys – TΔS_sys    ......(1)

This is one of the important equations of thermodynamics. Difference in G decides the spontaneity of any reaction, let’s find out how.

We know that total entropy change for a spontaneous reaction must be more than zero. Total entropy means the entropy of the system and surrounding.

ΔS_total = ΔS_sys  + ΔS_surr       ......(2)

For an exothermic reaction ΔH heat is released from the system to the surrounding. It means when -ΔH_sys heat is lost from the system, it is added to the surrounding as ΔH_sur.

ΔS_sys = -ΔH_sys / T

Then,

ΔS_sur = ΔH_sur / T

Since ΔH_sur = - ΔH_sys

ΔS_sur = ΔH_sur / T = - ΔH_sys / T

On putting the values of ΔS_sur in equation 2 we get,

ΔS_total  = ΔS_sys + (-ΔH_sys / T)

TΔS_total  = TΔS_sys  - ΔH_sys

TΔS_total  =  - (ΔH_sys  - TΔS_sys)

For spontaneity ΔS_total must be greater than zero, so

- (ΔH_sys  - TΔS_sys) > 0

On comparing it with equation 1 we can say that for spontaneity,

-ΔG_sys > 0

- ΔG_sys can be more than zero only when ΔG_sys is negative,

- (-ΔG_sys)  > 0

Or

ΔG_sys  < 0

So, for spontaneity ΔG_sys must be negative.  
 

Gibbs Free Energy and Spontaneity

We know that ΔG_sys = ΔH_sys – TΔS_sys, where ΔH_sys is the enthalpy of the system and TΔS_sys is the energy required to be released or absorbed by the system and by subtracting them you will get the remaining energy ΔG which can be used to do any work. That’s why ΔG is known as free energy which is available to do useful work. For any spontaneous reaction ΔG is negative which means all energy has been used to carry the reaction forward (spontaneous reactions are irreversible). Reactions for which ΔG is negative are known as ‘exergonic reactions’. ‘Endergonic reactions’ have positive ΔG.

Let’s see how we can predict the spontaneity of any reaction by using equation of ΔG:

A + B ⟶ C + D ...... ΔH = (+ve) and ΔS = (+ve)

On applying Gibbs equation

 ΔG_sys = ΔH_sys – TΔS_sys

At lower temperature ΔG will be positive so that above reaction will be non-spontaneous at lower temperature. But if temperature is higher enough to overrule the ΔH, then it can be spontaneous.

Now we understand that if ΔG < 0 then reaction will be spontaneous, and if ΔG > 0 then it will be non-spontaneous. So what happens if ΔG = 0? Reaction will be at equilibrium if ΔG = 0.

ΔG = ΔG^ө + RT lnQ

At equilibrium reaction quotient Q becomes equal to equilibrium constant K and ΔG = 0

ΔG = ΔG^ө + RT lnK = 0

ΔG^ө =- RT lnK

Or

 ΔG^ө =- 2.303RT logK

lnK = -ΔG^ө/ RT

Taking antilog of both side,

K = e ^(-ΔGө/RT)

where ΔG^ө is the standard gibbs energy of the reaction.

When ΔG^ө < 0 then  (-ΔG^ө/ RT) will be positive and e ^{(-ΔGө /RT)} will be more than zero. That means K > 1 which means forward reaction is favoured and equilibrium shifts forward. Here product concentration exceeds reactant concentration.

When ΔG^ө > 0 then  (-ΔG^ө/ RT) will be negative and e ^{(-ΔGө /RT)} will be less than zero. That means K < 1 which means backward reaction is favoured and equilibrium shifts backward and product formation is suppressed.

I hope these posts have been helpful in understanding the thermodynamic concepts. The next post of thermodynamics will focus on the applicability of these concepts where we will try to apply these concepts to solve some numerical problems.

We have discussed all about feasibility and spontaneity of any reaction but how do we decide the speed of the reaction? On which factors it depends? Can we accelerate its speed? We will find the way to solve this mystery in the coming posts of chemical kinetics.

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Qualitative analysis of Group III(A) cations

In the previous post of salt analysis we have learnt why we have to remove interfering radicals before 3^rd group analysis. After removing the interfering radicals, we will proceed to analyse the 3^rd group cations in the filtrate. To separate the radicals of 3^rdgroup we need to get them precipitated. The group reagent of 3^rd group is ammonium sulphide solution or hydrogen sulphide gas in the presence of ammonia and ammonium chloride. When we add group reagent to the filtrate we will get precipitate of 3^rd gr cations.

Cations of this group are cobalt(II) Co²⁺, Nickel(II) Ni²⁺, iron(II) Fe²⁺, iron(III) Fe³⁺, chromium(III) Cr³⁺, aluminium(III) Al³⁺, zinc(II) Zn²⁺, manganese(II) Mn²⁺, manganese(VII) Mn⁷⁺,  

Iron, aluminium and chromium and sometimes Manganese are precipitated as hydroxide by ammonia solution in presence of ammonium chloride while others are precipitated with ammonium sulphide in the form of sulphides. That’s why 3^rd group is further divided into A and B group.

III(A) is known as Iron group and it consists of iron(III) Fe³⁺, chromium(III) Cr³⁺, aluminium(III) Al³⁺.

III(B) is known as Zinc group and it consists of cobalt(II) Co²⁺, Nickel(II) Ni²⁺, zinc(II) Zn²⁺, manganese(II) Mn²⁺, manganese(VII) Mn⁷⁺.

Now let's continue from the group II filtrate and discuss the steps we have to follow to separate III(A) cations

Qualitative analysis of Group III(A) cations 

Step 1: Precipitation of III(A) cations as Hydroxide - To the filtrate, add few drops of nitric acid HNO₃ to convert ferrous Fe²⁺ into ferric Fe³⁺. Then add solid ammonium chloride NH₄Cl₃ and ammonia NH₃ solution and boil. On boiling you will get precipitate of III(A) cations which may contains Fe(OH)₃, Cr(OH)₃, Al(OH)₃ and a little MnO(OH)₂. Solubility product of iron(III) hydroxide is very small so it precipitates completely.

Fe³⁺ +3NH₃ + 3H₂O ⟶ Fe(OH)₃↓ + 3NH₄⁺

Ammonium chloride should be added in excess otherwise III(B) cations may get precipitated here. But avoid too much excess of ammonium chloride and ammonia otherwise Cr³⁺ may not precipitate and Al³⁺ may form colloidal solution. .

Cr³⁺ + 3NH₃ + 3H₂O ⟶ Cr(OH)₃↓ + 3NH₄⁺

Cr(OH)₃↓ + 6 NH₃ ⟶ [Cr(NH₃)₆]³⁺ + 3OH^-

Al(OH)₃↓ + OH^- ⟶ AlO₂^- + 2H₂O

Step 2: Separation of Fe(III) and Mn(II) from Al(III) and Cr(III) – Wash the precipitate with hot 

water and transfer to a test tube. Add a little water and add 1-2g of sodium peroxide or sodium hydroxide solution and bromine water/ hydrogen peroxide H₂O₂. Boil and filter in hot. On boiling, hydroxides of aluminium and chromium dissolve by forming meta- aluminate and chromate ion, and iron and manganese remain in the solution.

Al(OH)₃↓ + OH^- ⟶ AlO₂^- + 2H₂O

2Cr(OH)₃↓ + 2O₂^2- ⟶ 2CrO₄^2- + 2 OH^- + 2H₂O

Filter the residue to test for iron and manganese, and keep the filtrate for the test of aluminium and chromium.

Step 3: Test for iron and manganese – if residue is black, it means iron and manganese both are present and if it is reddish brown then only iron is present. Divide the residue into two parts.

Part 1 confirmatory test for Fe³⁺- dissolve the precipitate in dil HCl.

Fe(OH)₃ + H⁺ ⟶ Fe³⁺ + 3H₂O

Add few drops of potassium hexacyanoferrate (II) K₄[Fe(CN)₆] solution, intense blue precipitate of iron(III) hexacyanoferrate will appear.

4Fe³⁺ + 3 [Fe(CN)₆]^4- ⟶ Fe₄[Fe(CN)₆]₃ ↓

Excess of reagent dissolves it completely and intense blue solution is obtained and on adding sodium hydroxide solution to this blue solution, a red precipitate of iron(III) hydroxide is obtained.

Fe₄[Fe(CN)₆]₃ + 12OH^- ⟶ 4Fe(OH)₃ ↓ + 3[Fe(CN)₆]^4-

Part 2 confirmatory test for Mn²⁺- dissolve the precipitate in 1ml concentrated nitric acid HNO₃, if necessary add 1-2 drops of H₂SO₃. Add 0.05-0.1g sodium bismuthate NaBiO₃ and shake. A violet solution of permanganate will appear.

2Mn²⁺ + 5NaBiO₃ + 14H⁺ ⟶ 2MnO^-₄ + 5Bi³⁺ + 5Na⁺ + 7H₂O

Step 4: Test for aluminium and chromium – The filtrate from step 2 may contain CrO^-₄ and [Al(OH)₄]^-. If it is yellow, it means chromium might be present but if it is colourless then chromium is absent and you don’t need to test for it.

Divide the filtrate into 3 parts, we will test for chromium in 2 parts and in the third part we will test for aluminium.

Part 1 Test for Cr (III): acidify with dilute acetic acid CH₃COOH and add 0.25M lead acetate Pb(CH₃COO)₂ solution, a yellow precipitate of lead chromate is formed.

2CrO₄^2- + Pb²⁺ ⟶ PbCrO₄↓

This yellow precipitate is soluble in sodium hydroxide solution. it conforms the presence of Cr(III).

PbCrO₄↓ + 4OH^- ⟶ PbO₂^2- + CrO₄^2- + 2H₂O

Part 2 Test for Cr (III): acidify it with dilute nitric acid HNO₃, cool thoroughly and add 1ml amyl alcohol (2-methyl-butane-4-ol) and 3-4 drops of 3% hydrogen peroxide solution H₂O₂. Shake well and allow the two layers to separate. Upper blue layer contains perchromic acid (chromium pentoxide).

2 CrO₄^2- + 2H⁺ + 3 H₂O₂ ⟶ 2CrO₅ + 4 H₂O

Part 3 Test for Al (III): To test for aluminium acidify the filtrate with dilute HCl (test with litmus paper) then add 2M ammonia solution NH₃ until just alkaline. Heat till boiling. White gelatinous precipitate of aluminium hydroxide is obtained.

Al³⁺ + 3NH₃ + 3H₂O ⟶ Al(OH)₃↓ +3NH₄⁺

Filter this precipitate and dissolve a small portion of it into 1ml hot dil HCl. Cool and add 1ml 6M ammonium acetate solution and 0.5ml of 1M aluminon reagent (0.1g tri-ammonium arurine- tricarboxylate O(COONH₄)C₆H₃=C[C₆H₃(OH) COONH₄]₂dissolved in 100ml water), stir it and add ammonium carbonate solution. a red coloured precipitate confirms the presence of Al(III). In this post we discussed analysis of the III(A) group cations. In the next post of salt analysis we will continue with the test test for III(B) cations in the filtrate of III(A).

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