Chemical Kinetics - Class 12 Medical Chemistry - Extra Questions

Every chemical reaction has its own characteristic rate. It is affected by some important factors. The most important factors are            1           , temperature,            2           ,            3            and surface area out of these factors rise/decrease in temperature, light quantity, rise in the concentration of reactants and surface area            4            the rate of reaction. Whereas as            5            alter the rate of reaction.

$$1\rightarrow $$concentration of reactant
$$2\rightarrow $$ligh quantity
$$3\rightarrow $$catalysts
$$4\rightarrow $$increases
$$5\rightarrow $$catalysts
The rate of chemical reaction increases with increase in the concentration of reactant. Also the physical state effect the rate of chemical reaction. gaseous state increases. The rate of chemical reaction. catalyst reduced the activation energy and thereby increase the rate by providing a shorter path.


A gas is formed by the reaction between marble pieces and dilute $$HCl$$.
(a) Which gas is formed as a result of this reaction?
(b) Write the balanced chemical equation for this reaction.
(c) Suggest two ways of increasing the speed of the chemical reaction.

(a) Carbon dioxide gas
(b) $$Ca{CO}_{3}+2HCl\rightarrow Ca{Cl}_{2}+{CO}_{2}+{H}_{2}O$$
(c) Following are the ways of increasing the speed of the chemical reaction (Choose any 2):
(i) The speed of the reaction can be increased by cutting line stone into smaller pieces. When solid substances are split into small pieces, the surface area increases, resulting in an increase in the possibility of collision between reactant molecules. This is why the speed of the reaction increases.
(ii) The speed of the reaction can be increased the strength of the $$HCl$$ solution.
(iii) The speed of the reaction can also be increased by heating the contents (limestone in $$HCl$$). When heated, the kinetic energy of the molecules increases resulting in an increase in the number of effective collisions. This is why the speed of the reaction increases. In general, the speed of chemical reactions increases considerably on increasing the temperature.


Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrocloric acid is added to test tube A, while acetic acid is added to test tube B. The amount and concentration taken for both the acids are same. In which test tube does the reaction occur more vigourously and why?

Since, $$HCl$$ acid is much stronger than acetic acid $$(CH_3COOH)$$, therefore in test tube A reaction occurs more vigorously because due to strong acidic nature of $$HCl,Mg$$ can be easily displaced as compared with the reaction of $$Mg$$ and acetic acid.


The temperature coefficient of the rate of a reaction is $$2.3.$$ How many times will the rate of the reaction increase if the temperature is raised by $$25 K$$?
$$(8.02 times)$$

$$C=$$ temperature coefficient $$=2.3$$
If Temperature is $$T$$, let the rate is $$R$$.
TempRate
$$T\ K$$$$R$$
$$(T+10)K$$$$CR$$
$$(T+20)K$$$$C^2R$$
$$(T+30)\ K$$$$C^3R$$
If $$T$$ is increased by $$20\ K$$, $$Rate_1=(2.3)^2R$$
If $$T$$ is increased by $$30\ K$$, $$Rate_2=(2.3)^3R$$
At intermediate temperature, temperature is increased by $$25\ K$$
Rate $$=\dfrac{Rate_1+Rate_2}{2}=\dfrac{(2.3)^2R+(2.3)^2R}{2}$$
$$=\dfrac{5.29R+12.167R}{2}=8.72R$$
$$\therefore $$ Rate will be increased by $$8.72$$ times


What will be the initial rate of a reaction if its rate constant is $${10^{ - 3}}{\min ^{ - 1}}$$ and the concentration of reaction is $$0.2$$ $$mold{m^{ - 3}}$$.How much of reactant will be converted into products in $$200$$ min.


1312772_1377744_ans_7c342eb236084f69a29a4548c0ecaf05.jpg


In the following schematic diagram for the preparation of hydrogen as as shown in given figure, what would happen if the following changes are made?
(a) In place of zinc granules, same amount of zinc dust is taken in the test tube.
(b) Instead of dilute sulphuric acid, dilute hydrochloric acid is taken
(c) In place of zinc, copper turnings are taken
(d) Sodium hydroxide is taken in place of dilute sulphuric acid and the tube is heated.
1773731_603dd8b3c1bf4d638f2dd1337eb37470.png

(a) If in the place of zinc granules, the same amount of zinc dust is taken in the test tube then the rate of the reaction will increase and the reaction will take place more quickly. This is because the surface area per unit volume of powder is more than that of the zinc granules and hence the powder will provide more surface area for the reaction to take place.

(b) If in place of dilute sulphuric acid, we use dilute hydrochloric acid then the reaction will take place as usual because in both cases we are mixing a strong acid with a metal.

(c) If in the place of zinc, copper turning is taken then also the reaction will proceed in a similar way and would lead to the evolution of hydrogen gas as here we are only replacing one metal with another metal in the reactant side of the reaction. In this case, also we are mixing a strong acid with a metal.

(d) If sodium hydroxide is taken in place of dilute sulphuric acid and the tube is heated then the metal will react with the base to form the corresponding salt and will evolve hydrogen gas as per the following reaction:

$$Zn+2NaOH\rightarrow {Na}_{2}Zn{O}_{2}+{H}_{2}$$


Why does the rate of a reaction increase with rise in temperature ? 

As on increasing temperature, the energy of gases increases, so a larger fraction of colliding particles can cross the energy barrier (i.e., the activation energy ), which leads to an increase in rate.


What is the effect of temperature on the reactions?
(i) $$N_2 (g) + 3H_2 (g) \ \overset{\rightharpoonup}{\leftharpoondown} \ 2 NH_3 (g) + Heat$$
(ii) $$N_2 (g) + O_2 (g) \ \overset{\rightharpoonup}{\leftharpoondown} \ 2 NO_3 (g) - Heat$$

(i) The increase in temperature will favour backward reaction because the reaction is exothermic.
(ii) The increase in temperature favours forward reaction because the reaction is endothermic.


Why does the rate of any reaction generally decrease during the course of the reaction ? 

The rate of reaction depends on the concentration of reactants. As the reaction progress , reactant start getting converted to products so the concentration of reactants decreases hence the rate of reaction decreases. 


The rate of which reaction increases when the concentration of nitrogen is increased? Forward reaction / Backward reaction.

Forward reaction


The graph showing the progress of the reaction $$N_{2}+3H_{2}\rightleftharpoons 2NH_{3}$$ is given
Identify the reactions represented by $$B$$ and $$C$$?
1835930_a89a2dcecb49488c9336d7c36221c2b4.png

$$B-$$ Forward reaction, $$C-$$ Backward reaction


Find out the relation between Temperature and rate of a reaction?
Select the materials from the list above
Sodium the sulphate, Test tube, dil $$HCl,Cu,Mg$$ ribbon, Beaker, Water, Spirit lamp, boding Tube

Essential Materials:- Sodium Thiosu-phate, Hydrochloric Acid, Water Boiling Tube, Spirit Lamp. Prepare a so;ution of dilute sodium Thiosulphate in a beaker. Take equal amount of this solution in two Boiling Tubes. Heat one of the boiling tubes for same time. Add amount of dilute $$HCl$$ to both the boiling tubes.


A graph given below deals with the reversible reaction.
What inference can be drawn about the concentration of the reactants and products at the point $$D$$ and $$E$$?
1835943_405c708d707b4e748dd4830ad3ad3e65.png

There id no change in the concentration, of reactants and products at the points $$D$$ and $$E$$. Because at the point $$'C'$$ the process attains equilibrium.


Describe an experiment to prove that nature of the reactants affects the  rate of chemical reaction.

Materials required: Zn, Mg , dil. HCl arid test tubes.
Procedure: Take equal volume of dil. HCl in two test tubes. Add Zn to one and Mg of same mass to the other. Hydrogen gas is produced in both the test tubes. Rate of reaction is faster in the test tube containing Mg.


Is there any difference in the mass of the marble ?
1838608_160d73998b8247bbae09c8abca67cce4.png

Mass of the marble will be same.


What happens if the concentration of ammonia is increased?

Rate of backward reaction increases


A graph given below deals with the reversible reaction.
What happen to the forward and backward reactions as time passes?
1835973_1b423e3dfd6d4979982ea6d10fb2abf0.png

As time passes the Speed of forward reaction decreases and backward reaction
increases.


Why rate of reaction increases when concentration increases ?

As the concentration of reactants increases, the number of molecules per unit volume and the number of effective collisions increase. Consequently the rate of reaction increases.


What about the concentration of acids in both the reactions ?
1838607_62ab961709e644a4ac0014c69e2affc9.png

 The concentration of acids in both the reactions are same.


Some materials available in the laboratory are given below. Magnesium ribbon, marble powder , marble pieces , dilute HCl , concentrated HCl.
Which materials will you choose for the preparation of more $$CO_2$$ is less time ?

Marble powder and concentrated $$HCl$$


Giving reasons in brief, indicate whether the following statement is TRUE or FALSE:
The rate of an exothermic reaction increases with increasing temperature.

False because in an exothermic reaction, heat is evolved. Increase of temperature will shift the equilibrium in the backward direction i.e. the rate of reaction decreases.


The experiment conducted by two students are given below.
Experiment 1 : $$2\,mL$$ of sodium thiosulphate solution is taken in a test tube, heated and to it $$2mL $$ of Hg solution is added.
Experiment 2 : $$2\,mL $$ of sodium thiosulphate solution is taken in a test tube and to it $$2\,m$$ of $$HCl$$ solution is added.
In which experiment is the precipitate formed quickly ? Justify your answer.

In heated test tube.
When temperature increases more molecules get threshold energy and rate of reaction increases.


The experiment conducted by two students are given below.
Experiment 1 : $$2\,mL$$ of sodium thiosulphate solution is taken in a test tube, heated and to it $$2mL $$ of Hg solution is added.
Experiment 2 : $$2\,mL $$ of sodium thiosulphate solution is taken in a test tube and to it $$2\,m$$ of $$HCl$$ solution is added.
Write the balanced equation for the reaction.

$$Na_2SO_3 + 2HCl \rightarrow 2NaCl + SO_2 + S $$ 


Mention the factors that affect the rate of a chemical reaction.

Following factors affect the rate of a chemical reaction:
1) Reactant concentration
2) Reaction temperature
3) Nature of reactants and products
4) Presence of a catalyst
5) Surface area
6) Radiation exposure


For a given reaction $$A+B\to P$$, the ordes w.r.t $$A$$ and $$B$$ are $$1$$ and $$2$$ respectively. Fill in the blanks from the following data:

Rate $$(M s^{-1})$$$$[A]$$$$[B]$$
$$0.10$$$$1.0\ M$$$$0.20\ M$$
...$$2.0\ M$$$$0.20\ M$$
...$$2.0\ M$$$$0.40\ M$$

For $$A+B \to P$$

Rate $$=K[A]^1[B]^2$$

If $$[A]$$ is doubled, rate get doubled 

$$\therefore R_2=2\times 0.10=0.20\ M/s$$

If $$[A]$$ and $$[B]$$ is doubled, rate increase by 8 times.

$$\therefore R_2=0.10\times 2\times 2^2=0.80\ M/s$$


One mole of a gas $$A$$ and two moles of a gas $$B$$ are introduced into one vessel and $$2$$ moles of $$A$$ and $$1$$ mole of $$B$$ a second vessel having the same capacity. The temperature is the same in both vessels. Will the rate of reaction $$A$$ and $$B$$ in these vessel differ if its expressed by the equation.
rate $$=k[A][B]$$ 

In first vessel
$$[A]=\dfrac{1}{V}$$

$$[B]=\dfrac{2}{V}$$

In $$2^{nd}$$ vessel
$$[A]\dfrac{2}{V}$$

$$[B]=\dfrac{1}{V}$$

(i) If Rate $$=K[A][B]$$

Rate $$1=K \left( \dfrac{1}{V}\right)\left( \dfrac{2}{V}\right)=\dfrac{2K}{V^2}\therefore R_1=R_2$$

$$Rate_2=K(2/V)(1/V)=2K/V^2\therefore$$ Rate are same in two vessel


The rate law of the reaction,
$$CH_3COOC_2H_5 +H_2O \xrightarrow []{[H^+]} CH_3COOH+C_2H_5OH$$ is
$$\dfrac {dx}{dt}=k[CH_3COOC_2H_5] [H_2O]^o$$
What will be the effect on the rate if the concentration of $$H^+$$ is tripled?

$$CH_3COOC_2H_5 +H_2O \xrightarrow {[H^+]}CH_3COOH+C_2H_5OH$$

$$\dfrac{dx}{dt}=K[CH_3COOC_2H_5][H_2O]^o$$

On concentration of $$H^+$$, rate doesn't depend & so there will no change in rate as conc of $$H^+$$ changes


The rate law of the reaction,
$$CH_3COOC_2H_5 +H_2O \xrightarrow []{[H^+]} CH_3COOH+C_2H_5OH$$ is
$$\dfrac {dx}{dt}=k[CH_3COOC_2H_5] [H_2O]^o$$
What will be the effect on the rate if the concentration of the ester is doubled?

$$CH_3COOC_2H_5 +H_2O \underrightarrow {[H^-]}CH_3COOH+C_2H_5OH$$
$$\dfrac{dx}{dt}=K[CH_3COOC_2H_5][H_2O]^o$$
Rate is directly proportional to concentration of $$[CH_3COOC_2H_5]$$ & hence Rate doubles when concentration of easter doubles.


How many times will the rate of the reaction $$2A+B\to A_2B$$ change if the concentration of substance $$A$$ is doubled and that of substance $$B$$ is halved?

$$2A_1+B\to A_2B$$

$$Rate_1=K[A]^2[B]$$

If concentration of $$A$$ is doubled & that of $$B$$ is halved

Then $$Rate_2=K[2A]^2.\dfrac{B}{2}=2K[A]^2.B$$

$${Rate}_2=2\ {Rate}_1$$ 

Ans-$$2$$


One mole of a gas $$A$$ and two moles of a gas $$B$$ are introduced into one vessel and $$2$$ moles of $$A$$ and $$1$$ mole of $$B$$ a second vessel having the same capacity. The temperature is the same in both vessels. Will the rate of reaction $$A$$ and $$B$$ in these vessel differ if its expressed by the equation.
rate $$=k[A]^2 [B]$$ ?

In first vessel
$$[A]=\dfrac{1}{V}$$

$$[B]=\dfrac{2}{V}$$

In $$2^{nd}$$ vessel
$$[A]\dfrac{2}{V}$$

$$[B]=\dfrac{1}{V}$$

$$Rate =K[A]^2[B]$$,

In $$1^{st}$$ vessel

$$Rate_1=K\left( \dfrac{1}{V}\right)^2\left( \dfrac{2}{V}\right)=\dfrac{2K}{V^3}$$
In $$2^{nd}$$ vessel

$$Rate_2K\left( \dfrac{2}{V}\right)^2\left( \dfrac{1}{V}\right)=\dfrac{4K}{V^3}$$
So, $$Rate_1 \neq Rate_2$$

Hence in this case rate is not equal


For the reaction: $$2A+B_2+C\rightarrow A_2B+BC$$, the rate law expression has been determined experimentally to be $$R=k[A]^2[C]$$ with $$k=3.0\times 10^{-4}M^{-2}min^{-1}$$.
Determine the rate after $$0.04$$ mole per litre of A has reacted.


1737275_1738337_ans_d51b2e918fbf4adf88c0df76854e7176.JPG


From the following data for the reaction between A and B:

$$[A]$$, mol $$L^{-1}$$$$[B]$$, mol $$L^{-1}$$Initial rate, mol $$L^{-1}s^{-1}$$, atInitial rate, mol $$L^{-1}s^{-1}$$, at
$$300$$ K$$320$$K
$$2.5\times 10^{-4}$$$$3.0\times 10^{-5}$$$$5.0\times 10^{-4}$$$$2.0\times 10^{-3}$$
$$5.0\times 10^{-4}$$$$6.0\times 10^{-5}$$$$4.0\times 10^{-3}$$-
$$1.0\times 10^{-3}$$$$6.0\times 10^{-5}$$$$1.6\times 10^{-2}$$-

Calculate the rate constant at $$300$$K.


1614947_1739956_ans_bc60e1c5ba224390b7ea20abcd77f452.jpg


The rate constant for an isomerisation reaction $$A\rightarrow B$$ is $$4.5\times 10^{-3}$$ $$min^{-1}$$. If the initial concentration of A is $$1$$M, calculate the rate after $$1$$ h.


1611434_1740732_ans_6b7582cb197b48d49cc22a72d952f662.jpg


The activation energy of a certain uncatalysed reaction at $$300\ K$$ is $$76\ kJ\ mol^{-1}$$. The activation energy is lowered by $$19\ kJ\ mol^{-1}$$ by the use of catalyst. By what factor, the rate of catalysed reaction is increased?


1744083_1742586_ans_64bfc93fc3ac4a3aa155a5a704a2ceeb.jpg


The order of the reaction $$2A+B+C\rightarrow Product$$, is found to be 1, 2 and 0 w.r.t A, B and C respectively. If the concentration of each reactant is increased by two times, what will be the effect on the rate of the reaction?                            


1606775_1753360_ans_5c8f339e16b549c495641fc0f595a276.jpeg


Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction . Explain with the help of one example . 


1659682_1778252_ans_ce89d173a576448ea77fa709a9caff12.jpg


A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant doubled? 

Four times 

$$ Rate_{1} = K\left [ C_{A} \right ] ^{2} $$ 

$$ C_{A} = 2{C_A} $$

$$ Rate _{2} = K\left [ 2C_{A} \right ]^{2} = 4K\left [ C_{A} \right ]^{2} = 4 Rate_{1} $$ 


For the reaction :
$$ 2A + B \rightarrow A_{2}B $$
The rate $$ = k \left [ A \right ]\left [ B \right ]^{2} $$  $$ k= 2.0 \times 10 ^{-6} $$  $$ mol^{-2}   L ^{2} s^{-1} $$ . 

Calculate the initial  rate of the reaction when $$ \left [ A \right ] = 0.1 mol L^{-1} , \left [ B \right ] = 0.2 molL^{-1}$$  

Initial rate =   $$k\left [ A \right ] \left [ B \right ]^{2} $$

            $$ =2.0 \times 10^{-6} mol^{-2}L^{2}S^{-1}\times 0.1  molL^{-1} \times (0.2) ^{2}mol^{2} L^{-2} $$

            $$ =  8\times  10^{-9} mol L^{-1} S^{-1} $$ 





How many types of reactions are there on the basis of speed?

Slow Reaction: Chemical reactions that occur very slowly and can take a long time for completion are called slow reactions. 

Fast Reaction: Chemical reactions that complete in a very short time, such as less than 10 -6 seconds, they are called fast reactions.

Rusting of iron and burning of fuel both are oxidation reactions. The first one is a typical slow reaction while the second one is a fast reaction. 


Mention the factors that affect the rate of a chemical reaction.

1 ) Concentration of reactants 
2) Temperature 
3) Nature of reactants 
4) Exposure to light (Radiation)
5 ) Presence of catalyst.


In a reaction between A and B, the initial rate of reaction ($$r_0$$) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B?

1828238_daaba23094c24519a8a015202ef62b25.png

Let the rate law be , $$ r = K \left [ A \right ]^{x}\left [ B \right ]^{y} $$ 

$$ r_{1} = K \left [ 0.2 \right ]^{x} \left [ 0.3 \right ]^{y} = 5.07 \times 10^{-5} - \left (1  \right ) $$

$$ r_{2} = K\left [ 0.2 \right ]^{x} \left [ 0.1 \right ]^{y} = 5.07 \times 10^{-5} -\left ( 2 \right ) $$

$$ r_{3} = K\left [ 0.4 \right ]^{x} \left [ 0.05 \right ]^{y} = 1.43 \times 10^{-4} -\left ( 3 \right ) $$

$$ \dfrac{\left ( 1 \right )}{\left ( 2 \right )} = \dfrac{r_{1}}{r_{2}} = \dfrac{K\left [ 0.3 \right ]^{y}}{K\left [ 0.1 \right ]^y} = 1 $$ 

$$ 3^{y} = 1 $$ 

$$ \Rightarrow  y = 0 $$

$$ \dfrac{\left ( 3 \right )}{\left ( 1 \right )} = \dfrac{r_{3}}{r_{1}} = \dfrac{K\left [ 0.4 \right ]^{x}\left [ 0.05 \right ]^{y}}{K \left [ 0.2 \right ]^{x}\left [ 0.3 \right ]^{y}} = \dfrac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} $$ 

$$ = 2 ^{x} =2.82$$

$$ x = 1.5  approx$$ 


A reaction first order in A and second order in B.
How is the rate affected on increase the concentration of B three times ?

$$Rate = k [A]^1[B]^2$$

When concentration of B is tripled 

$$Rate = k [A]^1[3B]^2= 9 k [A]^1[B]^2$$

When concentration of B is tripled , rate of reaction increases by 9 times .


The reaction  between A and B is first order with respect to A and Zero order with respect to B .
Fill in the blanks in the following table : 
1828305_e360e40701034140ace5a2178ae9897e.PNG

$$ rate law = K\left [ A \right ] $$
$$ 2.0 \times 10^{-2} = K 0.1 $$
$$ \Rightarrow  K = 2.0 \times 10^{-1} min ^{-1} $$
from II $$  4.0 \times 10^{-2} = 0.2 \times 10^{-31}  X $$
$$ K = 2.0 \times 10^{-1}  mol L ^{-1}  $$ 
$$ \Rightarrow  Z = 0.1 mol ^{-1} $$ 


Write the expression for rate of reaction in terms of each reactant and product for the reaction, $$N_2 +3H_2 \rightarrow 2NH_3$$


1910712_1828342_ans_59903c1403a24946a8e22351d9c9c0f8.JPG


What is the effect of a temperature on the rate constant of a reaction? How can this temperature affect on rate constant be represented quantitatively. 

Increasing temperature on decreasing the activation energy will result in an increase in the rate of reaction and an exponential increase in the rate constant . 

On increasing the temperature the fraction of molecules which collide with energy greater than Ea increase and hence the rate constant (exponentially)

$$ K = A ^{-ea/RT} $$, quantitative representation of temperature effect on rate constant .  


Name the factors on which the rate of a particular reaction depends.

1) Nature of the reactants :

2) Concentration of the reactants:
Greater the concentration of the reactants, faster is the reaction. Conversely, as the concentrations of the reactants decrease, the rate of reaction also decreases.

3. Temperature:
The rate of reaction increases with increase of temperature. In most of the cases, the rate of reaction becomes nearly double for 10K rise of temperature. In some cases, reactions do not take place at room temperature but take place at higher temperature.

4. Presence of catalyst:
A catalyst generally increases the speed of a reaction without itself being consumed in the reaction. In case of reverse reactions, a catalyst helps to attain the equilibrium quickly without disturbing the state of equilibrium.

5. Surface area of the reactants:
For a reaction involving a solid reactant or catalyst, the smaller is the practical size, greater is the surface area, and the faster is the reaction.

6. Presence of light:
Some reactions do not take place in the dark but can take place in the presence of light like photosynthesis or photochemical reactions.


For an elementary reaction,
$$2 A + B \rightarrow 3 C$$
the rate of appearance of C at time 't' is $$1.3 \times 10^{-4} mol\ L^{-1} s^{-1}$$. Calculate at this time
1) Rate of the reaction.

$$  \dfrac {1}{2} \dfrac{d\left [ A \right ]}{dt} = \dfrac{-d\left [ B \right ]}{dt} = \dfrac{+1}{3} \dfrac{d\left [ C \right ]}{dt} $$

$$ \dfrac{d\left [ C \right ]}{dt} = 1.3 \times 10^{-4} mol L^{-1} S^{-1} $$

$$ \therefore  Rate\  of\  reaction = \dfrac {1}{3} \dfrac{d\left [ C \right ]}{dt} =\dfrac{1}{3}\times1.3\times 10^{-4}= 4.33 \times 10^{-5} mol L^{-1} S^{-1}$$ 


A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant reduced to half?

Four times 

$$ Rate_{1} = K\left [ C_{A} \right ] ^{2} $$ 

$$ C_{A} = \dfrac {1}{2}{C_A} $$

$$ Rate _{2} = K\left [ \dfrac{1}{2}C_{A} \right ]^{2} = \dfrac{1}{4}K\left [ C_{A} \right ]^{2} = \dfrac{1}{4} Rate_{1} $$ 

Therefore, the rate of the reaction would be reduced to ¼th


A reaction first order in A and second order in B.
How is the rate affected when the concentration of a both A and B are doubled ?

$$Rate = k [A]^1 [B]^2$$

When concentration of both A and B are doubled

$$Rate = k [2A]^1[2B]^2  = 8k [A]^1[B]^2$$

Rate of reaction increase by 8 times .


Coldwater is taken in one test tube an hot water in another one. Mg ribbon with same size is dropped in each of the test tubes.
Which factor influences the rate of reaction? Explain the reason.

Temperature, When temperature increases the number of molecules with thresh-old energy increases. As a result number of effective collisions increases.


The rate law for a reaction is found to be.
Rate $$= K [NO^{-}_{2}][I^{-}][H^{+}]^{2}$$

How would the rate of reaction change when
Concentration of $$H^{+}$$  is doubled

Suppose initially the concentration  are $$ \left [ NO_2^{-} \right ] = a mol \  L^{-1} , \left [ I^{-} \right ] = b mol L^{-} and \left [ H^{+} \right ] = c mol L^{-1} $$ 

$$ \therefore Rate  = K abc^2 $$
  
New $$ \left [ H^{+} \right ]  = 2c $$

$$\therefore New \ rate  = 4ab \left ( 2c \right )^{2} = 4abc^2 $$

                                 = 4 times 


Prepare a write up for finding the relation between temperature and rate of a reaction.

The Essential material/Materials Required $$Mg$$ Ribbon, Cone $$HCl$$, dil $$HCl$$
The Magnesium Ribbons of equal mass in two test tubes. Add dil $$HCl$$ to one test tube and dilute $$HCl$$ to the other in equal volumes.


The rate law for a reaction is found to be,

Rate $$= K [NO^{-}_{2}][I^{-}][H^{+}]^{2}$$ 

How would the rate of the reaction change when the concentration of $$H^{+}$$ is doubled?


1962559_1828369_ans_088834eca7634d82aa2c96739b90a77f.jpg


For an elementary reaction,
$$2 A + B \rightarrow 3 C$$
the rate of appearance of C at time 't' is $$1.3 \times 10^{-4} mol\ L^{-1} s^{-1}$$. Calculate at this time
a) Rate of disappearance of A.

$$- \dfrac{d\left [ A \right ]}{dt} =\dfrac{ 2}{3} \dfrac{d\left [ C \right ]}{dt} $$

$$= \dfrac {2}{3} \times 1.3 \times 10^{-4}  mol L ^{-1} S^{-1} $$

$$ = 8.66 \times 10^{-5} mol L^{-1} S^{-1} $$


The rate constant for a first order reactions is $$ 60 s^{-1}$$ . How much time will it take to reduce the initial concentration of the reactant to its 1/16  the value  ?

$$ \left [ A \right ] = \dfrac{1}{16} \left [ A \right ]_{o} \Rightarrow  \dfrac{\left [ A \right ]_{o}}{\left [ A \right ]} = 16 $$


$$ t = \dfrac{2.303}{K} \log  \dfrac{\left [ A \right ]_{o}}{\left [ A \right ]}  $$ 


$$ t = \dfrac{2.303}{60} \log (16)  $$ 

 = $$0.0462\  sec$$.


The rate law for a reaction is found to be.
Rate $$= K [NO^{-}_{2}][I^{-}][H^{+}]^{2}$$ How would the rate of reaction change when
Concentration of each of $$NO_{2}, I^{-}$$ and $$H^{+}$$ are tripled?

New $$ \left [ NO^{-}_{2} \right ] = 3a , \left [ I^{-} \right ] = 3b , \left [ H^{+} \right ] = 3c $$
$$Rate = K\left ( 3a \right )\left ( 3b \right )\left ( 3c \right )^{2} $$


What happens to the rates of forward and backward reactions as time progresses?

As the times passes, rate of forward reaction decreases and rate of backward reaction increases.


What will be the effect of removing ammonia continuously from the system ?

The effect of removing ammonia continuously from the system results in increase in rate of forward reaction.


Why rate of a reaction increase when temperature increases?

Energy as well as the speed of molecules increases when reactants are heated. That is as temperature increases the number of molecules with three hold energy increases. As a result, the number of effective collisions increases and these rate of reaction also increases.


Write an experiment to prove that concentration of reactants affect the rate of reactions.

Materials required: Mg , dil. HCl , Con, HCl and test tubes.
Procedure: Take magnesium ribbons of equal mass in two test tubes. Add concentrated HCl to one test tube and dilute HCl to the other in equal volume.


Complete the table writing the effect of change in concentration in the system at equilibrium.
ActionChange of concentrationChange in rate
$$\bullet$$ More hydrogen is added$$\bullet$$ Increase the concentration of reactant.$$\bullet$$ Rate the forward reaction increases.
$$\bullet$$ More ammonia is added$$\bullet$$ Increase the concentration of product.$$\bullet$$ ..............
$$\bullet$$ Ammonia is removed$$\bullet$$ Decrease the concentration of product.$$\bullet$$ ..............
$$\bullet$$ More nitrogen is added$$\bullet$$ Increase the concentration of reactant.$$\bullet$$ ................

ActionChange of concentration Change in rate
$$\bullet$$ More hydrogen is added    $$\bullet$$ Increase the concentration of reactant.$$\bullet$$ Rate of forward reaction increases.
$$\bullet$$ More ammonia is added$$\bullet$$ Increase the concentration of product. $$\bullet$$ Rate of Backward reaction increases.
$$\bullet$$ Ammonia is removed$$\bullet$$ Decrease the concentration of product.   $$\bullet$$ Rate of forward reaction increases.
$$\bullet$$ More nitrogen is added$$\bullet$$ Increase the concentration of reactant.$$\bullet$$ Rate of forward reaction increases.


Write your observation
Test tube 1 : ............
Test tube 2 : ............

1838603_ca466037d62c4d5281b060969b2e34a8.png

Test tube 1: Reaction is faster in con. HCl
Test tube 2: Reaction rate is slow


Mention the factor that affect the rate of a chemical reaction . 

The factor that affect the rate of  a chemical reaction are -

Concentration : 
On increasing concentration of reactants, the probability of collisions of molecules increases hence rate of reaction increases.

Temperature : 
 On increasing the temperature , the kinetic energy of molecules increases , hence, the number of collision increases . Therefore , the rate of reaction also increases .

Pressure 
On increasing pressure , the molecules of gases come closer to each other. As a result their collisions increases and hence rate of reaction increases.
 
Surface Area of Reactants:
On increasing surface area reactants, the rate of reaction increases . For example , the powdered  metals reacts faster than the metals in a lump .

Nature of reactants 
If the reactants are ionic in nature than the rate of reaction is faster than those in which reactants are molecular in nature .



A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactants is doubled?

Rate = k $$ [A] ^{2}$$ if concentration = a then , 

 Rate = k $$ (a)^{2} $$ 

 If concentration is double then , 

[A] = 2 a 

 $$(rate )_{1} = k (a) ^{2} $$ 

$$(rate )_{2} = k (2a)^{2}$$ 

$$\dfrac{(rate )_{1} }{(rate )_{2} }=\dfrac{k (a)^{2}}{k (2a)^{2}}$$

 $$\dfrac{(rate )_{2} }{(rate )_{1} }=4$$ 

$$(rate )_{2}=4\times (rate )_{1}$$

If concentration of reactants doubled the rate increases by 4 times


Is there any difference in the rate of reaction in the two beakers ?
1838606_417d59e5cb534e8390614c27687b60d2.png

Reaction rate is greater when powered marble is used.


Write an experiment to prove the relation between temperature and rate of reaction.
In which of the following tubes is the precipitate formed faster ?
What is the color of the precipitate formed ?

1. Materials required: Sodium thiosulphate , hydrochloric acid, water , boiling tube, spirit lamp. Procedure: Prepare dilute solution of sodium thiosulphate in a beaker. Take equal volumes of the solution in two boiling tubes. Heat one boiling tube for some time. Add dilute hydrochloric acid in equal amounts in both the boiling tubes.
Observations: Reaction is faster in heated test tube.
2. In heated test tubes.
3. Pale yellow


Some materials available in the laboratory are given below. Magnesium ribbon, marble powder , marble pieces , dilute HCl , concentrated HCl.
Write the balanced chemical equation of the reaction.

$$ CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 $$ 


A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactants is reduced to $$\dfrac{1}{2}$$?

Rate = k $$ [A] ^{2}$$ if concentration = a then , 

 Rate = k $$ (a)^{2} $$ 

If concentration is reduced to half 

[A] $$=\dfrac{a}{2}$$ 

$$\dfrac{(rate)_{1}}{(rate)_{2}}=\dfrac{k(a)^{2}}{k\left ( \dfrac{a}{2} \right )^{2}}$$ 

$$\dfrac{(rate)_{1}}{(rate)_{2}}=\dfrac{4}{1} $$ 

$$\dfrac{(rate)_{2}}{(rate)_{1}}=\dfrac{1}{4} $$ 

$$ (rate )_{2}=\dfrac{1}{4}\times (rate)_{1}$$


Some apparatus and chemical are given.
Zn , Mg , dilute HCl, $$ CaCO_3$$ , test tube , water.
(a) Design an experiment to prove that the nature of reactants can influence the rate of reaction.
(b) Write the equations for the chemical reactions.
(c) Write the expression for the rate of the reaction.

(a) Take a volume of dil $$HCl$$ in two test tubes. Add a piece of $$Mg$$ m to one of them . Add a piece of $$Zn$$ of equal mass to the other.
Observation : $$Mg$$ reacts faster with dil. $$HCl$$ than $$Zn$$. This is due to the difference in the chemical nature of $$Mg$$ and $$Zn$$.
(b) $$ Mg + 2HCl \rightarrow MgCl_2 + H_2 $$ 
$$ Zn + 2HCl \rightarrow ZnCl_2 + H_2 $$ 

(c) Rate of a chemical reaction 
$$ = \dfrac{{\text{amount of reactants used up}}}{{\text{time}}} $$ 

$$ = \dfrac{{\text{amount of product formed}}}{{\text{time}}} $$


Sulfur pieces do not react with cold concentrated nitric acid. But sulfur powder reacts. 
Explain the reason why the rate of chemical reaction is increased here ?

When surface area increases rate of effective collision increases. So the rate of reaction increases.


A reaction is first order with respect to A and second order with respect B.

How is the rate affected when the concentration of both A and B is doubled.

$$ \left ( rate \right ) = k [A][B]^{2}$$

$$\left ( rate \right )_{1} = k\times a\times b^{2}$$

$$\left ( rate \right )_{2} = k\times (2a) \times (2b)^{2}$$

$$ \dfrac{(rate)_{2}}{(rate)_{1}} = \dfrac{k\times 2a\times 4 \times b^{2}}{k\times a\times b^{2}}$$

$$\dfrac{(rate)_{2}}{(rate)_{1}} = 8 $$

$$(rate)2 = 8 (rate)_{1}$$

The rate becomes eight times.


Sulfur pieces do not react with cold concentrated nitric acid. But sulfur powder reacts. 
Suppose you want to increase the rate of reaction again. Which way you would choose ? Give reason.

Heat the mixture.
When temperature increases more molecules will get threshold energy. So rate of effective collision and rate of reaction increases.


A reaction is first order with respect to A and second order with respect B.
How is the rate affected when the concentration of B is tripled ?

Let [A] = a , [B] = b

If [B] is increase three times
[B] = 3b

then,

Rate $$ = k[A] [B]^{2}$$

$$rate_{1} = k\times a \times b^{2} ......(1)$$

$$rate_{2} = k\times a\times (3b)^{2} ......(2)$$

$$from\  eq (1)and(2)$$

$$ \dfrac{(rate)_{2}}{(rate)_{1}} = \dfrac{k\times a\times (3b)^{2}}{k\times a\times b^{2}}$$

$$ (rate)_{2} = 9 \times (rate_{1})$$

The rate becomes nine times 


How will the rate of the reaction be affected when
(a) Surface area of the reactant is reduced,
(b) Catalyst is added in a reversible reaction, and
(c) Temperature of the reaction is increased?

1. Rate of reaction increases
2. Rate of forward reaction generally increases
3. Rate of reaction increases


Class 12 Medical Chemistry Extra Questions