NCERT Solutions for Class 12 Science Chemistry Chapter 4 - Chemical Kinetics

Answer:

The role of catalyst is to change the activation energy of the reaction. It lowers the activation energy and hence, increases the rate of the reaction. 

Hence, the correct answer is option (iii). 

Answer:

In the presence of a catalyst, heat evolved or absorbed during the reaction remains unchanged as a catalyst only increases the rate at which the reaction takes place and does not participate in the reaction. 

Hence, the correct answer is option (iii). 

Answer:

Activation energy can be calculated using the Arrhenius equation.

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303\mathrm{R}}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ 

E_a = Activation energy
k₂ = Rate constant at temperature T₂
k₁ = Rate constant at temperature T₁
R = Universal gas constant 

Therefore, activation energy can be determined by determining the rate constants at two temperatures.

Hence, the correct answer is option (ii).


 

Answer:

Activation energy of forward reaction = E₁ + E₂
Since the energy of reactants is less than products, the product is less stable than the reactant.

Hence, the correct answer is option (i).

Answer:

Consider a reaction,

$$\begin{gathered} A_{\left(g\right)}\to B_{\left(g\right)}+C_{\left(g\right)} \\ \mathrm{Initial} \mathrm{pressure}: P_{i }0 0 \\ \mathrm{Pressure} \mathrm{after}: P_i-x x x \\ \mathrm{time} \mathrm{t} \\ {P}_t= \mathrm{Total} \mathrm{pressure} \mathrm{at} \mathrm{time} \mathrm{t} \\ {P}_t= (P_i-x)+x+x= P_i+x atm \\ x= P_t-P_i \\ \mathrm{Pressure} \mathrm{of} \mathrm{A} \mathrm{after} \mathrm{time} \mathrm{t} = P_i-x= P_i-P_t+P_i= 2P_i-P_t \\ K=\frac{2.303}{t}\log \frac{A_0}{A} \\ =\frac{2.303}{t}\log \frac{P_i}{2P_i-P_t} \end{gathered}$$

Hence, the correct answer is option (ii). 

Answer:

According to Arrhenius equation, 
$$\begin{gathered} k=A{e}^{\raisebox{1ex}{$-E_a$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.} \\ \mathrm{Taking} \ln \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get}, \\ \ln k=\ln A-\frac{E_a}{RT} \\ \ln k= -\frac{E_a}{R}\times \frac{1}{T}+\ln A....\left(1\right) \\ \mathrm{Equation} \left(1\right) \mathrm{relates} \mathrm{to} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{straight} \mathrm{line} \mathrm{i}.\mathrm{e}. y=mx+\mathrm{c} \\ \mathrm{Therefore}, \mathrm{a} \mathrm{plot} \mathrm{between} \ln k \mathrm{and} \mathrm{I}/T \mathrm{is} \mathrm{a} \mathrm{straight} \mathrm{line} \mathrm{with} \mathrm{a} \mathrm{negative} \mathrm{slope} \raisebox{1ex}{$-E_a$}\!\left/ \!\raisebox{-1ex}{$R$}\right. \mathrm{and} \mathrm{intercept} \ln A \end{gathered}$$


Hence, the correct answer is option (i).

Answer:

According to the Arrhenius equation, $k=\mathrm{A} {\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}$, the value of rate constant, i.e., k increases exponentially with decrease in activation energy, E_a and increase in temperature, T. 
$$\begin{gathered} \mathrm{As} E_a \mathrm{decreases}, \frac{-E_a}{\mathrm{R}T} \mathrm{increases} \mathrm{and} \mathrm{hence}, k \mathrm{increases} \\ \mathrm{As} T \mathrm{increases}, \frac{-E_a}{\mathrm{R}T} \mathrm{increases} \mathrm{and} \mathrm{hence}, k \mathrm{increases} \end{gathered}$$

Hence, the correct answer is option (iv). 

Answer:

According to the graph of volume of hydrogen released vs time,

Average rate upto 40 seconds= $\frac{V_3-0}{40-0}= \frac{V_3}{40}$

Hence, the correct answer is option (iii).

Answer:

Order of reaction is not always equal to sum of the stoichiometric coefficients of reactants in the balanced chemical equation. For a reaction it may or may not be equal to sum of stoichiometric coefficient of reactants.

Hence, the correct answer is option (iii). 

Answer:

Instantaneous rate is defined as the rate of the reaction at a particular instant of time. 
According to the graph, $\frac{V_4-V_2}{50-30}$ does not show instantaneous rate of reaction at 40^th second.  

Hence, the correct answer is option (ii). 
 

Answer:

The rate of a reaction depends on the concentration of reactant. As the concentration of reactant decreases, rate of the reaction also decreases.

Hence, the correct answer is option (i).
 

Answer:

$5{\mathrm{Br}}^{–}\left(\mathrm{aq}\right) + {\mathrm{BrO}}_3^{-}\left(\mathrm{aq}\right) + 6{\mathrm{H}}^{+}\left(\mathrm{aq}\right) \to 3{\mathrm{Br}}_2\left(\mathrm{aq}\right) + 3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
Rate law expression for the given reaction can be written as:
$\frac{-1}{5}\frac{∆\left[B{r}^{-}\right]}{∆t}= -\frac{∆\left[Br{O_3}^{-}\right]}{∆t}= \frac{-1∆\left[H^{+}\right]}{6∆t}= \frac{1∆\left[B{r}_2\right]}{3∆t}$

$\frac{∆\left[{\mathrm{Br}}^{-}\right]}{∆\mathrm{t}}=\frac{5}{6}\frac{∆\left[{\mathrm{H}}^{+}\right]}{∆\mathrm{t}}$

Hence, the correct answer is option (iii).

Answer:

For an exothermic reaction, activation energy of product should be greater than the activation energy of reactant. Therefore, (a) represents exothermic reaction.

Hence, the correct answer is option (i).

Answer:

Rate constant of a reaction depends only on temperature and not on the concentration of reactants. Therefore, when the concentration of B is doubled keeping the concentration of A constant, value of rate constant remains the same.

Hence, the correct answer is option (i).
 

Answer:

 According to the collision theory, conditions for any reaction to occur are:
a) Molecule should collide with sufficient threshold energy.
b) Their orientation must be proper.
c) The collision must be effective.

Hence, the correct answer is option (iii). 

Answer:

A first order reaction never goes for completion i.e. the reaction is never 100% complete. This can happen only at an infinite time which cannot be calculated.

Hence, the correct answer is option (iv).

 

Answer:

$$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{experiment}= \mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{x}}{\left[\mathrm{B}\right]}^{\mathrm{y}} \mathrm{where} \mathrm{x}= \mathrm{order} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{A} \\ \mathrm{y}= \mathrm{order} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{B} \\ {\mathrm{R}}_1=\mathrm{k}{(0.30)}^{\mathrm{x}}{(0.30)}^{\mathrm{y}} \\ 0.10=\mathrm{k}{(0.30)}^{\mathrm{x}}{(0.30)}^{\mathrm{y}}.....\left(1\right) \\ {\mathrm{R}}_2=\mathrm{k}{(0.30)}^{\mathrm{x}}{(0.60)}^{\mathrm{y}} \\ 0.40=\mathrm{k}{(0.30)}^{\mathrm{x}}{(0.60)}^{\mathrm{y}}..... \left(2\right) \\ {\mathrm{R}}_3=\mathrm{k}{(0.60)}^{\mathrm{x}}{(0.30)}^{\mathrm{y}} \\ 0.20=\mathrm{k}{(0.60)}^{\mathrm{x}}{(0.30)}^{\mathrm{y}}.....\left(3\right) \\ \mathrm{Dividing} \left(1\right) \mathrm{by} \left(2\right) \\ \frac{0.10=\mathrm{k}{(0.30)}^{\mathrm{x}}{(0.30)}^{\mathrm{y}}}{0.40=\mathrm{k}{(0.30)}^{\mathrm{x}}{(0.60)}^{\mathrm{y}}} \\ \mathrm{On} \mathrm{solving}, \mathrm{we} \mathrm{get}, \\ \mathrm{y}=2 \mathrm{i}.\mathrm{e}., \mathrm{order} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{B} \mathrm{is} 2 \\ \mathrm{Similarly}, \mathrm{on} \mathrm{dividing} \left(1\right) \mathrm{and} \left(3\right), \mathrm{we} \mathrm{get}, \\ \frac{0.10=\mathrm{k}{(0.30)}^{\mathrm{x}}{(0.30)}^{\mathrm{y}}}{0.20=\mathrm{k}{(0.60)}^{\mathrm{x}}{(0.30)}^{\mathrm{y}}} \\ \mathrm{On} \mathrm{solving}, \mathrm{we} \mathrm{get}, \\ \mathrm{x}=1 \mathrm{i}.\mathrm{e}., \mathrm{order} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{A} \mathrm{is} 1 \\ \mathrm{Therefore}, \mathrm{Rate}=\mathrm{R}= \mathrm{k}{\left[\mathrm{A}\right]}^1{\left[\mathrm{B}\right]}^2 \end{gathered}$$

Hence, the correct answer is option (ii).
 

Answer:

A catalyst can only alter the rate of the reaction. Gibbs free energy ($∆G$)  depends on the concentration of reactants and products and therefore, a catalyst cannot alter $∆G$ of a reaction.

Hence, the correct answer is option (ii). 

Answer:

In a pseudo first order reaction, one of the reactants is present in excess and its concentration does not get altered much during the course of the reaction. Therefore, the rate of reaction is affected by concentration of reactants present in small amount only. For example, during the hydrolysis of ethyl acetate, water is taken in large excess.
Rate = 
The concentration of water does not change much during the course of the reaction. Thus, the term [H₂O] can be taken to be constant.
 Thus, the value of rate constant depends on the concentration of reactants present in excess. 

Hence, the correct answer is option (ii).

Answer:

As the reaction proceeds, concentration of reactant decreases and the concentration of product increases exponentially with time. It can be depicted as follows:

Hence, the correct answer is option (ii). 

Answer:

Rate law can be determined from a balanced chemical equation only if it is an elementary reaction. 

Hence, the correct answers are options (i), (iii) and (iv).
 

Answer:

For an elementary reaction, the order and molecularity are same. Molecularity of a reaction can never be zero.

Hence, the correct answers are options (i) and (iv).
 

Answer:

 For any unimolecular reaction, only one reactant is involved in rate determining step and the order and molecularity of rate determining or slowest step are one.

Hence, the correct answers are options (i) and (ii). 
 

Answer:

For a complex reaction, order of overall reaction is same as the molecularity of slowest step.
As rate of overall reaction depends upon total number of molecules involved in slowest step of the reaction i.e. the rate determining step. Hence, molecularity of the slowest step is equal to order of overall reaction.
Since the completion of any chemical reaction is not possible in the absence of reactants. Hence, slowest step of any chemical reaction must
contain at least one reactant. Thus, molecularity of the slowest step is never zero or non-integer.

Hence, the correct answers are options (i) and (iv).

Answer:

Since the given reaction is a zero order reaction, the rate of the reaction is independent of the concentration of ammonia. Therefore, the rate law can be expressed as follows:
Rate of reaction = Rate constant [NH₃]⁰ = Rate constant
At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration.

Hence, the correct answers are options (i) and (iii).
 

Answer:

Activated complex is formed at the highest energy level of the system. It is unstable and hence, decomposes to give products with release of energy. Reactants may be formed if the reaction is reversible.

Hence, the correct answers are options (i) and (iv).
 

Answer:

According to the Maxwell Boltzmann distribution of energy, as the temperature increases:
a) Most probable kinetic energy increases
b) Fraction of molecules possessing most probable kinetic energy decreases. 

Hence, the correct answers are options (i) and (iii). 

Answer:

Area under the curve remains the same with increase in temperature since total probability must be one at all times. And as the temperature increases, curve broadens and shifts to the right side such that there is a greater proportion of molecules with much higher energies.


Hence, the correct answers are options (i) and (iv). 

Answer:

For every 10^o rise in temperature, rate of the reaction almost doubles, i.e., rate of the reaction increases with the increase in temperature.
According to the Arrhenius equation (​), rate of the reaction increases with a decrease in the activation energy.

Hence, the correct answers are options (i) and (ii). 

Answer:

When a catalyst is added to a reaction medium, it increases the rate of the reaction by decreasing the activation energy of the molecule. Therefore, it follows an alternative pathway. It does not alter enthalpy change of the reaction.

Hence, the correct answers are options (ii) and (iv).


 

Answer:

For a zero order reaction,
$$\begin{gathered} \left[\mathrm{R}\right]={\left[\mathrm{R}\right]}_0-kt....\left(1\right) \\ \left[\mathrm{R}\right]= \mathrm{concentration} \mathrm{of} \mathrm{the} \mathrm{reactant} \mathrm{at} \mathrm{time} t \\ {\left[\mathrm{R}\right]}_0= \mathrm{concentration} \mathrm{of} \mathrm{the} \mathrm{reactant} \mathrm{at} \mathrm{time} t=0 \\ k= \mathrm{rate} \mathrm{constant} \\ \mathrm{Comparing} \mathrm{equation} \left(1\right) \mathrm{with} \mathrm{the} \mathrm{straight} \mathrm{line} \mathrm{equation}, \mathrm{i}.\mathrm{e}., y\mathit{=}mx+\mathrm{c} \end{gathered}$$
If we plot [R] against t, we get a straight line with slope = –k and intercept equal to [R]₀.

Also for a zero order reaction,
Rate = k [R]⁰ = k
This means that for a zero order reaction, rate of the reaction is independent of the concentration of reactant and remains constant with time till reactant is present. Therefore, graph is a straight line parallel to x axis.

Hence, the correct answers are options (i) and (iv).

Answer:

$$\begin{gathered} \mathrm{For} \mathrm{a} \mathrm{first} \mathrm{order} \mathrm{reaction}, \\ {t}_{1/2}=\frac{0.693}{k}, \mathrm{Rate} =k\left[\mathrm{R}\right] \\ \left[\mathrm{R}\right]= \mathrm{concentration} \mathrm{of} \mathrm{reactant} \\ k= \mathrm{rate} \mathrm{constant} \\ {t}_{1/2}= \mathrm{half} \mathrm{life} \mathrm{of} \mathrm{the} \mathrm{reaction} \\ \mathrm{Since} t_{1/2} \mathrm{is} \mathrm{independent} \mathrm{of} \mathrm{the} \mathrm{concentration} \mathrm{of} \mathrm{reactant}, \mathrm{we} \mathrm{get} \mathrm{a} \mathrm{straight} \mathrm{line} \mathrm{parallel} \mathrm{to} x \mathrm{axis}. \\ \mathrm{Also}, \mathrm{for} \mathrm{a} \mathrm{first} \mathrm{order} \mathrm{reaction} \\ \log \frac{\left[{\mathrm{R}}_0\right]}{\left[\mathrm{R}\right]}=\frac{kt}{2.303} \\ \mathrm{Curve} \mathrm{between} \log \frac{\left[{\mathrm{R}}_0\right]}{\left[\mathrm{R}\right]} \mathrm{versus} t, \mathrm{we} \mathrm{get} \mathrm{a} \mathrm{straight} \mathrm{line} \mathrm{with} \mathrm{slope} k/2.303 \end{gathered}$$


Hence, the correct answers are options (i) and (iv).

Answer:

Consider a biomolecular reaction:
$A + B \to Product$
Rate = k[A] [B]
When concentration of [B] is taken in large excess, rate law will become Rate = k’ [A]
where, k’ = k [B] 
Hence, the order of reaction will be equal to one.

Answer:

For the reaction, 2A + B → C
For a zero order reaction, rate law equation can be written as, R = k[A]⁰[B]⁰ = k

Answer:

The rate law for the reaction can be determined experimentally. Initially, the concentration of one of the reactants (e.g. O₂) is taken in large excess and the reaction rate is determined with respect to NO(g). Then the other reactant, i.e., NO is taken in large excess and the reaction rate is determined with respect to O₂ (g).
The rate law expression for the reaction may be expressed as Rate = k[NO]^a[O₂]^b
Where a and b represent the order of the reaction with respect to NO and O₂ respectively. 

Answer:

An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state. For elementary reactions, order and molecularity have the same value. 

Answer:

Rate of the reaction can be expressed as follows:
R = k[A]^x .....(1)
where x represents the order of the reaction
According to question,
27R = k[3A]^x ....(2)
Dividing equation (1) and (2), we get,
$$\begin{gathered} \frac{\mathrm{R}}{27\mathrm{R}}={\left(\frac{\mathrm{A}}{3\mathrm{A}}\right)}^x \\ \mathrm{On} \mathrm{solving}, \mathrm{we} \mathrm{get}, x=3 \\ \mathrm{Hence}, \mathrm{order} \mathrm{of} \mathrm{the} \mathrm{reaction} \mathrm{is} 3. \end{gathered}$$

 

Answer:

For a zero order reaction,
[A] = [A]₀ − kt
A = concentration of reactant at time t
A₀ = concentration of reactant at time t = 0
k = rate constant
At the completion of the reaction,  A becomes zero or A = 0
∴ 0 = [A]₀ − kt
[A]₀ = kt
$t=\frac{{\mathrm{A}}_0}{k}$
Hence, the time required for completion of a zero order reaction is given by $t=\frac{{\mathrm{A}}_0}{k}$.

Answer:

Rate = k [A][B]^3/2
Overall order of the reaction = $1+\frac{3}{2}=\frac{5}{2}$
Order of an elementary reaction cannot be fractional.

Hence, the given reaction cannot be an elementary reaction. 

Answer:

Though the reacting molecules may be having energy more than threshold energy, yet they may not be effective due to lack of proper orientation. The rate of reaction is very slow due to lack of proper orientation and hence, less effective collisions between molecules. 
 

Answer:

No, the molecularity can never be zero or a fractional number as it shows the number of reactants taking part in a reaction , which can never be zero.

Answer:

(i) For a zero order reaction, [A] = [A]₀ − kt
This equation can be compared to the equation of straight line, i.e., y = mx + c with a negative slope.
Therefore, it is a zero order reaction
(ii) For a zero order reaction, [A] = [A]₀ − kt
      Comparing this equation to the equation of straight line, i.e., y = mx + c, we get,
      Slope = m = −k 
(iii) Rate = k[A]⁰ = k
      Unit of rate = mol L⁻¹ s⁻¹ = unit of k
      Hence, the unit of rate constant, k is mol L⁻¹ s⁻¹.

Answer:

At room temperature, the activation energy is very high and it becomes difficult for the H₂ (g) and O₂ (g) bonds to break and overcome the energy barrier for the formation of water. Therefore, the reaction between H₂ (g) and O₂ (g) does not lead to the formation of water at room temperature. 

Answer:

On increasing temperature, a larger fraction of colliding particles can cross the energy barrier, i.e., the activation energy, which leads to an increase in rate of the reaction.

Answer:

Combustion of fuel in presence of oxygen is a chemical reaction and every chemical reaction takes place only after reacting molecules cross the threshold energy value, i.e., activation energy. The activation energy for combustion reactions of fuels is very high at room temperature, therefore, they do not bum by themselves.

Answer:

Molecularity is defined as number of molecules that come together to react in an elementary reaction and is equal to the sum of stoichiometric coefficients of reactants in this elementary reaction. The probability of more than three molecules colliding simultaneously is very small. Therefore, the possibility of molecularity being three is very low.

Answer:

The rate of a reaction depends on the concentration of reactants. As the reaction proceeds, the concentration of reactants decreases because the reactant converts into products. Hence, the rate of the reaction decreases.

Answer:

Thermodynamics feasibility of a reaction depends on Gibbs free energy change, i.e., ∆G must be negative for spontaneous process. Kinetic feasibility depends on the activation energy of reaction, the lesser is the activation energy, the greater is the feasibility of reaction.
For example, Diamond → Graphite, ∆G = −ve
This process is thermodynamically feasible but it is very slow due to its high activation energy.

Answer:

The reaction between KMnO₄ and oxalic acid is very slow. Heating leads to an increase in temperature and as the temperature increases, the rate of reaction also increases.

Answer:

Molecularity is the number of molecules that come together to react in an elementary reaction and is equal to the sum of stoichiometric coefficients of reactants in this elementary reaction. Therefore, it can never be equal to zero. 

Answer:

A complex reaction comprises of several elementary reactions. Numbers of molecules involved in each elementary reaction may be different i.e., the molecularity of each step may be different. Therefore, the molecularity of overall complex reaction cannot be determined. Order of a complex reaction is experimentally determined by the slowest step in its mechanism i.e., the rate determining step and therefore, it is applicable even in the case of complex reactions.

Answer:

Balanced chemical equations of complex reactions which comprises of various elementary steps may lead to incorrect order or rate law. The order of such reaction is determined by the slowest step in the reaction mechanism. Order is determined experimentally and gives the actual dependence of observed rate of reaction on the concentration of reactants.

Answer:

The correct match between column I and column II is:

(i)$\to$(a)
(ii)$\to$ (b)
(iii)$\to$(b)
(iv)$\to$(a)

Answer:

The correct match between column I and column II is as follows:

(i)(c)
(ii)(a)
(iii)$\to$(d)
(iv)$\to$(f)
(v)$\to$(b)
(vi)$\to$(e)
 

Answer:

The correct match between column I and column II is as follows:

(i)$\to$(b)
(ii)$\to$(a)
(iii)$\to$(c)
 

Answer:

The correct match between column I and column II is as follows:

(i)$\to$(b)
(ii)$\to$(a)
(iii)$\to$(d)
(iv)$\to$(c)

Answer:

Order of a reaction may be zero or fractional. Order of  a reaction cannot be determined from a balanced chemical equation and is determined experimentally by the rate law expression. Therefore, both assertion and reason are correct but reason does not explain assertion.

Hence, the correct answer is option (ii).
 

Answer:

Order and molecularity may not be necessarily same. Order is determined experimentally but molecularity is calculated using stoichiometric coefficient of rate determining elementary step. Therefore, assertion is incorrect but reason is correct.

Hence, the correct answer is option (v).
 

Answer:

Enthalpy of a reaction i.e., $∆H$ = Activation Energy of forward reaction – Activation Energy of backward reaction.
Catalyst does not alter heat of reaction because it affects activation energy of forward and backward reactions equally.
Therefore, both assertion and reason are correct and the reason is correct explanation of assertion.

Hence, the correct answer is option (i).
 

Answer:

All collisions between reactant molecules does not lead to the formation of product. Only effective collisions due to proper orientation and sufficient kinetic energy lead to compound formation. Therefore, assertion is incorrect but reason is correct.

Hence, the correct answer is option (v).
 

Answer:

Arrhenius equation () is almost accurate for single as well as complex reaction. Also, for the reactant molecules to undergo a chemical change, orientation plays a very important role. Therefore, assertion is correct but reason is incorrect.

Hence, the correct answer is option (iii).

 

Answer:

Only effective collision leads to the formation of products. The collisions in which molecules collide with sufficient kinetic energy and proper orientation leads to a chemical change because it facilitates the breaking of old bonds between reactant molecules and formation of the new bonds, i.e., in products.
For e.g., formation of methanol from bromomethane depends upon the orientation of the reactant molecules.

The proper orientation of reactant molecules leads to the formation of products. Improper orientation of reactant molecules gives no product.
To account for effective collisions, another factor P, called the probability or steric factor is introduced. It takes into account the fact that in a collision, molecules must be properly oriented i.e.,  Rate =
Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction.



 

Answer:

All the molecules in the reacting species do not have the same kinetic energy. The distribution of kinetic energy may be described by plotting the fraction of molecules with a given kinetic energy vs kinetic energy. The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximum fraction of molecules. There are decreasing number of molecules with energies higher or lower than this value.

When the temperature is raised, the maxima of the curve moves towards the higher energy value and the curve broadens. 

This implies that there is a greater proportion of the molecules with higher energies. Most probable kinetic energy increases with increase in temperature.

In the Arrhenius equation, the factor corresponds to the fraction of molecules that have kinetic energy greater than E_a. As the temperature increases, the fraction of molecules, which collide with energies greater than E_a increases. This leads to a decreases in activation energy which will result in the increase in the rate of reaction.

Answer:

A catalyst is a substance which increases the speed of a reaction without itself undergoing any chemical change.
According to the activated complex theory, reactants first combine with the catalyst to form an intermediate complex or an activated complex which is short-lived and hence, decomposes to form the products and the catalyst is regenerated.

The activated complex formed in the presence of catalyst is much lower as compared to the complex formed in the absence of catalyst. 
Thus, the presence of a catalyst lower the potential energy barrier and the reaction follows a new alternate pathway which requires less activation energy.
Lower the activation energy, faster is the reaction because more reactant molecules can cross the energy barrier and convert into products.
Enthalpy of reaction, i.e., difference in energy between reactants and product is constant, which is clear from potential energy diagram drawn below.

Answer:

Average Rate

• Average rate of a reaction is defined as the change in concentration of reactants or products in unit time taken.
• It can be expressed in terms of the rate of decrease in concentration of any one of the reactants, or the rate of increase in concentration of any one of the products.
• Average rate = $\frac{-∆R}{∆t}=\frac{+∆P}{∆t}$
• Average rate is calculated for a longer interval of time. 

• When the average rate is considered at the smallest interval of time i.e., dt, then, instantaneous rate is obtained. 
• For an infinitesimally small interval of time dt, Instantaneous rate = $\frac{-dR}{dt}=\frac{+dP}{dt}$

Answer:

A reaction in which one reactant is present in large excess and its concentration does not get altered during the course of the reaction, it behaves as a first order reaction. Such reaction is known as pseudo first order reaction.
For example, hydrolysis of ethyl acetate is a second order reaction in reality and concentration of both ethyl acetate and water affect the rate of the reaction. 

Rate = 
But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction. Thus, the rate of reaction is affected by concentration of ethyl acetate only. Thus, the term [H₂O] can be taken to be constant.