4 Marks Questions

Question 14 Marks

Derive integrated rate equation for first order reaction and also explain how than rate constant can be determine with help of graph.

Answer

$\rightarrow$ "First order reaction means that the rate reaction is proportional to the first power the concentration of the reactant $R.$ "
$\rightarrow$ For example, consider the following reaction
$R \rightarrow P$
$\rightarrow$ Rate of reaction for this reaction can be expressed as
Rate $=-\frac{ d [ R ]}{ dt }= k [ R ]$
Or $\frac{ d [ R ]}{[ R ]}=- kdt$
\rightarrow Integrating this equation, we get
$\ln [R]=-k t+I .. ..Eq. (1)$
where, $I$ is the constant of integration and its value can be determined easily.
$\rightarrow$ When $t =0, R =[ R ]_0$, where $[ R ]_0$ is the initia concentration of the reactant.
$\rightarrow$ Therefore, equation $(1)$ can be written as
$\ln [R]_0=- k \times 0+ I$
$\ln [ R ]_0= I$
$\rightarrow$ Substituting the value of $I$ in equation $(1)$
$\ln [ R ]=- kt +\ln [ R ]_0 \ldots \ldots \text { Eq. (2) }$
$\rightarrow $ Rearranging this equation
$\ln \frac{[ R ]}{[ R ]_0}=- kt$
or $k =\frac{1}{ t } \ln \frac{[ R ]_0}{[ R ]} ....Eq. (3)$
$\rightarrow$ At time $t_1$ from equation $(1)$
$\ln [R]_1=-k t_1+\ln [R]_0 \ldots \ldots$ Eq. $(4)$
At time $t _2$
$\ln [R]_2=- kt _2+\ln [ R ]_0 \ldots \ldots$ Eq. $(5)$
where, $[ R ]_1$ and $[ R ]_2$ are the concentration the reactants at time $t_1$ and $t_2$ respectively.
$\rightarrow \rightarrow \rightarrow$ Subtracting Eq. $(5)$ from $(4)$
$\ln [ R ]_1-\ln [ R ]_2=- kt _1-\left(- kt _2\right)$
$\ln \frac{[ R ]_1}{[ R ]_2}= k \left( t _2- t _1\right)$
$k =\frac{1}{\left( t _2- t _1\right)} \ln \frac{[ R ]_1}{[ R ]_2} \ldots \ldots \text { Eq. (6) }$
$\rightarrow$ So, equation $(2)$ can also be written as
$\ln \frac{[ R ]}{[ R ]_0}=- kt$
Taking antilog of both the side
$[ R ]=[ R ]_0 e ^{- kt }.. ...$ Eq. $(7)$
$\rightarrow$ Comparing equation $(2)$ with $y = mx + c,$ if we plot $\ln [R]$ against $t, $ we get a straight line with slope $=- k$ and intercept equal to $\ln [ R ]_0$

$\rightarrow $ The first order rate equation $(3)$ can also be written in the form
$k =\frac{2.303}{ t } \log \frac{[ R ]_0}{[ R ]} \ldots \ldots \text { Eq. (8) }$
or $\log \frac{[ R ]_0}{[ R ]}=\frac{ kt }{2.303}$
$\rightarrow$ If we plot a graph between $\log \frac{[R]_0}{[R]}$ vs $t,$ the slope $=\frac{ k }{2.303}$

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Question 24 Marks

Explain how the molecularity of a complex reaction determined? And give the points showing difference between molecularity of a reaction and order of a reaction.

Answer

→ Term molecularity is only applicable to elementary reactions, even though to determine molecularity of a complex reaction it is necessary to know the mechanism of that reaction.
→ Through mechanism of a reaction we can have information about slowest step of reaction.
→ The overall rate of the reaction is controlled by the slowest step in a reaction called the rate determining step.
→ Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium $2 H _2 O _2 \xrightarrow[\text { Alkaline medium }]{ I ^{-}} 2 H _2 O + O _2$
→ The rate equation for this reaction is found to be Rate $=\frac{- d \left[ H _2 O _2\right]}{ dt }= k \left[ H _2 O _2\right]\left[ r ^{-}\right]$
→ This reaction is first order with respect to both $H _2 O _2$ and $I ^{-}$.
→ Mechanism involve two steps.
(1) $H _2 O _2+ I ^{-} \xrightarrow{\text { Slow Step }} H _2 O + IO ^{-}$
(2) $H _2 O _2+ IO ^{-} \longrightarrow H _2 O + I ^{-}+ O _2$
→ The first step, being slow, is the rate determining step.
→ Thus, the rate of formation of intermediate will determine the rate of this reaction.
→ Thus, it can be said that, for complex reaction, order is given by the slowest step and molecularity of the slowest step is same as the order of the overall reaction.
Difference between molecularity and order of a reaction:
(i) Order of a reaction is an experimental quantity. It can be zero and even a fraction but molecularity cannot be zero or a non integer.
(ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions. For complex reaction molecularity has no meaning.
(iii) For complex reaction, order is given by the slowest step and generally molecularity of the slowest step is same as the order of the overall reaction.

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Question 34 Marks

Write a note on collision theory of chemical reaction. ### What is meant by collision theory? Explain.

Answer

→ Collision theory, which was developed by Max Trautz and William Lewis in 1916-18, provides greater insight into the energetic and mechanistic aspects of reaction.
→ It is based on kinetic theory of gases.
→ According to this theory, the reactant molecules are assumed to be hard spheres and reaction is postulated to occur when molecules collide with each other.
→ "The number of collision per second per unit volume of the reaction mixture is known as collision frequency (Z)."
→ Another factor which affects the rate of chemical reaction is activation energy.
→ For a bimolecular elementary reaction
→ A+B → Products
→ Rate of reaction can be expressed as
→ Rate $=Z_{A B} e^{-\frac{E_a}{R T}} \quad \ldots$ Eq. (1)
→ where $Z_{A B}$ represents the collision frequency of reactants $A$ and $B$, and $e^{-\frac{E_a}{R T}}$ represents the fraction of molecules with energies equal to or greater than $E _{ a }$.
→ Comparing equation (1) with Arrhenius equation, we can say that A is related to collision frequency.
→ Equation (1) predicts the value of rate constant fairly accurately for the reaction that involve atomic species or simple molecule but for complex molecules significant deviations are observed.
→ The reason could be that all collision do not lead to the formation of products.
→ "The collision in which molecules collide with sufficient kinetic energy (called threshold energy) and proper orientation, so as to facilitate breaking of bonds between reacting species and formation of new bonds to form products are called as effective collision."
→ For example, formation of methanol from bromomethane $CH _3 Br + OH ^{-} \rightarrow CH _3 OH + Br ^{-}$depends upon the orientation of reactant molecule as shown in figure.

→ The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no product are formed.
→ To account for effective collision, another factor P, called the probability factor or steric factor is introduced.
→ It takes into account the fact that in a collision, molecules be properly oriented i.e.
→ Rate $=P Z_{A B} e^{-\frac{E_{ a }}{R T}}$
→ 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 chemical reaction.

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Question 44 Marks

Derive the formula for determining activation energy from Arrhenius equation using rate constant at different temperatures. And also explain how activation energy can be determined using graph.

Answer

Activation energy using graph:
$\rightarrow$ Taking natural logarithm on both the side of Arrhenius equation $k = Ae ^{-\frac{ E _{ a }}{ RT }}$
$\rightarrow \ln k =-\frac{ E _{ a }}{ RT }+\ln A \ \ \ \ldots \ldots$ Eq. $(1)$
$\rightarrow$ The plot of $\ln k$ vs $1 / T$ gives a straight line as shown in figure.

$\rightarrow$ In figure, slope $=-\frac{E_{ a }}{R}$ and intercept $=\ln A$.
So, we can calculate activation energy $( E _{ a } )$ and Arrhenius constant $A$ using these values.
Formula of activation energy:
$\rightarrow$ It has been found from Arrhenius equation that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in rate constant.
$\rightarrow$ Thus, at temperature $T_1$, equation $(1)$ is
$\ln k _1=-\frac{ E _{ a }}{ RT _1}+\ln A ........Eq. (2)$
$\rightarrow$ at temperature $T _2,$ equation $(1)$
is $\ln k _2=-\frac{ E _{ a }}{ RT _2}+\ln A ........ Eq. (3)$
$($since $A$ is constant for given reaction$)$
$k _1$ and $k _2$ are rate constant at temperatures $T _1$ and $T _2$ respectively.
$\rightarrow$ Subtracting equation $(2)$ from $(3),$ we obtain
$\ln k _2-\ln k _1=\frac{- E _{ a }}{ RT _2}+\frac{ E _{ a }}{ RT _1}$
$\ln \frac{ k _2}{ k _1}=\frac{ E _{ a }}{ R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\log \frac{ k _2}{ k _1}=\frac{ E _{ a }}{2.303 R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\log \frac{ k _2}{ k _1}=\frac{ E _{ a }}{2.303 R }\left[\frac{ T _2- T _1}{T_1 T_2}\right]$
$\rightarrow$ From above formula activation energy can be calculate using measured values of rate constants at different temperatures.

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Question 54 Marks

What is rate of a chemical reaction? Explain average and instantaneous rate of reaction with help of graph and also give unit of rate of reaction.

Answer

$\rightarrow$ Rate of a chemical reaction$:$ "The change in concentration of reactant or product in unit time is called rate of a chemical reaction."
$\rightarrow$ To be more specific, it can be expressed in terms of $:$
$(i)$ The rate in decrease in concentration of any one of the reactants, or
$(ii)$ The rate in increase in concentration of any one of the products.
$\rightarrow$ Consider a hypothetical reaction, assuming that the volume of the system remains constant.
$R$ $\rightarrow$ $P$
$\rightarrow$ One mole of the reactant $R$ produces one mole of the product $P.$
$\rightarrow$ If $[ R ]_1$ and $[ P ]_1$ are the concentrations of $R$ and $P$ respectively at time $t_1$ and $[ R ]_2$ and $[ P ]_2$ are their concentrations at time $t _2$ then,
$\rightarrow$ $\Delta t = t _2- t _1$
$\rightarrow$ $\Delta[R]=[R]_2-[R]_1$
$\rightarrow$ $\Delta[ P ]=[ P ]_2-[ P ]_1$ (square bracket expressingmolar concentration.)
$\rightarrow$ Rate of disappearance of
$R=\frac{\text { Decrease in concentration of } R}{\text { Time taken }}=-\frac{\Delta[R]}{\Delta t}$
(as concentration of reactants is decreasing $\Delta[R]$ is a negative quantity, it is multiplied with $-1$ to make the rate of the reaction a positive quantity.)
Rate of appearance of
$P=\frac{\text { increase in concentration of } P}{\text { Time taken }}=+\frac{\Delta[P]}{\Delta t}$
$\rightarrow$ Above given equations represent the average rate of a reaction $r_{a v}$
$\rightarrow$ Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur.

Instantaneous rate of reaction $:$
$\rightarrow$ Average rate cannot be used to predict the rate of a reaction at a particular instant as it would be constant for the time interval for which it is calculated.
$\rightarrow$ So, to express the rate at a particular moment of time we determine instantaneous rate.
$\rightarrow$ Instantaneous rate is obtained when we consider the average rate at the smallest time interval say dt(when At approaches zero).
$\rightarrow$ Hence, mathematically for an infinitesimally small dt instantaneous rate is given by
$r _{ av }=\frac{-\Delta[ R ]}{\Delta t}=\frac{\Delta[ P ]}{\Delta t}$
As $\Delta_t \rightarrow 0$ or $r_{\text {inst }}=\frac{- d [ R ]}{ d t}=\frac{ d [ P ]}{ d t}$
Unit of rate of a reaction $:$
$\rightarrow$ Units of rate are concentration time ${ }^{-1}$.
$\rightarrow$ If concentration is in $mol\ L ^{-1}$ and time is in seconds then the units will be $mol\ L\ ^{-1} \sec ^{-1}$.
$\rightarrow$ However, in gaseous reactions, when the concentration of gases is expressed in terms of their partial pressur then the units of the rate equation will be atm $s^{-1.}$

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