Intext Questions — Chemical Kinetics

1. 3.13 marksNCERT Intext Questions, Chapter 3

   For the reaction R $\to$ P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

   Hint. Average rate is the change in concentration divided by the time interval, with a minus sign for a reactant so the rate comes out positive.

   For a reactant the average rate carries a minus sign, so that the rate itself is positive:

   $r_{\text{av}}=-\frac{\Delta [\text{R}]}{\Delta t}=-\frac{[\text{R}{]}_2-[\text{R}{]}_1}{t_2-t_1}$

   Substituting the given values:

   $r_{\text{av}}=-\frac{0.02-0.03}{25}=\frac{0.01}{25}=4\times 10^{-4}{\text{mol L}}^{-1}{\text{min}}^{-1}$

   Converting the time unit to seconds means dividing by 60:

   $r_{\text{av}}=\frac{4\times 10^{-4}}{60}=6.67\times 10^{-6}{\text{mol L}}^{-1}{\text{s}}^{-1}$

   NCERT's answer key prints $6.66\times 10^{-6}$ M s${}^{-1}$, which is the same number rounded down.

   Since the stoichiometry is one to one, no dividing by a coefficient is needed here, which is exactly what changes in the next question.

   ✦ $4\times 10^{-4}$ mol L${}^{-1}$ min${}^{-1}$, which is $6.67\times 10^{-6}$ mol L${}^{-1}$ s${}^{-1}$

2. 3.23 marksNCERT Intext Questions, Chapter 3

   In a reaction, 2A $\to$ Products, the concentration of A decreases from 0.5 mol L${}^{-1}$ to 0.4 mol L${}^{-1}$ in 10 minutes. Calculate the rate during this interval.

   Hint. The coefficient of A is 2, so the rate of the reaction is half the rate at which A disappears.

   The rate at which A disappears is

   $-\frac{\Delta [\text{A}]}{\Delta t}=\frac{0.5-0.4}{10}=0.01{\text{mol L}}^{-1}{\text{min}}^{-1}$

   But the balanced equation is $2\text{A}\to \text{Products}$, so two molecules of A vanish for each act of reaction. The rate of the reaction is therefore

   $\text{Rate}=-\frac{1}{2}\frac{\Delta [\text{A}]}{\Delta t}=\frac{1}{2}\times 0.01=0.005{\text{mol L}}^{-1}{\text{min}}^{-1}$

   The factor of $\frac{1}{2}$ is the entire point of this question, and it is what makes the rate of reaction a single number no matter which species is monitored.

   A note on the printed answer key: it reads "Rate of reaction = rate of disappearance of A = 0.005 mol litre${}^{-1}$ min${}^{-1}$". The number 0.005 is the rate of the reaction, but it is not the rate of disappearance of A, which is 0.010. The two are equal only when the coefficient is 1.

   ✦ Rate of reaction $=0.005$ mol L${}^{-1}$ min${}^{-1}$, while the rate of disappearance of A is $0.010$ mol L${}^{-1}$ min${}^{-1}$

3. 3.32 marksNCERT Intext Questions, Chapter 3

   For a reaction, A + B $\to$ Product, the rate law is given by $r=k[\text{A}{]}^{1/2}[\text{B}{]}^2$. What is the order of the reaction?

   Hint. Order is the sum of the exponents in the experimentally determined rate law, and nothing else.

   The order of a reaction is the sum of the powers to which the concentrations are raised in the rate law:

   $\text{order}=\frac{1}{2}+2=2.5$

   The reaction is of order $2.5$ overall, of order $0.5$ with respect to A and of order $2$ with respect to B.

   Two things are worth noticing. First, the exponents bear no relation to the stoichiometric coefficients, which are both 1 here; the rate law is an experimental result and can never be read off the balanced equation. Second, order may be fractional, as it is here, whereas molecularity may not, because molecularity counts actual colliding particles in a single elementary step.

   ✦ Order $=2.5$

4. 3.42 marksNCERT Intext Questions, Chapter 3

   The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y?

   Hint. Second order means the rate depends on the square of the concentration.

   Second order kinetics for the conversion $\text{X}\to \text{Y}$ means

   $\text{Rate}=k[\text{X}{]}^2$

   If the concentration is tripled to $3[\text{X}]$, the new rate is

   ${\text{Rate}}^{\prime }=k(3[\text{X}]{)}^2=9k[\text{X}{]}^2=9\times \text{Rate}$

   The rate of formation of Y increases nine times.

   The general pattern is worth committing to memory: multiplying a concentration by a factor $f$ multiplies the rate by $f^n$, where $n$ is the order with respect to that species. So a threefold increase gives 3 times for first order, 9 times for second and 27 times for third.

   ✦ The rate becomes nine times its original value

5. 3.53 marksNCERT Intext Questions, Chapter 3

   A first order reaction has a rate constant $1.15\times 10^{-3}$ s${}^{-1}$. How long will 5 g of this reactant take to reduce to 3 g?

   Hint. For a first order reaction the integrated rate law works with any quantity proportional to concentration, so masses may be used directly.

   The integrated first order rate law is

   $k=\frac{2.303}{t}\log \frac{[\text{R}{]}_0}{[\text{R}]}$

   Because the ratio of two concentrations is involved, any quantity proportional to concentration may be substituted, so the masses can be used as they stand without converting to moles.

   Rearranging for $t$:

   $t=\frac{2.303}{k}\log \frac{5}{3}=\frac{2.303}{1.15\times 10^{-3}}\log (1.667)$

   $t=2002.6\times 0.2218=444\text{s}$

   This matches NCERT's printed answer key exactly. As a rough check, the half-life is $0.693/k=603$ s, and reducing 5 g to 3 g is somewhat less than halving, so a time somewhat under 603 s is what should be expected.

   ✦ $t=444$ s

6. 3.62 marksNCERT Intext Questions, Chapter 3

   Time required to decompose SO${}_2$Cl${}_2$ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

   Hint. For a first order reaction the half-life and rate constant are linked by a single constant, independent of the starting amount.

   For a first order reaction the half-life is independent of the initial concentration and is given by

   $t_{1/2}=\frac{0.693}{k}\text{so}k=\frac{0.693}{t_{1/2}}$

   The answer is wanted in reciprocal seconds, so convert the half-life first:

   $t_{1/2}=60\text{min}=3600\text{s}$

   $k=\frac{0.693}{3600}=1.925\times 10^{-4}{\text{s}}^{-1}$

   This matches NCERT's printed answer key. Had the time been left in minutes, the answer would have been $1.155\times 10^{-2}$ min${}^{-1}$, which is the same rate constant expressed in a different unit.

   ✦ $k=1.925\times 10^{-4}$ s${}^{-1}$

7. 3.73 marksNCERT Intext Questions, Chapter 3

   What will be the effect of temperature on rate constant?

   Hint. State the observed rule of thumb first, then give the quantitative law that explains it.

   The rate constant increases with temperature. As a rough rule, a rise of 10 K roughly doubles the rate constant for most reactions near room temperature.

   Quantitatively the dependence is given by the Arrhenius equation:

   $k=A{e}^{-E_a/RT}$

   where $A$ is the Arrhenius or pre-exponential factor, $E_a$ the activation energy and $R$ the gas constant. Taking logarithms:

   $\ln k=\ln A-\frac{E_a}{RT}$

   so a plot of $\ln k$ against $1/T$ is a straight line of slope $-E_a/R$ and intercept $\ln A$.

   The reason for the increase lies in the Maxwell-Boltzmann distribution of molecular energies. The factor $e^{-E_a/RT}$ is the fraction of molecules with energy at least $E_a$. Raising the temperature broadens the distribution and shifts its maximum to higher energy, so that fraction grows sharply, and the rate constant grows with it.

   Note that this is the only one of the nine intext questions for which the book prints no answer.

   ✦ The rate constant rises with temperature, roughly doubling per 10 K, as described quantitatively by the Arrhenius equation $k=A{e}^{-E_a/RT}$

8. 3.83 marksNCERT Intext Questions, Chapter 3

   The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate $E_a$.

   Hint. Use the two-temperature form of the Arrhenius equation with $k_2/k_1=2$.

   The two-temperature form of the Arrhenius equation is

   $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

   Here $T_1=298$ K, $T_2=308$ K and $k_2/k_1=2$, so $\log (k_2/k_1)=0.3010$.

   $0.3010=\frac{E_a}{2.303\times 8.314}\times \frac{10}{298\times 308}$

   $E_a=\frac{0.3010\times 2.303\times 8.314\times 298\times 308}{10}$

   $E_a=52894{\text{J mol}}^{-1}=52.9{\text{kJ mol}}^{-1}$

   NCERT's key gives $52.897$ kJ mol${}^{-1}$, agreeing to the last digit that the rounding allows.

   The result explains the familiar rule of thumb quoted in section 3.4: a reaction whose activation energy is around 50 kJ mol${}^{-1}$ will roughly double its rate for a 10 K rise near room temperature. Reactions with much larger $E_a$ are far more temperature-sensitive than that.

   ✦ $E_a=52.9$ kJ mol${}^{-1}$

9. 3.93 marksNCERT Intext Questions, Chapter 3

   The activation energy for the reaction $2\text{HI(g)}\to {\text{H}}_2+{\text{I}}_2(g)$ is 209.5 kJ mol${}^{-1}$ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

   Hint. The exponential factor in the Arrhenius equation is itself that fraction.

   In the Arrhenius equation $k=A{e}^{-E_a/RT}$, the exponential factor is precisely the fraction of molecules whose energy equals or exceeds the activation energy:

   $x=e^{-E_a/RT}$

   Taking logarithms to base ten to keep the arithmetic manageable:

   $\log x=-\frac{E_a}{2.303RT}=-\frac{209500}{2.303\times 8.314\times 581}$

   $\log x=-\frac{209500}{11122.5}=-18.836$

   $x=10^{-18.836}=1.47\times 10^{-19}$

   This matches NCERT's printed answer of $1.471\times 10^{-19}$.

   The number is worth pausing on: fewer than two molecules in $10^{19}$ carry enough energy to react at 581 K, which is why the reaction is slow despite the collisions being enormously frequent. It also shows why lowering $E_a$ with a catalyst has such a dramatic effect, since the fraction depends exponentially on it.

   ✦ $x=1.47\times 10^{-19}$

Solutions written by the tuition.in editorial team and checked against lech103.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.