CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 10

Question 1. Which of the following expressions is correct for the rate of reaction given below?
\( 5\text{Br}^-(aq) + \text{BrO}_3^-(aq) + 6\text{H}^+(aq) \rightarrow 3\text{Br}_2(aq) + 3\text{H}_2\text{O}(l) \)
(a) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = 5 \frac{\Delta[\text{H}^+]}{\Delta t} \)
(b) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{6}{5} \frac{\Delta[\text{H}^+]}{\Delta t} \)
(c) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{5}{6} \frac{\Delta[\text{H}^+]}{\Delta t} \)
(d) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = 6 \frac{\Delta[\text{H}^+]}{\Delta t} \)
Answer: (c) \( \frac{\Delta[\text{Br}^-]}{\Delta t} = \frac{5}{6} \frac{\Delta[\text{H}^+]}{\Delta t} \)
In simple words: To compare the speeds of two parts of a reaction, you relate them using their chemical equation numbers. For these two, dividing by their coefficients gives the correct match as option (c).
Exam Tip: Remember to always include the negative sign for reactants when writing individual rate expressions, and then solve for the desired rate term algebraically.

Question 2. For the reaction \( 3A \rightarrow 2B \), rate of reaction \( \frac{d[B]}{dt} \) is equal to
(a) \( -\frac{3}{2} \frac{d[A]}{dt} \)
(b) \( -\frac{2}{3} \frac{d[A]}{dt} \)
(c) \( -\frac{1}{3} \frac{d[A]}{dt} \)
(d) \( +\frac{2d[A]}{dt} \)
Answer: (b) \( -\frac{2}{3} \frac{d[A]}{dt} \)
In simple words: Since 3 molecules of A turn into 2 molecules of B, the rate of B appearing is two-thirds of the speed at which A is disappearing.

Exam Tip: Pay close attention to the signs; reactant concentrations decrease over time (negative sign), while product concentrations increase (positive sign).

Question 4. Which of the following can increase the rate of a chemical reaction?
P. Increasing the temperature
Q. Increasing the concentration of products
R. Adding a catalyst
S. Increasing the concentration of reactants
(a) Q and S
(b) P and Q
(c) P, R and S
(d) P, Q, R and S
Answer: (c) P, R and S
In simple words: Heating up the mix, adding a helper substance (catalyst), or adding more starting materials makes reactions go faster. Adding more final products does not speed it up.

Exam Tip: Focus on factors that increase collision frequency or lower activation energy to determine what accelerates reaction rates.

Question 6. For the reaction, \( A + 2B \rightarrow C + D \), the order of reaction is
(a) 1 with respect to A
(b) 2 with respect to B
(c) can't be predicted as order is determined experimentally.
(d) 3
Answer: (c) can't be predicted as order is determined experimentally.
In simple words: You cannot tell how fast a reaction will change just by looking at the written recipe. You have to run laboratory experiments to find the actual order.

Exam Tip: Always state that order is an experimental quantity and cannot be deduced from a balanced chemical equation alone unless it is defined as elementary.

Question 7. Which of the following statement is true?
(a) Molecularity of a reaction can be zero or a fraction.
(b) Molecularity has no meaning for complex reactions.
(c) Molecularity of a reaction is an experimental quantity.
(d) Reactions with molecularity three are very rare but are fast.
Answer: (b) Molecularity has no meaning for complex reactions.
In simple words: Since complex reactions happen in many separate steps, talking about the overall molecularity makes no sense. Only individual steps have molecularity.

Exam Tip: Remember that molecularity is a theoretical value for elementary steps, whereas reaction order is an experimental value for the overall process.

Question 8. The rate of a reaction increases sixteen times when the concentration of the reactant increases four times. The order of the reaction is
(a) 2.5
(b) 2.0
(c) 1.5
(d) 0.5
Answer: (b) 2.0
In simple words: If making the reactant four times larger causes the speed to jump by sixteen times, the reaction order is two, because four squared equals sixteen.

Exam Tip: Use the formula \( r \propto [A]^n \) where \( 16 = 4^n \) to quickly solve for the exponent \( n = 2 \).

Question 10. Match the Column I with Column II and choose the correct option.

Codes:
(a) A-(i), B-(ii), C-(ii)
(b) A-(ii), B-(iii), C-(i)
(c) A-(iii), B-(ii), C-(i)
(d) A-(ii), B-(i), C-(iii)
Answer: (c) A-(iii), B-(ii), C-(i)
In simple words: Sugar inversion is pseudo-first-order because of excess water, \( \text{N}_2\text{O} \) breakdown is a regular first-order reaction, and HI splitting on metal is zero-order.

Exam Tip: Metal surface reactions are generally zero-order as active sites saturate, while reactions with excess solvent are typically pseudo-first-order.

Question 11. The following experimental rate data were obtained for a reaction carried out at 25°C: \( A(g) + B(g) \rightarrow C(g) + D(g) \)

What are the orders with respect to A(g) and B(g)?
(a) Zero, Second
(b) First, Zero
(c) Second, Zero
(d) Second, First
Answer: (c) Second, Zero
In simple words: When we double B, the speed does not change, meaning B has a zero-order effect. Doubling A quadruples the speed, so A has a second-order effect.

Exam Tip: Find experimental runs where one reactant concentration is constant to isolate the relationship between the other reactant's concentration and the overall rate.

Question 12. The slope in the plot of \( \ln[R] \) vs time for a first order reaction is
(a) \( \frac{+k}{2.303} \)
(b) \( -k \)
(c) \( \frac{-k}{2.303} \)
(d) \( +k \)
Answer: (b) \( -k \)
In simple words: Plotting the natural logarithm of reactant remaining against time yields a falling straight line, and the steepness of this fall equals \( -k \).

Exam Tip: Pay close attention to the log base; natural logarithm \( \ln[R] \) plots have a slope of \( -k \), while common logarithm \( \log[R] \) plots have a slope of \( \frac{-k}{2.303} \).

Question 13. The half-life for a zero order reaction equals
(a) \( \frac{2k}{[R]_0} \)
(b) \( \frac{1}{2[R]_0^2k} \)
(c) \( \frac{[R]_0^2}{2k} \)
(d) \( \frac{[R]_0}{2k} \)
Answer: (d) \( \frac{[R]_0}{2k} \)
In simple words: For zero-order reactions, the time needed for half the reactant to disappear is simply the starting concentration divided by two times the speed constant.

Exam Tip: Know that zero-order half-life is directly proportional to initial concentration, whereas first-order half-life is entirely independent of it.

Question 14. Rate law for the reaction \( A + 2B \rightarrow C \) is found to be Rate = \( k[A][B] \). Concentration of reactant 'B' is doubled, keeping the concentration of 'A' constant, the value of rate constant will be.....
(a) the same
(b) doubled
(c) quadrupled
(d) halved
Answer: (a) the same
In simple words: The rate constant is a fixed number at a specific temperature. Changing the concentration speeds up the reaction, but the constant value \( k \) stays unchanged.

Exam Tip: Be sure not to confuse rate of reaction with the rate constant. The constant only changes with temperature or catalyst presence.

Question 15. A first order reaction is 50% completed in \( 1.26 \times 10^{14}\text{ s} \). How much time would it take for 100% completion?
(a) \( 1.26 \times 10^{15}\text{ s} \)
(b) \( 2.52 \times 10^{14}\text{ s} \)
(c) \( 2.52 \times 10^{28}\text{ s} \)
(d) Infinite
Answer: (d) Infinite
In simple words: Since first-order reactions keep slowing down as the starting material is used up, they theoretically take forever to reach complete 100% consumption.

Exam Tip: Remember that first-order reactions decay exponentially and only reach complete exhaustion at an infinite time interval.

Question 16. The rate of reaction \( A + B \rightarrow \text{Products} \), is given by the equation \( r = k[A][B] \). If B is taken in large excess, the order of reaction would be
(a) 2
(b) 0
(c) 1
(d) Cannot be predicted
Answer: (c) 1
In simple words: When you have a massive excess of reactant B, its concentration barely changes at all. The speed then depends only on A, reducing the order to 1.

Exam Tip: Reactions with one reactant in massive excess behave as pseudo-first-order reactions because the excess reactant's concentration remains constant.

Question 17. In effective collisions the colliding molecules must have:
(a) proper orientation only.
(b) a certain minimum amount of activation energy.
(c) threshold energy only.
(d) threshold energy and proper orientation both.
Answer: (d) threshold energy and proper orientation both.
In simple words: For a collision to cause a reaction, the molecules must crash with enough kinetic energy and also hit each other in the correct alignment.

Exam Tip: Collision theory states that both energetic (threshold energy) and steric (proper orientation) barriers must be overcome for a reaction to occur.

Question 18. Activation energy of a chemical reaction can be determined by
(a) determining the rate constant at standard temperature
(b) determining the rate constant at two different temperatures
(c) determining probability of collision
(d) using catalyst
Answer: (b) determining the rate constant at two different temperatures
In simple words: By measuring how much faster the reaction proceeds at two different temperatures, we can calculate the activation energy using the Arrhenius equation.

Exam Tip: Memorize the log form of the Arrhenius equation as it is frequently tested in numerical questions relating rates to temperature changes.

Question 19. What is the activation energy for a reaction, if its rate doubles when the temperature is raised from 20°C to 35°C? (\( R = 8.314\text{ J mol}^{-1}\text{ K}^{-1} \))
(a) \( 342\text{ kJ mol}^{-1} \)
(b) \( 269\text{ kJ mol}^{-1} \)
(c) \( 34.7\text{ kJ mol}^{-1} \)
(d) \( 15.1\text{ kJ mol}^{-1} \)
Answer: (c) \( 34.7\text{ kJ mol}^{-1} \)
In simple words: Using the temperature values in Kelvin, and placing them in the Arrhenius equation along with the rate doubling, we calculate the activation energy as \( 34.7\text{ kJ mol}^{-1} \).

Exam Tip: Be sure to convert Celsius temperatures to Kelvin (\( K = ^\circ C + 273 \)) before inserting them into chemical formulas.

Question 21. The diagram below shows the potential energy variation for a reaction using a catalyst and for the same reaction without a catalyst. Which of the following represents the enthalpy change (\( \Delta H \)) and activation energy (\( E_a \)) for the reaction with a catalyst?

(a) P
(b) Q
(c) R
(d) S
Answer: (a) P
In simple words: The energy difference from start to finish is \( z \), which is \( \Delta H \). The catalyst creates a lower peak, making the new activation energy \( x+z \).

Exam Tip: Remember that a catalyst changes the activation energy pathway but has no effect on the starting or ending energy levels, meaning \( \Delta H \) remains constant.

Question 22. Which of the following statements is/are correct?
I. A catalyst lowers the activation energy of a reaction.
II. A catalyst allows the same rate of reaction to be achieved at a lower temperature.
III. A catalyst mixes with reactants and increases the overall concentration of reactants in the rate equation.
(a) Only I
(b) I and II
(c) II and III
(d) I, II and III
Answer: (b) I and II
In simple words: A catalyst works by lowering the activation energy barrier, which means you can get the same reaction speed at a cooler temperature. It does not change reactant concentrations.

Exam Tip: Remember that catalysts provide an alternative pathway with lower activation energy but never alter concentrations in the rate equation.

Assertion-Reason

Question 24. Assertion (A) Rate of reaction is a measure of change in concentration of reactant with respect to time.
Reason (R) Rate of reaction is a measure of change in concentration of product with respect to time.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
In simple words: Both statements are correct facts, as you can measure reaction rates using either reactants or products. However, one fact is not the cause or explanation for the other.

Exam Tip: For Assertion-Reason questions, first check the factual accuracy of both statements independently before looking for a logical explanation link.

Question 25. Assertion (A) For a zero order reaction, the unit of rate constant and rate of reaction are same.
Reason (R) Rate of reaction for zero order reaction is independent of concentration of reactant.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
In simple words: In zero-order reactions, the speed is completely independent of concentration, making the rate equal to the rate constant. Therefore, they share the exact same unit.

Exam Tip: The rate of reaction always has the unit \( \text{mol L}^{-1}\text{ s}^{-1} \); for a zero-order reaction, \( \text{Rate} = k \), so \( k \) has this exact same unit.

Question 26. Assertion (A) For complex reactions, molecularity and order are not same.
Reason (R) Order of reaction may be zero.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
In simple words: Complex reactions proceed in multiple steps, so molecularity does not match the overall order. Also, reaction order can be zero, but this separate fact does not explain the difference.

Exam Tip: Remember that molecularity is always a positive integer, while reaction order can be zero, fractional, or an integer.

Question 27. Assertion (A) Rate constant determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.
Reason (R) Reactant molecules undergo chemical change irrespective of their orientation during collision.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (c) Assertion is true, but Reason is false.
In simple words: The Arrhenius equation calculates rates reasonably well, but molecules definitely need the correct orientation to react. They cannot react if they hit the wrong way.

Exam Tip: For complex molecules, collision theory incorporates a probability or steric factor \( P \) to account for orientation requirements.

Question 28. Assertion (A) Order and molecularity of a reaction are always same.
Reason (R) Complex reactions involve a sequence of elementary reactions and the slowest step is rate determining.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (d) Assertion is false, but Reason is true.
In simple words: Order and molecularity are often different. In multi-step reactions, the overall speed is determined by the slowest individual step.

Exam Tip: If an Assertion statement is clearly false, you can immediately identify the correct key option as (d).

Question 29. Assertion (A) A catalyst increases the rate of reaction.
Reason (R) Catalyst also changes the equilibrium constant.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (c) Assertion is true, but Reason is false.
In simple words: A catalyst speeds up the reaction process, but it does not shift the final balance of chemical equilibrium, so the equilibrium constant remains unchanged.

Exam Tip: Catalysts increase both the forward and reverse reaction rates equally, which is why the overall equilibrium constant remains constant.

Question 30. Assertion (A) Hydrolysis of an ester follows first order kinetics.
Reason (R) Concentration of water remains nearly constant during the course of the reaction.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
In simple words: Acid hydrolysis of ester is a pseudo-first-order reaction because water is present in such massive amounts that its level does not change in a noticeable way.

Exam Tip: Reactions where one reactant is in large excess are classic examples of pseudo-first-order kinetics.

Question 31. Assertion (A) The unit of \( k \) is independent of order of reaction.
Reason (R) The unit of \( k \) for zero order reaction is \( \text{mol L}^{-1}\text{ s}^{-1} \).
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (d) Assertion is false, but Reason is true.
In simple words: The unit for the rate constant changes depending on the reaction order. For a zero-order reaction, the unit is moles per liter per second.

Exam Tip: Use the general dimensional formula \( (\text{mol L}^{-1})^{1-n}\text{ s}^{-1} \) to quickly find the rate constant unit for any reaction order \( n \).

Question 32. Assertion (A) Reactions of higher order are rare.
Reason (R) The chances of simultaneous multimolecular collision are extremely small.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
In simple words: Reactions with orders higher than three are rare because it is highly unlikely for four or more molecules to crash together at the exact same moment and angle.

Exam Tip: Write that the probability of simultaneous collision decreases exponentially with an increasing number of reacting molecules.

Question 33. Assertion (A) Rate of reaction increase with increase in temperature.
Reason (R) Number of collisions increases with increase in temperature.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
In simple words: Heating makes molecules speed up and crash more often, which is why chemical reactions happen much faster at higher temperatures.

Exam Tip: Although collision frequency increases, the main reason for the rate increase with temperature is the dramatic rise in the fraction of molecules with energy greater than the activation energy.

Question 34. Assertion (A) All collisions of reactant molecules lead to product formation.
Reason (R) Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Answer: (d) Assertion is false, but Reason is true.
In simple words: Many collisions do not result in a reaction. Only the high-energy crashes with correct spatial orientation produce a chemical change.

Exam Tip: Clearly state that only "effective collisions" (those overcoming both activation energy and orientation barriers) lead to products.

Case-Study Passage for Questions 35
Arrhenius equation for first order reactions are useful for modeling the timing and temperature of petroleum generation from organic rich shales and kerogens at laboratory scales (e.g. per minute), based on pyrolysis 52 peak data is measured at multiple heating rates, and for burial histories occurring over geological time scales (e.g. per millions of years). First order reactions are accompanied by second order reactions and physical changes impacting the kerogens as they thermally mature, so simple Arrhenius equation, first order reaction modeling can only provide an approximation of the complete petroleum generation process. Thermally immature shale and/or kerogen samples yield less complex pyrolysis S2-peaks than thermally mature sample and provide more credible kinetic distributions when modeled at multiple heating rates with the Arrhenius equation. More credible kinetic fits to the S2-peak pyrolysis data are derived by allowing the activation energy (\( E \)) and the frequency factor (\( A \)) to vary for each reaction to derive good-fit kinetic distributions than by applying the commonly used assumption that the \( A \) value remain constant for the set of parallel reactions included in the distribution.

Question (i) The plot of \( \ln k \) versus \( \frac{1}{T} \) gives a straight line. What is the value of slope and intercept in this graph?
Answer: Comparing the Arrhenius equation \( \ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A \) with \( y = mx + c \), the values are:
Slope = \( -\frac{E_a}{R} \)
Intercept = \( \ln A \)
In simple words: When graphing natural log of speed against inverse temperature, the slope of the line equals \( -\frac{E_a}{R} \) and the vertical axis intercept is \( \ln A \).

Exam Tip: Always state both the mathematical values of the slope and the intercept along with their physical terms to score full marks.

Question (ii) If the activation energy for a reaction is zero and rate constant is \( 1.72 \times 10^5\text{ s}^{-1} \) at 300 K. Calculate the rate constant at 310 K.
Answer: According to the Arrhenius relation, \( \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left[\frac{T_2 - T_1}{T_1 T_2}\right] \). Since the activation energy \( E_a = 0 \), the entire right side of the equation becomes zero:
\( \log\left(\frac{k_2}{k_1}\right) = 0 \)
\( \implies \frac{k_2}{k_1} = 10^0 = 1 \)
\( \implies k_2 = k_1 = 1.72 \times 10^5\text{ s}^{-1} \)
Therefore, the rate constant at 310 K remains \( 1.72 \times 10^5\text{ s}^{-1} \).
In simple words: If a reaction has zero activation energy, its speed does not change with temperature. The rate constant stays exactly the same.

Exam Tip: Remember that temperature only influences the rate constant when there is a positive activation energy barrier to overcome.

Question (iii) What will be the SI unit for Arrhenius factor in Arrhenius equation of a second order reaction?
Answer: The Arrhenius factor \( A \) shares the exact same units as the rate constant \( k \) because the exponential term \( e^{-E_a/RT} \) is a dimensionless number. For a second-order reaction, the unit of \( k \) is \( \text{L mol}^{-1}\text{ s}^{-1} \). Thus, the SI unit for the Arrhenius factor \( A \) is \( \text{L mol}^{-1}\text{ s}^{-1} \).
In simple words: The unit for the Arrhenius pre-exponential factor is always the same as the rate constant. For second-order, it is liters per mole-second.

Exam Tip: Always write down units clearly, ensuring you match the specific order of the reaction requested.

Question (iii) [OR] Give the relation between rate constant and activation energy.
Answer: The logarithmic relationship between rate constant and activation energy is given by the equation:
\( \log k = \log A - \frac{E_a}{2.303 RT} \)
where \( k \) is the rate constant, \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
In simple words: The formula shows that a higher activation energy leads to a smaller rate constant, meaning the reaction goes slower.

Exam Tip: When presenting equations, always define each symbol clearly to ensure you earn full points.

Case-Study Passage for Question 36
The rate of a chemical reaction is expressed either in terms of decrease in the concentration of reactants or increase in the concentration of a product per unit time. Rate of the reaction depends upon the nature of reactants, concentration of reactants, temperature, presence of catalyst, surface area of the reactants and presence of light. Rate of reaction is directly related to the concentration of reactant. Rate law states that, the rate of reaction depends upon the concentration terms on which the rate of reaction actually depends, as observed experimentally. The sum of powers of the concentration of the reactants in the rate law expression is called order of reaction while the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of the reaction.

Question (i) (a) What is a rate determining step?
Answer: In a complex chemical reaction that proceeds through multiple steps, the slowest individual elementary step is called the rate-determining step, as it controls the overall speed of the reaction.
In simple words: The slowest step in a chemical process acts as a bottleneck, determining how fast the entire reaction can finish.

Exam Tip: State clearly that the overall reaction rate is equal to the rate of the slowest elementary step in the mechanism.

Question (i) (b) Define complex reaction.
Answer: A complex reaction is a chemical process that does not finish in a single step, but instead proceeds through a sequence of two or more elementary reaction steps.
In simple words: A complex reaction is a multi-step reaction that takes place in sequence rather than all at once.

Exam Tip: Contrast complex reactions with elementary reactions to make your definition more robust.

Question (ii) What is the effect of temperature on the rate constant of a reaction?
Answer: The rate constant of a chemical reaction increases exponentially as the temperature rises, according to the Arrhenius relation. For most reactions, the rate constant approximately doubles with every 10°C increase in temperature.
In simple words: As the temperature goes up, the rate constant increases quickly, roughly doubling for every ten-degree rise.

Exam Tip: Mention the temperature coefficient rule (rate doubles for every 10°C rise) to demonstrate a complete understanding.

Question (ii) [OR] Why is molecularity applicable only for elementary reactions whereas order is applicable for elementary as well as complex reactions?
Answer: Molecularity measures the number of reacting molecules colliding simultaneously in a single elementary reaction step, which has no meaning for overall complex processes with multiple steps. On the other hand, the reaction order is determined experimentally based on the overall rate law, making it applicable to both single-step and multi-step reactions.
In simple words: Molecularity counts molecules colliding at once in one step, so it is useless for multi-step reactions. Order is calculated from lab tests of the overall reaction speed.

Exam Tip: Clearly emphasize that molecularity is a theoretical concept, while order of reaction is an experimental quantity.

Question (iii) The conversion of molecule X to Y follows second order kinetics. If concentration of X is increased 3 times, how will it affect the rate of formation of Y?
Answer: The rate law for second-order kinetics is written as \( \text{Rate} = k[X]^2 \). If the concentration of reactant X is tripled, we substitute \( 3[X] \) into the equation:
\( \text{Rate}' = k[3X]^2 = 9 k[X]^2 = 9 \times \text{Rate} \)
Therefore, the rate of formation of Y will increase by 9 times.
In simple words: Since the reaction depends on the concentration squared, tripling the reactant makes the speed nine times faster because three squared is nine.

Exam Tip: Always show the rate equations before and after the concentration change to clearly justify your final calculated factor.