Questions: Chemical Kinetics

Q1 Define rate of reaction, order of reaction and molecularity. How can a reaction have a fractional order?

Answer (English): A reaction rate records how fast a reactant disappears or a product appears during a given interval. In a rate law found from measurements, the exponents on the concentrations add up to the overall order. Molecularity is the number of reacting species that take part in an elementary step, so it is always a whole number. Fractional order arises in multi-step reactions whose rate law contains a term from an equilibrium before the slow step, so the power can become a fraction even though every elementary collision involves whole molecules.

Hinglish: Rate concentration ke change per unit time hai. Order rate law ke powers ka sum hota hai aur experiment se milta hai. Molecularity elementary step ke species ki ginti hai, isliye whole number hoti hai. Fractional order tab aata hai jab mechanism mein slow step se pehle equilibrium ho jaye.

Q2 For a reaction A + B giving products, doubling the concentration of A doubles the rate while tripling B multiplies the rate by nine. Write the rate law and find the overall order.

Answer (English): Doubling A doubles the rate, so the order in A is one. Tripling B multiplies the rate by nine, which is three squared, so the order in B is two. The rate law is rate equals k times the concentration of A times the square of the concentration of B. The overall order is the sum of the powers, which is three. The rate constant has units of L^2 mol^-2 s^-1 because the overall order is three, and it is found by substituting one measured rate and the corresponding concentrations.

Hinglish: A ka double karne par rate double, isliye A mein order 1. B ka triple karne par rate 9 guna, isliye B mein order 2. Rate = k[A][B]^2 aur overall order 3. k ka unit L^2 mol^-2 s^-1 hoga.

Q3 How does the half-life of a zero order reaction differ from that of a first order reaction?

Answer (English): For a zero order reaction the half-life is the initial concentration divided by twice the rate constant, so it is directly proportional to the initial concentration. The reaction takes longer to fall from 1 M to 0.5 M than from 0.5 M to 0.25 M. For a first order reaction the half-life is 0.693 divided by the rate constant and does not depend on the starting concentration. Every successive half-life is equal, and after n half-lives the amount left is the initial amount divided by two to the power n. The constancy of the half-life is a quick test for first order behaviour.

Hinglish: Zero order mein t1/2 = [R]0/2k, isliye starting concentration badhne par half-life badhti hai. First order mein t1/2 = 0.693/k, initial concentration se independent; har half-life barabar hoti hai aur n half-lives ke baad [R]0/2^n bachta hai.

Q4 A first order reaction is 50 percent complete in 20 minutes. Calculate the time needed for 75 percent completion.

Answer (English): For a first order reaction the half-life is 20 minutes because 50 percent completion means the concentration has halved. After one half-life half the reactant remains; to reach 75 percent completion only 25 percent may remain, which is a further halving. Therefore the time for 75 percent completion is two half-lives, or 40 minutes. In general the time for 75 percent completion is twice the half-life, and the time for 87.5 percent completion is three half-lives.

Hinglish: 50 percent completion matlab t1/2 = 20 minute. 75 percent completion ke liye ek aur half-life chahiye kyunki 25 percent bachta hai, isliye total time 40 minutes hota hai. General rule: 75 percent ke liye 2 t1/2, 87.5 percent ke liye 3 t1/2.

Q5 How can you decide whether a reaction is first order or second order from a graph of concentration against time?

Answer (English): Plot the raw concentration against time. If the graph is a straight line with a negative slope, the reaction is zero order. If it is a curve, test the integrated forms: for a first order reaction a plot of log concentration against time is a straight line, and for a second order reaction a plot of reciprocal concentration against time is a straight line. The plot that is straight selects the order. A further quick check uses the half-life: a constant half-life means first order, while a half-life that doubles as the starting concentration doubles means zero order.

Hinglish: Raw concentration vs time graph straight ho to zero order. Curve ho to log[R] vs time aur 1/[R] vs time dono plot karo; jo straight aaye wahi order hai. t1/2 constant ho to first order, aur t1/2 initial concentration ke proportional ho to zero order.

Q6 Explain how temperature and a catalyst affect the rate of a reaction.

Answer (English): A rise in temperature increases the average kinetic energy of the molecules and shifts the Maxwell-Boltzmann distribution to higher energies, so a much larger fraction of collisions has energy above the activation barrier. The rate constant therefore increases, exponentially according to the Arrhenius equation. A catalyst acts differently: it provides an alternative reaction pathway with a lower activation energy, so a larger fraction of the collisions at the same temperature can succeed. A catalyst speeds up both the forward and backward reactions and does not change the equilibrium constant or the enthalpy change; it only shortens the time needed to reach equilibrium.

Hinglish: Temperature badhne par Maxwell-Boltzmann curve right shift hoti hai aur activation energy se upar wali fraction badh jaati hai, isliye k badhta hai. Catalyst naya pathway deta hai jiski Ea kam hoti hai, dono directions ki speed badhta hai, lekin equilibrium constant nahi badalta.

Q7 The rate constant of a reaction doubles when the temperature rises from 300 K to 310 K. Calculate the activation energy. R = 8.314 J K^-1 mol^-1.

Answer (English): Use the two temperature form, log of k2 by k1 equals Ea divided by 2.303R times the product of the temperatures divided by their difference. Here k2 by k1 is 2, so log 2 is 0.301. The temperature factor is 300 times 310 divided by 10, which is 9300 K. Then Ea equals 19.15 times 9300 times 0.301, which is about 53.6 kJ mol^-1. The value is typical of many organic reactions, where a moderate temperature rise produces a large increase in rate.

Hinglish: log(k2/k1) = (Ea/2.303R)(T2-T1)/(T1T2). Yahan log 2 = 0.301, temperature factor = 300×310/10 = 9300 K, isliye Ea = 19.15 × 9300 × 0.301 ≈ 53.6 kJ/mol.

Q8 The half-life of a radioactive isotope is 10 years. What fraction of the original sample remains after 30 years?

Answer (English): Thirty years is three half-lives. After one half-life one half remains, after two half-lives one quarter remains, and after three half-lives one eighth remains. The fraction left is 1 divided by 2 to the power 3, which is 0.125, or 12.5 percent. The calculation uses the fact that radioactive decay follows first order kinetics, so the half-life is constant and the amount falls by a fixed fraction in every equal time interval.

Hinglish: 30 saal matlab 3 half-lives. Har half-life ke baad aadha bachta hai, isliye fraction = 1/2^3 = 1/8 = 0.125, yani 12.5 percent. Radioactive decay first order kinetics follow karta hai, isliye half-life constant rehti hai.

Q9 Explain, using collision theory, why a small rise in temperature causes a large increase in the rate of a reaction.

Answer (English): Collision theory says that only collisions with energy above the activation energy can lead to products. The Maxwell-Boltzmann distribution shows that the fraction of molecules with energy above a high barrier lies in the tail of the curve and is very small. When the temperature rises, the peak of the distribution moves to higher energy and the curve broadens, so the area beyond the barrier increases sharply. Because the rate depends on that area, a rise of ten degrees can double or triple the rate even though the average energy rises only modestly. Orientation and collision frequency also matter, but their change with temperature is far smaller.

Hinglish: Sirf wahi collisions react karti hain jinme activation energy se zyada energy ho. Maxwell-Boltzmann curve ki high energy tail bahut chhoti hoti hai; temperature badhne par tail ka area tezi se badh jaata hai, isliye rate kai guna badh jaata hai.

Q10 What is a pseudo first order reaction? Give one example and explain why its molecularity differs from its order.

Answer (English): A pseudo first order reaction is one that is actually bimolecular but follows first order kinetics because one reactant is present in large excess. The acid catalysed hydrolysis of ethyl acetate is the standard example: water is present in such excess that its concentration changes by a negligible fraction, so it is treated as constant and absorbed into the rate constant. The experimental rate law then contains only the ester concentration and appears first order. The molecularity of the elementary step is two because one ester molecule and one water molecule react, while the observed order is one.

Hinglish: Pseudo first order reaction mein ek reactant excess mein hota hai, jaise ester ka acid catalysed hydrolysis mein paani. Paani ki concentration constant maan li jaati hai, isliye rate law sirf ester par depend karta hai. Molecularity 2 hoti hai lekin observed order 1.

The chapter quiz is attached to this page and has five questions on rate laws, half-life, the units of the rate constant and the Arrhenius equation. Attempt it after writing out the ten answers above, because each quiz explanation repeats one step from these answers.