Chemical Kinetics: common doubts, answered
The questions students ask most often about Chemical Kinetics, each with a short answer. For the full chapter, read the Chemical Kinetics notes.
About the chapter
How is Chemical Kinetics usually tested in NEET?
Expect questions on units of the rate constant, finding order from rate data or from units, first-order calculations using half-lives and logarithms, and activation energy from rate constants at two temperatures. Conceptual items contrast order with molecularity and ask what a catalyst does and does not change. Careful handling of 2.303, kelvin temperatures and joules versus kilojoules saves many marks.
Rate of a reaction
Read this section in the notes →What is the difference between average rate and instantaneous rate?
Average rate is the change in concentration over a finite time interval, the slope of a straight line joining two points on the concentration-time graph. Instantaneous rate is the rate at a single moment, given by the slope of the tangent to the curve at that time. Since the rate usually slows as reactants are used up, the two values generally differ.
How do you write the rate of a reaction with unequal coefficients, like 2N₂O₅ → 4NO₂ + O₂?
Divide each rate of change by that species' coefficient, so every species gives the same single rate. Here rate = −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt, with a minus sign for the reactant because its concentration falls. So NO₂ appears twice as fast as N₂O₅ disappears, and four times as fast as O₂ appears.
Rate law and rate constant
Read this section in the notes →Why can't the rate law be written from the balanced equation?
The balanced equation shows only the overall change, not the steps by which it actually happens. The rate depends on the real mechanism, especially its slowest step, which may involve different species and numbers of molecules. So the powers in the rate law have to be found by experiment; for CHCl₃ + Cl₂ → CCl₄ + HCl, the rate depends on [Cl₂] raised to ½.
What is the difference between rate of reaction and rate constant?
The rate tells how fast concentrations are changing at a given moment, and it drops as reactants are consumed. The rate constant k is the proportionality constant in the rate law, numerically equal to the rate when every concentration is 1 mol L⁻¹. It stays fixed for a reaction at a given temperature and changes only with temperature or a catalyst, never with concentration.
Order of a reaction and units of k
Read this section in the notes →What is the difference between order and molecularity of a reaction?
Order is the sum of the powers of the concentration terms in the experimentally found rate law, and it can be zero, a fraction or a whole number. Molecularity is the number of particles that collide in one elementary step, so it is always 1, 2 or 3. Order describes the overall reaction; molecularity belongs only to individual steps of a mechanism.
How do you find the units of the rate constant from the order?
For order n the unit is (mol L⁻¹)^(1 − n) s⁻¹. That gives mol L⁻¹ s⁻¹ for zero order, s⁻¹ for first order and L mol⁻¹ s⁻¹ for second order. Questions often run the other way: a rate constant quoted in s⁻¹ tells you the reaction is first order, whatever its balanced equation looks like.
Can the order of a reaction be zero or a fraction?
Yes. Order is measured, not predicted, so it can take any value the data demand. In a zero-order reaction the rate does not depend on concentration, as when ammonia decomposes on a hot platinum surface at high pressure and the surface is fully covered. Fractional orders, such as 1.5 for the reaction of chloroform with chlorine, come from multi-step mechanisms.
Molecularity and mechanism
Read this section in the notes →Why is molecularity never more than three?
Molecularity counts the particles that must collide at the same instant in a single step. A three-body collision is already rare, and four or more particles meeting at once with enough energy and the right orientation is practically impossible. So elementary steps have molecularity 1, 2 or occasionally 3, and reactions that seem to need more particles proceed through a series of simpler steps.
What is the rate-determining step?
In a reaction that happens in several steps, the slowest step limits how fast the whole reaction can proceed, much as the narrowest point of a road limits traffic. This rate-determining step decides the form of the rate law. If the slow step involves one molecule of A and one of B, the rate depends on [A] and [B], whatever the overall equation shows.
Zero-order reactions
Read this section in the notes →What does a zero-order reaction look like and when does it occur?
In a zero-order reaction the rate stays constant and equals k, so the concentration falls in a straight line, [R] = [R]₀ − kt, and a plot of [R] against time has slope −k. It occurs when something other than reactant concentration limits the rate, such as a catalyst surface that is completely covered, as with ammonia decomposing on hot platinum at high pressure.
First-order reactions
Read this section in the notes →What is the integrated rate equation for a first-order reaction?
It is k = (2.303/t) log([R]₀/[R]), which is the same as ln[R] = ln[R]₀ − kt. So a graph of ln[R] against time is a straight line with slope −k, and a graph of log([R]₀/[R]) against t is a straight line through the origin with slope k/2.303. Remember that ln x = 2.303 log x when moving between the two forms.
Half-life and pseudo first order
Read this section in the notes →Why is the half-life of a first-order reaction independent of concentration?
Putting [R] = [R]₀/2 into the first-order equation gives t½ = 0.693/k, which has no concentration term. So the time taken for half the reactant to go is the same whatever amount you start with, which is why radioactive decay has a fixed half-life. For a zero-order reaction, t½ = [R]₀/2k, so the half-life shortens as the starting concentration falls.
How many half-lives does a first-order reaction need to reach 99.9% completion?
About ten. After n half-lives the fraction left is (½)ⁿ, and (½)¹⁰ is 1/1024, just under 0.1%, so the reaction is about 99.9% complete. Reaching 99% takes about 6.6 half-lives. A first-order reaction never strictly finishes; it only approaches completion, so two half-lives leave a quarter of the reactant, not none.
Why is the acid hydrolysis of an ester called pseudo first order?
Water is a reactant in the hydrolysis of ethyl acetate, but it is present in such large excess that its concentration barely changes. That nearly constant value gets absorbed into the rate constant, so the rate seems to depend only on the ester concentration. A reaction that really involves two reactants then behaves as first order. The inversion of cane sugar is another example.
Temperature and the Arrhenius equation
Read this section in the notes →Why does a 10 °C rise in temperature roughly double the rate of many reactions?
A 10 °C rise increases the average kinetic energy only a little, but it sharply increases the fraction of molecules whose energy reaches the activation energy. For many reactions that fraction roughly doubles, so the rate constant roughly doubles as well. The effect comes from the exponential term e^(−Ea/RT), not from a big increase in how often molecules collide.
How do you calculate activation energy from rate constants at two temperatures?
Use log(k₂/k₁) = [Ea/(2.303R)] × (T₂ − T₁)/(T₁T₂). Insert the two rate constants and the two absolute temperatures, then solve for Ea; with R = 8.314 J K⁻¹ mol⁻¹ the answer comes out in J mol⁻¹. Temperatures must be in kelvin. Alternatively, a plot of ln k against 1/T gives a straight line of slope −Ea/R.
What does the term e^(−Ea/RT) in the Arrhenius equation mean?
It is the fraction of molecules, or of collisions, that carry energy equal to or greater than the activation energy. A high activation energy or a low temperature makes this fraction very small, so the reaction is slow. In k = A e^(−Ea/RT), the factor A reflects how often suitably placed molecules collide, while the exponential term picks out those energetic enough to react.
What is the difference between threshold energy and activation energy?
Threshold energy is the minimum total energy that colliding molecules must have for a reaction to take place. Activation energy is the additional energy they must gain beyond what they already possess on average, so threshold energy = activation energy + average energy of the reactants. Lowering the activation energy lets more molecules reach the threshold at the same temperature.
Effect of a catalyst
Read this section in the notes →How does a catalyst speed up a reaction without changing the equilibrium?
A catalyst opens an alternative pathway with a lower activation energy, so more molecules can react at the same temperature. It lowers the barrier for the forward and reverse reactions by the same amount, so ΔG, ΔH and the equilibrium constant all stay unchanged. It cannot make a non-spontaneous reaction go; it only helps a feasible reaction reach equilibrium sooner.
Collision theory
Read this section in the notes →Why doesn't every collision between molecules lead to a reaction?
A collision is effective only if the molecules have enough energy and also meet in a suitable orientation, so the atoms that need to bond actually come into contact. Many energetic collisions hit at the wrong angle and the molecules simply bounce apart. Collision theory allows for this with the steric or probability factor P, writing the rate as P Z_AB e^(−Ea/RT).
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