Chemistry of Solutions 202-SN2-RE • CEGEP, Montreal

Revision sheet: chemical kinetics (202-SN2-RE)

Kinetics answers a question that thermodynamics ignores: not whether a reaction is possible, but how fast it happens. Two tools are enough, the rate law and the Arrhenius equation, plus one idea: it is the slowest step of a mechanism that sets the pace of the whole.

This sheet fixes the three integrated rate laws and their graphical signatures, separates what comes from experiment from what comes from the equation, and gathers the pitfalls that come back from one exam to the next on half-life, on the Arrhenius equation and on mechanisms.

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The thread of the chapter

The orders of a rate law are EXPERIMENTAL: they are never read off the coefficients of the equation. Believing otherwise gives a wrong rate law and makes the rest of the question pointless.

This chapter is part of 202-SN2-RE, Chemistry of Solutions

The essentials

Rate, rate law and orders

• For $aA+bB\rightarrow cC+dD$ : $v=-\dfrac{1}{a}\dfrac{\Delta[A]}{\Delta t}=+\dfrac{1}{c}\dfrac{\Delta[C]}{\Delta t}$ . A SINGLE rate for the reaction, each species divided by its coefficient.
• Rate law: $v=k[A]^{m}[B]^{n}$ . The orders $m$ and $n$ are EXPERIMENTAL and have no connection with $a$ and $b$ .
• Method of initial rates: double ONE concentration at a time. A rate that stays the same gives order $0$ , a doubled rate order $1$ , a quadrupled rate order $2$ .
• The units of $k$ depend on the OVERALL order: $\text{mol·L}^{-1}\text{s}^{-1}$ at order $0$ , $\text{s}^{-1}$ at order $1$ , $\text{L·mol}^{-1}\text{s}^{-1}$ at order $2$ .
• The only exception: in an ELEMENTARY step of a mechanism, the orders coincide with the coefficients, because it describes an actual collision.

We double [A]₀: the initial rate stays the same at order

0

, doubles at order

1

, quadruples at order

2

. That is the whole method of initial rates.

The unit of $k$ is a free check: it reveals the overall order without any data. A constant in $\text{s}^{-1}$ announces an order $1$ reaction, and nothing else.

The three integrated rate laws and their signature

• Order $0$ : $[A]_{t}=[A]_{0}-kt$ . Straight line of $[A]$ versus $t$ . Half-life $t_{1/2}=\dfrac{[A]_{0}}{2k}$ , which DECREASES over time.
• Order $1$ : $\ln[A]_{t}=\ln[A]_{0}-kt$ . Straight line of $\ln[A]$ versus $t$ . Half-life $t_{1/2}=\dfrac{\ln 2}{k}$ , CONSTANT: this is the signature of order $1$ .
• Order $2$ : $\dfrac{1}{[A]_{t}}=\dfrac{1}{[A]_{0}}+kt$ . Straight line of $\dfrac{1}{[A]}$ versus $t$ . Half-life $t_{1/2}=\dfrac{1}{k[A]_{0}}$ , which INCREASES over time.
• The graph that gives a STRAIGHT LINE designates the order: this is the surest method when the problem supplies a table of concentrations.
• Order $1$ : the time to reach a given percentage of conversion does NOT depend on the initial concentration.

Each order has ITS line:

[A]

for order

0

\ln[A]

for order

1

\dfrac{1}{[A]}

for order

2

, the only one that RISES, with slope

+k

The half-life is the quickest test: measuring two successive half-lives is enough to decide among the three orders, without plotting any graph.

Arrhenius, catalysis and mechanisms

• $k=Ae^{-E_{a}/RT}$ , with $R=8.314$ J/(mol·K) and $T$ IN KELVINS. Two-temperature form: $\ln\dfrac{k_{2}}{k_{1}}=\dfrac{E_{a}}{R}\left(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}\right)$ .
• Graphically, $\ln k$ versus $\dfrac{1}{T}$ is a STRAIGHT LINE with slope $-\dfrac{E_{a}}{R}$ : this is how an activation energy is measured.
• A catalyst opens a new pathway with a lower activation energy. It speeds up BOTH directions equally, so it does not shift the equilibrium and does not change $\Delta H$ .
• A mechanism is a sequence of ELEMENTARY steps whose sum gives back the overall equation. The SLOW step imposes its rate law.
• An INTERMEDIATE is produced then consumed; a CATALYST is consumed then regenerated. Neither appears in the overall equation.

The slope is

-\dfrac{E_{a}}{R}

: the steeper line, the one with the larger activation energy, gains the most when you heat, that is, when you move to the left.

A rate law must never contain an intermediate: if one is left after writing the slow step, it must be eliminated using the pre-equilibrium of the fast step that precedes it.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading the orders off the stoichiometric coefficients

the whole question, 6 points when the orders differ from the coefficients

What not to write

“The equation is $2NO+O_{2}\rightarrow 2NO_{2}$ , so $v=k[NO]^{2}[O_{2}]$ .”

What to write

“The orders are EXPERIMENTAL: they must be determined by the method of initial rates. Here they do turn out to be $2$ and $1$ , but nothing in the equation guaranteed it.”

Why: The overall equation describes a balance of matter, not a mechanism. Only an ELEMENTARY step, which describes an actual collision between molecules, has orders equal to its coefficients.

2. Forgetting to divide by the coefficient in the rate of reaction

3 points, and every species-specific rate wrong

What not to write

“ $[O_{2}]$ increases by $0.020$ mol/L in $10$ s, so the rate of the reaction $2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2}$ is $2.0\times 10^{-3}$ mol/(L·s) for every species.”

What to write

“ $v=+\dfrac{\Delta[O_{2}]}{\Delta t}=2.0\times 10^{-3}$ , but then $-\dfrac{\Delta[N_{2}O_{5}]}{\Delta t}=2v=4.0\times 10^{-3}$ and $\dfrac{\Delta[NO_{2}]}{\Delta t}=4v=8.0\times 10^{-3}$ mol/(L·s).”

A single rate of reaction, three slopes:

N_{2}O_{5}

disappears twice as fast as the rate,

NO_{2}

appears four times as fast,

O_{2}

at the pace of the rate itself.

Why: There is only ONE rate of reaction, but as many rates of disappearance or appearance as there are species. The coefficients link them, and forgetting them amounts to violating the stoichiometry.

3. Believing the half-life is always constant

the whole question, 5 points

What not to write

“The first half-life is $45$ min, so a quarter remains after $90$ min, whatever the order.”

What to write

“The half-life is CONSTANT only at order $1$ . At order $2$ it DOUBLES each time, at order $0$ it is HALVED.”

Same first half-life of

45

min, three different continuations: at

90

min, a quarter remains at order

1

only, a third at order

2

and nothing at order

0

Why: The half-life of order $2$ is $\dfrac{1}{k[A]_{0}}$ : since the concentration has been halved, the next half-life is twice as long. This behaviour is precisely what identifies the order.

4. Leaving temperatures in degrees Celsius in the Arrhenius equation

the whole question, 5 points

What not to write

“ $\ln\dfrac{k_{2}}{k_{1}}=\dfrac{E_{a}}{R}\left(\dfrac{1}{25}-\dfrac{1}{45}\right)$ .”

What to write

“In kelvins: $\dfrac{1}{298.15}-\dfrac{1}{318.15}=2.11\times 10^{-4}$ , against $1.78\times 10^{-2}$ in degrees Celsius, which is $84$ times too large.”

The same heating from

25

45

°C: in degrees Celsius, the gap between the reciprocals is

84

times too large, and the exponential turns this factor into an absurdity.

Why: The exponential amplifies the error catastrophically: the speed-up factor would go from $10$ to a number with eighty-five digits. Fortunately the absurdity of the result is immediate.

5. Believing a catalyst shifts the equilibrium or changes the enthalpy

4 points on the reasoning question

What not to write

“The catalyst speeds up the forward reaction, so it increases the equilibrium yield.”

What to write

“It lowers $E_{a}$ in BOTH directions by the same amount, so it multiplies $k$ and $k'$ by the same factor: $K=\dfrac{k}{k'}$ is unchanged. It gets you to the same equilibrium faster.”

Both curves start from the SAME level and end at the SAME level: only the height of the bump changes. The enthalpy of reaction, hence the equilibrium constant, is not affected.

Why: The enthalpy of reaction is the difference between the levels of the reactants and the products, which a catalyst does not touch: it changes only the height of the barrier between them.

6. Keeping an intermediate in the final rate law

the whole question, 6 points

What not to write

“The slow step is $NO_{3}+CO\rightarrow NO_{2}+CO_{2}$ , so $v=k[NO_{3}][CO]$ .”

What to write

“ $NO_{3}$ is an INTERMEDIATE: it must be eliminated using the pre-equilibrium of the fast step that produces it, which gives a law in terms of the starting reactants only.”

NO_{3}

sits in the valley between the two bumps: formed in the fast step, consumed in the slow step, it is neither reactant nor product and must not stay in the rate law.

Why: A rate law must be experimentally checkable, so it must contain only species whose concentration can be measured. An intermediate exists only in tiny, fleeting amounts.

7. Getting the units of the rate constant wrong

2 points, but the question is almost systematic

What not to write

“ $k$ is always expressed in mol/(L·s).”

What to write

“The units depend on the OVERALL order, so that $k[A]^{m}[B]^{n}$ always has the units of a rate: $\text{s}^{-1}$ at order $1$ , $\text{L·mol}^{-1}\text{s}^{-1}$ at order $2$ .”

Why: The unit of $k$ carries the information on the overall order. It is also the quickest check of a result: a constant found in s $^{-1}$ for an order $2$ reaction signals an error.

8. Confusing average rate and instantaneous rate

3 points

What not to write

“The concentration goes from $1.00$ to $0.40$ mol/L in $60$ s: the rate at $t=60$ s is $1.0\times 10^{-2}$ mol/(L·s).”

What to write

“This calculation gives the AVERAGE rate over the interval. The instantaneous rate at $t=60$ s is the slope of the TANGENT to the curve at that point, smaller here since the reaction is slowing down.”

The red secant line gives the AVERAGE rate over

60

s; the green tangent, less steep, gives the instantaneous rate at

t=60

s, smaller since the reaction is slowing down.

Why: A reaction slows down as the reactants are used up: the secant line between two points always overestimates the rate at the end point. It is the same distinction as between average rate of change and derivative in differential calculus.

9. Believing a zero-order reaction has no rate constant

3 points

What not to write

“The order is zero in $[A]$ , so the rate depends on nothing and the reaction is instantaneous.”

What to write

“ $v=k$ : the rate is CONSTANT, equal to $k$ , as long as reactant remains. It does not depend on the concentration, which is not the same thing as not existing.”

Why: Zero order appears when a catalyst or a surface is saturated: the rate is then set by the catalyst, not by the concentration. The elimination of alcohol by the liver is an example.

10. Adding up the steps of a mechanism without checking the overall equation

3 points on the validation question

What not to write

“I propose two steps, the slow step gives the right rate law, so the mechanism is valid.”

What to write

“A mechanism must meet TWO conditions: the sum of its steps reproduces the overall equation, AND the slow step gives the experimental rate law.”

Why: A mechanism is never proved, it is only consistent with experiment. A balance that does not come back to the overall equation disqualifies the proposal at once, whatever its rate law.

Which method to choose

How to determine the order

The nature of the data supplied by the problem

The same measurements plotted three times: only the curve of

\dfrac{1}{[A]}

is a straight line, so the reaction is of order

2

If several experiments with initial concentrations and initial rates → method of initial rates: compare two runs in which ONE concentration only changes

Example: concentration doubled and rate quadrupled gives order $2$
If a table of concentration against time → test the three graphs and keep the one that gives a straight line

Example: $\ln[A]$ linear in $t$ designates order $1$
If two successive half-lives → equal gives order $1$ , doubled gives order $2$ , halved gives order $0$

Example: $45$ min then $45$ min: order $1$
If the problem gives the units of $k$ → they fix the overall order with no other calculation

Example: $\text{s}^{-1}$ gives overall order $1$
If the problem gives a MECHANISM and asks for the rate law → write the law of the slow step with its coefficients, then eliminate any intermediate

this is the only case where the coefficients directly give the orders

The direction of the deduction never reverses: experiment gives the orders, and the orders let you propose a mechanism. Starting from the mechanism to predict the law is legitimate only if the problem supplies that mechanism.

Which Arrhenius formula to use

What the problem supplies and what it asks for

Heating slides the point along ITS line; catalyzing makes it jump, at the same temperature, onto a less steep line. The equivalent temperature is read at the same height.

If two rate constants at two temperatures → $\ln\dfrac{k_{2}}{k_{1}}=\dfrac{E_{a}}{R}\left(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}\right)$ to obtain $E_{a}$

Example: a ratio of $10.3$ between $25$ and $45$ °C gives about $92$ kJ/mol
If $E_{a}$ and a constant at one temperature, and the other is asked for → same formula, solved for $k_{2}$

Example: $k$ multiplied by $10.3$ going from $25$ to $45$ °C
If the problem compares the same reaction with and without a catalyst AT THE SAME temperature → $\ln\dfrac{k'}{k}=\dfrac{E_{a}-E'_{a}}{RT}$ , the temperature cancelling out

Example: $17$ kJ/mol less at $25$ °C multiplies $k$ by $950$
If the problem gives a table of $k$ at several temperatures → plot $\ln k$ against $\dfrac{1}{T}$ and read $E_{a}$ from the slope, which equals $-\dfrac{E_{a}}{R}$

Example: a slope of $-11\,066$ K gives $E_{a}=92$ kJ/mol
If the problem asks for the temperature that would have the same effect as a catalyst → equate the two speed-up factors and solve for $T$

this is the most frequent synthesis question on this chapter

Always check the DIRECTION of the result: heating speeds things up, so $k_{2}>k_{1}$ if $T_{2}>T_{1}$ . A ratio below $1$ signals that the two temperatures were swapped.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Determining a rate law by initial rates

When to use it: The problem supplies a table of runs with the initial concentrations and the corresponding initial rates.

1 Pick two runs in which ONE concentration only changes, and say so explicitly: “between runs 1 and 2, only $[A]$ varies.”
2 Calculate the ratio of the concentrations and that of the rates, then deduce the order from $\dfrac{v_{2}}{v_{1}}=\left(\dfrac{[A]_{2}}{[A]_{1}}\right)^{m}$ .
3 Repeat for each reactant, with a suitable pair of runs.
4 Write the complete rate law, then calculate $k$ by substituting the data of ANY ONE run.
5 Give $k$ with its units, deduced from the overall order, and check on a second run.

Concluding sentence

“Between runs $1$ and $2$ , only $[A]$ doubles and the rate quadruples: the order with respect to $A$ is $2$ . Between runs $1$ and $3$ , only $[B]$ doubles and the rate doubles: the order is $1$ . The law is therefore $v=k[A]^{2}[B]$ with $k$ in $\text{L}^{2}\text{mol}^{-2}\text{s}^{-1}$ .”

The trap: Comparing two runs in which TWO concentrations change at once. The ratio of rates then mixes the two orders and no longer allows a conclusion on either.

Marking: Typically 2 points per order correctly justified by a pair of runs, 2 points for the value of $k$ , 1 point for its units.

Establishing the rate law of a pre-equilibrium mechanism

When to use it: The problem gives a mechanism whose first step is a fast equilibrium and whose second is the slow step.

1 Write the rate law of the SLOW STEP, using its coefficients as orders, since it is elementary.
2 Spot the intermediate present in this expression and name it.
3 Write the pre-equilibrium condition of the fast step: forward rate equals reverse rate, which gives the intermediate in terms of the reactants.
4 Substitute and gather all the constants into a single apparent constant $k_{obs}$ .
5 Compare the law obtained with the experimental law and conclude on the compatibility of the mechanism.

Concluding sentence

“The slow step gives $v=k_{2}[I][B]$ ; the pre-equilibrium of the first step imposes $[I]=\dfrac{k_{1}}{k_{-1}}[A]^{2}$ , hence $v=\dfrac{k_{1}k_{2}}{k_{-1}}[A]^{2}[B]=k_{obs}[A]^{2}[B]$ , in agreement with the experimental law.”

The trap: Forgetting to check that the sum of the steps gives back the overall equation. A mechanism that gives the right rate law but the wrong overall equation is rejected, and this check is often worth a point on its own.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The units of $k$ against the overall order

Check that $k[A]^{m}[B]^{n}$ does give moles per litre per second. This is a check that needs no data.

A $k$ in s $^{-1}$ for a law $v=k[A]^{2}$ : one of the two is wrong.
The direction of the temperature effect

Heating always speeds things up: the ratio $\dfrac{k_{2}}{k_{1}}$ must be greater than $1$ if $T_{2}>T_{1}$ .

A ratio of $0.097$ for a heating: the two temperatures were swapped inside the parentheses.
The consistency of successive half-lives

Check that the behaviour of the half-lives matches the order announced: constant at order $1$ , doubling at order $2$ .

An order $1$ announced with half-lives of $45$ then $90$ min: it is order $2$ .
The absence of an intermediate in the final law

Reread the rate law obtained and check that every species appearing in it also appears in the overall equation.

A law containing $NO_{3}$ when the overall equation does not mention it: the pre-equilibrium was not used.
The order of magnitude of an activation energy

An activation energy is typically $40$ to $200$ kJ/mol. A value of a few joules or several thousand kilojoules signals a unit error.

An $E_{a}=92$ J/mol: the factor of one thousand between joules and kilojoules was forgotten.

The typical problem, taken apart

Heating or catalyzing: two ways of speeding up the same reaction

A first-order reaction has a half-life of $45.0$ min at $25.0$ °C and an activation energy of $92.0$ kJ/mol.

Calculate its rate constant at $25.0$ °C, then the time needed to convert $90.0\%$ of the reactant. Next, determine the factor by which the constant is multiplied if the mixture is brought to $45.0$ °C, then if a catalyst that lowers the activation energy to $75.0$ kJ/mol is added at $25.0$ °C. Finally, at what temperature would one have to work without a catalyst to obtain the same speed-up?

Given: $R=8.314$ J/(mol·K).

The same first-order reaction in four situations: we heat from

25.0

45.0

°C, or add a catalyst at

25.0

°C, then look for the equivalent temperature.

Step 1

$k=\dfrac{\ln 2}{t_{1/2}}=\dfrac{0.693}{45.0}=1.54\times 10^{-2}$ min $^{-1}$ .

Why

The relation between half-life and constant is valid only at order $1$ , which the problem states. The units in min $^{-1}$ confirm overall order $1$ , incidentally.

Step 2

Conversion of $90.0\%$ : $10\%$ remains, so $\ln\dfrac{[A]_{0}}{[A]}=\ln 10=2.303=kt$ , hence $t=\dfrac{2.303}{1.54\times 10^{-2}}=150$ min.

Why

At order $1$ , this time does NOT depend on the initial concentration: this is what makes the notion of half-life usable, and what sets order $1$ apart from the other two.

Step 3

Heating: $\ln\dfrac{k_{2}}{k_{1}}=\dfrac{92\,000}{8.314}\left(\dfrac{1}{298.15}-\dfrac{1}{318.15}\right)=11\,066\times 2.108\times 10^{-4}=2.33$ .

Why

Both temperatures are converted to kelvins on this very line, and the gap between the reciprocals is of the order of $10^{-4}$ : this is the check that confirms the conversion was done.

Step 4

So $\dfrac{k_{2}}{k_{1}}=e^{2.33}=10.3$ : heating by $20$ degrees multiplies the rate by about $10$ .

Why

The rule of thumb that the rate doubles every ten degrees would give a factor of $4$ ; the difference comes from the activation energy, which is larger than that of common reactions.

Step 5

Catalysis at constant temperature: $\ln\dfrac{k'}{k}=\dfrac{E_{a}-E'_{a}}{RT}=\dfrac{92\,000-75\,000}{8.314\times 298.15}=\dfrac{17\,000}{2478.8}=6.86$ , so $\dfrac{k'}{k}=e^{6.86}=951$ .

Why

The temperature cancels since it is the same in both cases: the formula reduces to a difference of activation energies, which makes the calculation much shorter than the general form.

Step 6

The catalyst therefore multiplies the rate by $951$ , against $10.3$ for a $20$ -degree heating: it is nearly a hundred times more effective.

Why

The comparison is the real question. It explains why industry invests in catalysts rather than in heating, whose energy cost grows without limit.

Step 7

Equivalent temperature: $\ln 951=6.86=11\,066\left(\dfrac{1}{298.15}-\dfrac{1}{T}\right)$ , hence $\dfrac{1}{T}=3.354\times 10^{-3}-6.20\times 10^{-4}=2.734\times 10^{-3}$ and $T=366$ K, that is $92.5$ °C.

Why

We invert the Arrhenius equation rather than feeling our way by trial and error. The result gives a concrete measure of what the catalyst contributes: at room temperature, it does what would otherwise require heating to nearly a hundred degrees.

Step 8

Check: at $45$ °C the half-life falls to $\dfrac{45.0}{10.3}=4.4$ min, and with the catalyst at $25$ °C it falls to $\dfrac{45.0}{951}=2.8$ s.

Why

Translating the factors into half-lives makes the orders of magnitude immediately readable, and lets you check that the two speed-ups go in the same direction as the numbers calculated.

The conclusion, written out

“The constant is $1.54\times 10^{-2}$ min $^{-1}$ at $25$ °C and it takes $150$ min to convert $90\%$ of the reactant. Heating to $45$ °C multiplies the rate by $10.3$ , whereas the catalyst multiplies it by $951$ : to get the same effect without it, the mixture would have to be brought to $92.5$ °C.”

The classic mistake on this problem: Using $\dfrac{1}{25}-\dfrac{1}{45}$ in the Arrhenius equation. The gap between the reciprocals is then $84$ times too large, and the speed-up factor obtained exceeds anything physically conceivable: the absurdity of the result is the best alarm signal of the chapter.

Learn by heart

• A single rate for the reaction: each species divided by its COEFFICIENT.
• The orders are EXPERIMENTAL, except in an elementary step of a mechanism.
• Order $0$ : straight line in $[A]$ . Order $1$ : straight line in $\ln[A]$ . Order $2$ : straight line in $\dfrac{1}{[A]}$ .
• Half-life CONSTANT only at order $1$ : $t_{1/2}=\dfrac{\ln 2}{k}$ . It doubles at order $2$ .
• The units of $k$ give the overall order: s $^{-1}$ at order $1$ .
• Arrhenius: $\ln\dfrac{k_{2}}{k_{1}}=\dfrac{E_{a}}{R}\left(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}\right)$ , in KELVINS. Graphically, slope $-\dfrac{E_{a}}{R}$ .
• A catalyst lowers $E_{a}$ in both directions: it affects neither $\Delta H$ nor $K$ .
• The SLOW step imposes the rate law, and any intermediate must be eliminated from it.

Frequently asked questions

Can the orders of a reaction be read off its equation?

No, except for an elementary step of a mechanism. The overall equation describes a material balance, whereas the rate law describes what happens in the slowest step. Orders are determined experimentally, most often by the method of initial rates, and they can even be fractional or negative.

How do I recognize a first-order reaction?

By its constant half-life: it always takes the same time for the concentration to be halved, whatever the starting value. Graphically, it is the logarithm of the concentration that varies linearly with time. At second order the half-life doubles each time; at zero order it is halved.

Does a catalyst change the yield of a reaction?

No. It lowers the energy barrier in both directions by the same amount, so it speeds up the forward and reverse reactions equally. The equilibrium constant, which is the ratio of the two rate constants, stays unchanged. You reach the same equilibrium, just much faster.

Why do I need kelvins in the Arrhenius equation?

Because the temperature appears there as a reciprocal inside an exponential, which only makes sense on an absolute scale. In degrees Celsius, the gap between the reciprocals is several dozen times too large, and since it then goes through an exponential, the result becomes astronomical.

How do I find the rate law from a mechanism?

You write the rate law of the slow step, using its coefficients as orders since it is elementary. If this expression contains an intermediate, you eliminate it using the fast-equilibrium condition of the step that produces it. The final law must contain only species present in the overall equation.

What does a zero-order reaction mean?

That its rate does not depend on the concentration of the reactant: it stays constant as long as any remains. This typically happens when a catalyst or a surface is saturated, so that it is the catalyst, not the reactant, that limits the throughput. The elimination of alcohol by the liver is the classic example.

Practise it

Corrected exercises: Chemical kinetics, 202-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn2-chemical-kinetics. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: chemical kinetics (202-SN2-RE)

The essentials

Rate, rate law and orders

The three integrated rate laws and their signature

Arrhenius, catalysis and mechanisms

The mistakes that cost marks

1. Reading the orders off the stoichiometric coefficients

2. Forgetting to divide by the coefficient in the rate of reaction

3. Believing the half-life is always constant

4. Leaving temperatures in degrees Celsius in the Arrhenius equation

5. Believing a catalyst shifts the equilibrium or changes the enthalpy

6. Keeping an intermediate in the final rate law

7. Getting the units of the rate constant wrong

8. Confusing average rate and instantaneous rate

9. Believing a zero-order reaction has no rate constant

10. Adding up the steps of a mechanism without checking the overall equation

Which method to choose

How to determine the order

Which Arrhenius formula to use

How the answer is expected to be written

Determining a rate law by initial rates

Establishing the rate law of a pre-equilibrium mechanism

Check before you hand in

The typical problem, taken apart

Heating or catalyzing: two ways of speeding up the same reaction

Learn by heart

Frequently asked questions

Can the orders of a reaction be read off its equation?

How do I recognize a first-order reaction?

Does a catalyst change the yield of a reaction?

Why do I need kelvins in the Arrhenius equation?

How do I find the rate law from a mechanism?

What does a zero-order reaction mean?

Corrected exercises: Chemical kinetics, 202-SN2-RE

See also

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