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

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

Q: Why do solids not appear in the equilibrium constant?

Because the concentration of a pure solid is fixed by its density and does not change when some is added or removed. This constant value is already built into the equilibrium constant. This is why the decomposition of limestone does not depend on the amount of limestone present.

Q: Which disturbance changes the value of the equilibrium constant?

Only one: temperature. Adding a reactant, removing a product, compressing, diluting or adding a catalyst shifts the position of equilibrium without touching the constant. Heating or cooling, on the other hand, really does change it, in the direction that favours the endothermic reaction when you heat.

Q: When can you neglect x next to the initial concentration?

When it represents less than five percent of that concentration, which must be checked after calculating the approximate value. If the test fails, you have to start again with the full quadratic equation. The approximation is the safer the smaller the constant is compared with the initial concentration.

Q: Does a catalyst improve the yield of a reaction at equilibrium?

No. It speeds up the forward and the reverse reactions in the same proportion, so the ratio of the two rate constants, which is exactly the equilibrium constant, stays unchanged. The gain is a gain in time: you reach exactly the same equilibrium sooner.

Q: Why is a weak acid more dissociated when you dilute it?

Because dissociation produces more particles than it consumes. Diluting is equivalent to lowering the pressure, and the system responds by shifting toward the side with the most particles. The percentage dissociated therefore increases, even though the ion concentration decreases in absolute value.

Q: What is the difference between Kp and Kc?

Kc is written with concentrations, Kp with partial pressures. They are related by the product of the gas constant and the temperature, raised to the power of the change in the number of moles of gas. They coincide only if that change is zero, that is, if there are as many moles of gas on each side.

A chemical equilibrium is not a standstill: both reactions carry on at the same rate. The constant $K$ measures the position of this compromise, and the quotient $Q$ says where the system stands at a given moment. Comparing the two is enough to predict the direction of change, with no further calculation.

This sheet gives the expression of $K$ with its exclusions, the ICE method with its validity test, and the list of disturbances, separating those that shift the equilibrium from the only one that changes the constant.

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The thread of the chapter

Only one disturbance changes the VALUE of $K$ : temperature. Everything else, adding a reactant, changing the volume, a catalyst, shifts the position of equilibrium without touching the constant, and confusing the two costs the reasoning question on every exam.

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

The essentials

Writing $K$ , and comparing $Q$ with $K$

• For $aA+bB\rightleftharpoons cC+dD$ : $K_{c}=\dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$ , with the concentrations AT EQUILIBRIUM.
• Pure SOLIDS and LIQUIDS do NOT appear in it: their concentration is a constant already absorbed into $K$ .
• The quotient $Q$ has exactly the same expression, but with the CURRENT concentrations, at any moment.
• $Q<K$ : the system moves toward the PRODUCTS. $Q>K$ : toward the REACTANTS. $Q=K$ : equilibrium has been reached.
• $K$ says nothing about the RATE: a reaction can have an enormous $K$ and still not occur at room temperature.

The system ALWAYS moves toward

K

, never away from it. Comparing

Q

with

K

therefore gives the direction of change without any calculation of final concentrations.

Leaving out the solids is what makes the chemistry of precipitates and calcination possible: the position of equilibrium does not depend on the amount of solid present, only on its presence.

The ICE table, and the test that validates it

• Three rows: Initial, Change, Equilibrium. The change carries the sign $-$ for reactants and $+$ for products, multiplied by the COEFFICIENT.
• Substitute the Equilibrium row into the expression of $K$ , then solve for $x$ .
• Small- $x$ approximation: assume $[A]_{0}-x\approx[A]_{0}$ . It is ONLY VALID if $x<5\%$ of $[A]_{0}$ , which must be CHECKED afterwards.
• If the test fails, solve the full quadratic equation, with no approximation.
• Degree of dissociation $\alpha=\dfrac{x}{[A]_{0}}$ , often expressed as a percentage: it is exactly the quantity that the $5\%$ rule tests.

The approximation is the more legitimate the smaller $K$ is compared with the initial concentration. In practice, a $K$ below $10^{-4}$ for a concentration of the order of $0.1$ mol/L makes it almost always valid.

Le Chatelier: what shifts the equilibrium and what changes $K$

• ADDING a species: the system moves away from the added species. The position moves, $K$ does not.
• REDUCED volume, hence higher pressure: the system goes toward the side with FEWER moles of GAS. The position moves, $K$ does not.
• INERT GAS added at constant volume: nothing moves at all, since no concentration changes.
• CATALYST: equilibrium is reached sooner, the position and $K$ are unchanged.
• TEMPERATURE: it is the ONLY disturbance that changes the value of $K$ . Treat heat as a reactant if the reaction is endothermic, as a product if it is exothermic.

Heating means moving to the LEFT on this axis. The exothermic line goes down there, the endothermic line goes up: temperature really is the only disturbance that changes the value of

K

A DILUTION is equivalent to a drop in pressure: the system goes toward the side with the MOST moles of gas, hence toward dissociation. This is why a weak acid is more dissociated the more dilute it is.

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. Putting a pure solid or liquid into the expression of $K$

the whole question, 5 marks

What not to write

“For $CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)$ : $K_{c}=\dfrac{[CaO][CO_{2}]}{[CaCO_{3}]}$ .”

What to write

“Both solids are left out: $K_{c}=[CO_{2}]$ . The position of equilibrium therefore does not depend on the amount of limestone present.”

Little or a lot of limestone, at the same temperature, the same amount of carbon dioxide: solids do not enter into

K_{c}=[CO_{2}]

Why: The concentration of a pure solid is fixed by its density: it does not vary when more is added. This constant is already absorbed into the value of $K$ .

2. Using the initial concentrations in the expression of $K$

the whole question, 6 marks

What not to write

“ $0.100$ mol/L of $N_{2}O_{4}$ is introduced: $K_{c}=\dfrac{[NO_{2}]^{2}}{[N_{2}O_{4}]}=\dfrac{0}{0.100}=0$ .”

What to write

“ $K$ is written with the concentrations AT EQUILIBRIUM: you first need an ICE table to obtain them, here $[NO_{2}]=0.0204$ and $[N_{2}O_{4}]=0.0898$ mol/L.”

At the start,

0.100

and

0

: this is not equilibrium. It is the plateau,

0.0898

and

0.0204

mol/L, that goes into

K

Why: The initial state is almost never the equilibrium state: the gap between the two is precisely what the ICE table calculates. Confusing the two amounts to assuming that nothing happens.

3. Using the small- $x$ approximation without testing it

3 marks, plus the rigour mark for the test

What not to write

“ $\dfrac{4x^{2}}{0.100}=4.63\times 10^{-3}$ , so $x=1.08\times 10^{-2}$ , and that is the answer.”

What to write

“Test: $\dfrac{1.08\times 10^{-2}}{0.100}=10.8\%>5\%$ , so the approximation is NOT valid. Solving the full quadratic gives $x=1.02\times 10^{-2}$ .”

0.100

mol/L, the approximation gives

10.8\,\%

against

10.2\,\%

: the

5\,\%

test rejects it. At low concentrations it even exceeds

100\,\%

Why: The error here is only $5.5\%$ , so invisible without the test. With larger constants, it reaches tens of percent, and the $5\%$ rule is there precisely to settle the matter.

4. Believing that adding a reactant increases the equilibrium constant

4 marks on the reasoning question

What not to write

“Dinitrogen is added to the Haber mixture, so $K$ increases and the yield improves.”

What to write

“ $K$ is UNCHANGED: only the POSITION of equilibrium shifts toward the products. The amount of ammonia increases, but the ratio that defines $K$ returns to exactly its value.”

Doubling

N_{2}

divides

Q

by two; the system evolves until

Q

recovers the SAME value of

K

. Only the position has moved.

Why: $K$ depends only on temperature. An addition changes $Q$ , which is no longer equal to $K$ , and the system evolves until $Q$ is again equal to that same value of $K$ .

5. Forgetting the factor $(RT)^{\Delta n}$ between $K_{p}$ and $K_{c}$

the whole question, 5 marks

What not to write

“ $K_{p}=K_{c}$ , they are just two ways of writing the same thing.”

What to write

“ $K_{p}=K_{c}(RT)^{\Delta n_{\text{gas}}}$ . For $N_{2}O_{4}\rightleftharpoons 2NO_{2}$ , $\Delta n=1$ , so $K_{p}=4.63\times 10^{-3}\times 8.314\times 298.15=11.5$ (in kPa, with R in kPa·L/(mol·K)).”

Why: The two constants coincide only if $\Delta n_{\text{gas}}=0$ . The factor $RT$ is about $2479$ at room temperature in units of kPa·L, so the gap is huge as soon as $\Delta n$ is not zero.

6. Counting all the moles instead of only the moles of gas for the effect of pressure

4 marks

What not to write

“ $C(s)+CO_{2}(g)\rightleftharpoons 2CO(g)$ : two moles on the left and two on the right, so pressure has no effect.”

What to write

“Only GASES count: ONE mole of gas on the left against TWO on the right. Compressing therefore shifts the equilibrium to the left.”

The solid does not count: one mole of gas on the left, two on the right. Compressing therefore pushes to the left.

Why: Pressure acts only on gases, whose volume can change. The volume of a solid or a liquid is negligible and practically incompressible, so it takes no part in the effect.

7. Believing that an inert gas added at constant volume shifts the equilibrium

4 marks on the reasoning question

What not to write

“Argon is injected into the reactor: the total pressure rises, so the equilibrium shifts toward the side with fewer moles of gas.”

What to write

“At constant VOLUME, adding an inert gas changes NO concentration or partial pressure: $Q$ stays equal to $K$ and nothing moves.”

The argon is added, but at constant volume the

N_{2}O_{4}

and

NO_{2}

molecules keep their concentrations:

Q

stays equal to

K

Why: Le Chatelier's principle applies to the quantities that appear in $Q$ , that is, to concentrations or PARTIAL pressures. The total pressure does not appear in it, unless the volume actually changes.

8. Crediting the catalyst with a better yield at equilibrium

3 marks

What not to write

“The catalyst in the Haber process increases the amount of ammonia obtained at equilibrium.”

What to write

“It speeds up BOTH directions equally: equilibrium is reached sooner, but at the same position and with the same value of $K$ .”

With or without a catalyst, the plateau is the same: equilibrium is reached sooner, at the same position (sketch).

Why: $K=\dfrac{k_{\text{forward}}}{k_{\text{reverse}}}$ : multiplying both rate constants by the same factor leaves the ratio unchanged. The industrial gain is a gain in time, not in yield.

9. Getting it wrong when manipulating an equilibrium constant

the whole question, 5 marks

What not to write

“If I double the coefficients, I double $K$ ; if I reverse the equation, I change the sign of $K$ .”

What to write

“Doubling the coefficients SQUARES $K$ ; reversing the equation gives $\dfrac{1}{K}$ ; adding two equations MULTIPLIES their constants.”

Why: $K$ is a product of powers of concentrations: operations on the equation therefore become multiplicative operations on $K$ , never additive ones. A constant can never be negative anyway.

10. Believing that dilution does not change the degree of dissociation

4 marks

What not to write

“Diluting ten times divides all the concentrations by ten, so the percentage dissociated stays the same.”

What to write

“Diluting favours the side with the MOST particles: the degree of dissociation goes here from $10.2\%$ to $28.7\%$ .”

Same constant, same amount of matter, and yet almost three times as much dissociation after dilution: the system moves toward the side with the most particles.

Why: The ratio that defines $K$ is not preserved under a uniform division when the exponents of the numerator and the denominator differ. It is exactly the analogue of a drop in pressure.

Which method to choose

What does the system do after a disturbance

The exact nature of the disturbance described in the problem

Volume cut in half: the system goes toward the side with the FEWEST moles of gas, two

NO_{2}

recombine into one

N_{2}O_{4}

, and

K_{c}

stays the same.

If a species that appears in $K$ is ADDED → the system consumes it: it moves away from the side where it was added. $K$ unchanged

Example: adding $N_{2}$ to the Haber process shifts toward ammonia
If a species is REMOVED → the system re-forms it: it goes toward the side where it is found. $K$ unchanged

Example: condensing the ammonia as it forms increases the conversion
If the volume is REDUCED → toward the side with the FEWEST moles of gas. $K$ unchanged

Example: $N_{2}+3H_{2}\rightleftharpoons 2NH_{3}$ goes to the right
If you DILUTE, or increase the volume → toward the side with the MOST moles or particles. $K$ unchanged

Example: a weak acid is more dissociated when dilute
If you ADD AN INERT GAS at constant volume → nothing happens: no concentration or partial pressure changes

Example: injecting argon into the reactor
If you CHANGE THE TEMPERATURE → the only disturbance that changes $K$ : heating favours the ENDOTHERMIC direction

treat heat as a reactant if $\Delta H>0$ , as a product if $\Delta H<0$

Two questions to ask in order: does the position move, and does $K$ change? The second answer is YES in a single case, that of temperature. Papers that answer the two separately never lose these marks.

How to attack an equilibrium calculation

What the problem provides and what it asks for

If all the equilibrium concentrations are given → substitute directly into the expression of $K$ , no table

Example: calculating $K$ from an analyzed mixture
If the initial concentrations are given and $K$ is known → ICE table, then solve for $x$ , with the $5\%$ test

Example: dissociation of $N_{2}O_{4}$ from $0.100$ mol/L
If the problem asks only for the DIRECTION of change → calculate $Q$ and compare it with $K$ , no table is needed

Example: $Q=0$ at the start, so the change is toward the products
If the expression in $x$ gives a perfect square → take the square root of both sides rather than expanding

Example: $\dfrac{x^{2}}{(a-x)^{2}}=K$ gives $\dfrac{x}{a-x}=\sqrt{K}$
If the problem gives $K$ at two temperatures → van 't Hoff: $\ln\dfrac{K_{2}}{K_{1}}=-\dfrac{\Delta H^{\circ}}{R}\left(\dfrac{1}{T_{2}}-\dfrac{1}{T_{1}}\right)$

the sign of $\Delta H$ obtained must agree with the direction in which $K$ varies

Spotting a perfect square saves five minutes and removes the risk of an error on the discriminant. It is the case whenever the number of moles is the same on both sides.

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.

Taking an ICE table all the way through

When to use it: The problem gives initial concentrations and a constant, and asks for the equilibrium concentrations or a degree of dissociation.

1 Write the balanced equation and the expression of $K$ , explicitly leaving out pure solids and liquids.
2 Set up the three-row table, multiplying $x$ by the COEFFICIENT of each species and respecting the signs.
3 Substitute the Equilibrium row into $K$ and write the equation in $x$ before any simplification.
4 Try the small- $x$ approximation, then TEST it: if $x$ exceeds $5\%$ of the initial concentration, start again with the full quadratic.
5 Reject the negative root, giving the reason, then calculate all the equilibrium concentrations.
6 Check by recalculating $K$ from the concentrations obtained.

Concluding sentence

“The table gives $\dfrac{(2x)^{2}}{0.100-x}=4.63\times 10^{-3}$ ; the approximation would lead to $x=1.08\times 10^{-2}$ , that is $10.8\%$ of the initial concentration, so it is rejected; the full solution gives $x=1.02\times 10^{-2}$ mol/L.”

The trap: Forgetting to multiply $x$ by the coefficient. For $N_{2}O_{4}\rightleftharpoons 2NO_{2}$ , the change in nitrogen dioxide is $+2x$ , and forgetting the factor $2$ changes $K$ by a factor of $4$ .

Marking: Typically 1 mark for the expression of $K$ , 2 for the table, 2 for the solution, 1 for the $5\%$ test, 1 for the check.

Answering a Le Chatelier question

When to use it: The problem describes a disturbance and asks in which direction the equilibrium shifts, or what becomes of a given quantity.

1 Name the disturbance precisely: addition, removal, change of volume, inert gas, catalyst, temperature.
2 Say first whether $K$ changes: the answer is NO, except for temperature.
3 Determine the direction of the shift by applying the corresponding rule, and justify it by counting moles of gas when it is a matter of pressure.
4 Conclude on what exactly becomes of the quantity asked for, which does not always move in the intuitive direction.

Concluding sentence

“The reduction in volume is an increase in pressure: $K$ stays unchanged, and the equilibrium shifts toward the side with the fewest moles of gas, that is toward ammonia, whose amount increases.”

The trap: Concluding about the CONCENTRATION when the question is about the AMOUNT, or the reverse. After a compression, a concentration can increase even though the equilibrium has shifted the other way.

Check before you hand in

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

$K$ recalculated from the concentrations found

Substitute the equilibrium concentrations obtained into the expression of $K$ : you must land back on the value in the problem. It is the most complete check of the chapter.

A recalculated $K$ of $1.9\times 10^{-2}$ instead of $4.63\times 10^{-3}$ : the coefficient $2$ was forgotten in the table.
The five percent test written out in full

After any approximation, calculate the ratio of $x$ to the initial concentration and conclude explicitly. This test is often worth a mark on its own.

An $x$ at $10.8\%$ of the initial concentration: the approximation had to be rejected.
The sign and plausibility of the concentrations

No equilibrium concentration can be negative. A root that gives $[A]<0$ is to be rejected, and you must say so.

An $x=0.12$ for an initial concentration of $0.100$ : the root chosen is the wrong one.
Consistency between the initial $Q$ and the direction found

Calculate $Q$ in the initial state and check that the direction of change in the ICE table matches the comparison with $K$ .

A table that consumes products while $Q<K$ : the signs of the Change row are reversed.
The order of magnitude of $K$ against the position of equilibrium

A very small $K$ means that equilibrium lies far to the left: the concentrations of products must be small. A very large $K$ gives the opposite.

A $K=4.6\times 10^{-3}$ with more products than reactants at equilibrium: this is incompatible.

The typical problem, taken apart

The dissociation of dinitrogen tetroxide, then the effect of dilution

The equilibrium $N_{2}O_{4}(g)\rightleftharpoons 2\,NO_{2}(g)$ has the constant $K_{c}=4.63\times 10^{-3}$ at $25.0$ °C.

$0.100$ mol of $N_{2}O_{4}$ is introduced into an empty $1.00$ L flask. Determine the equilibrium concentrations and the degree of dissociation. Then calculate $K_{p}$ .

The experiment is repeated with the same amount of matter in a $10.0$ L flask: what becomes of the degree of dissociation? Given: $R=8.314$ kPa·L/(mol·K).

The two experiments of the example: the same amount of

N_{2}O_{4}

in a volume ten times larger.

Step 1

Expression: $K_{c}=\dfrac{[NO_{2}]^{2}}{[N_{2}O_{4}]}$ , both species being gases. ICE table, in mol/L: initial $0.100$ and $0$ ; change $-x$ and $+2x$ ; equilibrium $0.100-x$ and $2x$ .

Why

The coefficient $2$ of nitrogen dioxide appears TWICE: once in the Change row and once as an exponent in $K$ . It is the double use that papers miss.

Step 2

Substitution: $\dfrac{(2x)^{2}}{0.100-x}=\dfrac{4x^{2}}{0.100-x}=4.63\times 10^{-3}$ .

Why

Write the complete equation before any simplification. It is the one you will come back to if the five percent test fails, and having it in front of you saves redoing the table.

Step 3

Trying the approximation: $4x^{2}\approx 4.63\times 10^{-4}$ , so $x\approx 1.08\times 10^{-2}$ . Test: $\dfrac{1.08\times 10^{-2}}{0.100}=10.8\%$ , greater than $5\%$ : the approximation is REJECTED.

Why

The test is done afterwards, on the value found, and its conclusion is written down. An approximation properly rejected earns the rigour mark; an approximation silently kept costs the question.

Step 4

Full quadratic: $4x^{2}+4.63\times 10^{-3}x-4.63\times 10^{-4}=0$ , hence $x=\dfrac{-4.63\times 10^{-3}+\sqrt{2.14\times 10^{-5}+7.41\times 10^{-3}}}{8}=1.02\times 10^{-2}$ mol/L. The negative root is rejected, since a concentration cannot be negative.

Why

Rejecting the negative root can be justified in one sentence. It is worth a mark, and without it the marker sees two possible answers and keeps neither.

Step 5

Equilibrium concentrations: $[N_{2}O_{4}]=0.100-0.0102=0.0898$ mol/L and $[NO_{2}]=2\times 0.0102=0.0204$ mol/L. Degree of dissociation: $\alpha=\dfrac{0.0102}{0.100}=10.2\%$ .

Why

The degree of dissociation is exactly the quantity tested by the five percent rule: the two numbers are the same, which gives a free check on the calculation.

Step 6

Check: $\dfrac{(0.0204)^{2}}{0.0898}=4.63\times 10^{-3}$ , in agreement with the problem.

Why

Recalculating $K$ from the concentrations found validates the table, the solution and the rounding all at once. It is the only check that tests the whole chain.

Step 7

$K_{p}$ : $\Delta n_{\text{gas}}=2-1=1$ , so $K_{p}=K_{c}(RT)^{1}=4.63\times 10^{-3}\times 8.314\times 298.15=11.5$ kPa.

Why

The $\Delta n$ count is made on gases only, products minus reactants. With $RT$ close to $2500$ , the gap between the two constants is more than three orders of magnitude.

Step 8

Dilution to $10.0$ L: $[N_{2}O_{4}]_{0}=0.0100$ mol/L, hence $4x^{2}+4.63\times 10^{-3}x-4.63\times 10^{-5}=0$ and $x=2.87\times 10^{-3}$ mol/L, that is $\alpha=28.7\%$ .

Why

The constant has not changed, only the initial concentration has. The system shifts toward the side with the most particles, exactly as under a drop in pressure.

Step 9

Conclusion on the dilution: the degree goes from $10.2\%$ to $28.7\%$ , almost triple, while $K_{c}$ has stayed at $4.63\times 10^{-3}$ .

Why

This is the central distinction of the chapter, illustrated with numbers: the POSITION of equilibrium has moved a lot, the CONSTANT not at all.

The conclusion, written out

“At $1.00$ L, equilibrium is established with $[N_{2}O_{4}]=0.0898$ mol/L and $[NO_{2}]=0.0204$ mol/L, that is a degree of dissociation of $10.2\%$ , and $K_{p}=11.5$ kPa. Diluted ten times, the same system dissociates to $28.7\%$ , without the equilibrium constant having changed.”

The classic mistake on this problem: Keeping the small- $x$ approximation and answering $1.08\times 10^{-2}$ mol/L. The gap from the exact value is only $5.5\%$ , so perfectly invisible without the test: this is what makes it the most profitable trap of the chapter.

Learn by heart

• $K$ is written with the concentrations AT EQUILIBRIUM; pure SOLIDS and LIQUIDS are left out of it.
• $Q<K$ : toward the products. $Q>K$ : toward the reactants. $Q=K$ : equilibrium.
• ICE table: the change is $\pm$ the COEFFICIENT times $x$ .
• Small- $x$ approximation: valid only if $x<5\%$ of the initial concentration, AND it must be tested.
• $K_{p}=K_{c}(RT)^{\Delta n_{\text{gas}}}$ , with $\Delta n$ counted on gases only.
• Reversing gives $\dfrac{1}{K}$ ; multiplying the coefficients by $n$ gives $K^{n}$ ; adding two equations MULTIPLIES the constants.
• Pressure: toward the side with FEWER moles of GAS. Dilution: toward the side with more.
• TEMPERATURE is the only disturbance that changes the value of $K$ .

Frequently asked questions

Why do solids not appear in the equilibrium constant?

Because the concentration of a pure solid is fixed by its density and does not change when some is added or removed. This constant value is already built into the equilibrium constant. This is why the decomposition of limestone does not depend on the amount of limestone present.

Which disturbance changes the value of the equilibrium constant?

Only one: temperature. Adding a reactant, removing a product, compressing, diluting or adding a catalyst shifts the position of equilibrium without touching the constant. Heating or cooling, on the other hand, really does change it, in the direction that favours the endothermic reaction when you heat.

When can you neglect x next to the initial concentration?

When it represents less than five percent of that concentration, which must be checked after calculating the approximate value. If the test fails, you have to start again with the full quadratic equation. The approximation is the safer the smaller the constant is compared with the initial concentration.

Does a catalyst improve the yield of a reaction at equilibrium?

No. It speeds up the forward and the reverse reactions in the same proportion, so the ratio of the two rate constants, which is exactly the equilibrium constant, stays unchanged. The gain is a gain in time: you reach exactly the same equilibrium sooner.

Why is a weak acid more dissociated when you dilute it?

Because dissociation produces more particles than it consumes. Diluting is equivalent to lowering the pressure, and the system responds by shifting toward the side with the most particles. The percentage dissociated therefore increases, even though the ion concentration decreases in absolute value.

What is the difference between Kp and Kc?

Kc is written with concentrations, Kp with partial pressures. They are related by the product of the gas constant and the temperature, raised to the power of the change in the number of moles of gas. They coincide only if that change is zero, that is, if there are as many moles of gas on each side.

Practise it

Corrected exercises: Chemical equilibrium, 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

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

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

The essentials

Writing KKK, and comparing QQQ with KKK

The ICE table, and the test that validates it

Le Chatelier: what shifts the equilibrium and what changes KKK

The mistakes that cost marks

1. Putting a pure solid or liquid into the expression of KKK

2. Using the initial concentrations in the expression of KKK

3. Using the small-xxx approximation without testing it

4. Believing that adding a reactant increases the equilibrium constant

5. Forgetting the factor (RT)Δn(RT)^{\Delta n}(RT)Δn between KpK_{p}Kp​ and KcK_{c}Kc​

6. Counting all the moles instead of only the moles of gas for the effect of pressure

7. Believing that an inert gas added at constant volume shifts the equilibrium

8. Crediting the catalyst with a better yield at equilibrium

9. Getting it wrong when manipulating an equilibrium constant

10. Believing that dilution does not change the degree of dissociation

Which method to choose

What does the system do after a disturbance

How to attack an equilibrium calculation

How the answer is expected to be written

Taking an ICE table all the way through

Answering a Le Chatelier question

Check before you hand in

The typical problem, taken apart

The dissociation of dinitrogen tetroxide, then the effect of dilution

Learn by heart

Frequently asked questions

Why do solids not appear in the equilibrium constant?

Which disturbance changes the value of the equilibrium constant?

When can you neglect x next to the initial concentration?

Does a catalyst improve the yield of a reaction at equilibrium?

Why is a weak acid more dissociated when you dilute it?

What is the difference between Kp and Kc?

Corrected exercises: Chemical equilibrium, 202-SN2-RE

See also

Looking for a 202-SN2-RE tutor in Montreal?

Writing $K$ , and comparing $Q$ with $K$

Le Chatelier: what shifts the equilibrium and what changes $K$

1. Putting a pure solid or liquid into the expression of $K$

2. Using the initial concentrations in the expression of $K$

3. Using the small- $x$ approximation without testing it

5. Forgetting the factor $(RT)^{\Delta n}$ between $K_{p}$ and $K_{c}$