Solved Examples

How to Use This Section

The theory is finished. What follows is one long problem set — forty questions worked out in full, arranged from the simplest to the hardest. Between them they cover the in-text problems and the end-of-chapter exercises that boards, JEE Main and NEET keep recycling, in the same order the chapter builds them: equilibrium expressions first, then the manipulation and conversion of $K$, then ICE tables, then $Q$, $\Delta G^\circ$ and Le Chatelier, and finally the whole of ionic equilibrium from the pH of a strong acid to the prediction of a precipitate.

Work each question yourself before reading the answer. The working is written the way it should appear on an answer sheet: equation, expression, substitution, result.

The working method

1. Write the balanced equation before anything else. Every power in the equilibrium expression comes from a coefficient in that equation, and a wrong coefficient poisons everything downstream.
2. Decide whether the question is asking for $K_c$ or $K_p$, and note the units the data arrive in. Concentrations in $\mathrm{mol\,L^{-1}}$ go into $K_c$; partial pressures in bar or atm go into $K_p$.
3. Build an ICE table whenever initial amounts are given and equilibrium amounts are not. Moles must be converted to concentrations by dividing by the volume before they enter $K_c$.
4. Substitute, then look at the algebra. A perfect square on both sides lets you take a square root and avoid the quadratic entirely. A very small $K$ lets you drop $x$ against the initial concentration — but only after checking that $x$ is under about five per cent of it.
5. Convert between $K_p$ and $K_c$ only at the end, using $K_p = K_c(RT)^{\Delta n}$ with $\Delta n$ counted over gaseous species alone.
6. Finish by looking at the number. A negative concentration, a degree of dissociation above 1, or a pH of $-3$ for a dilute acid means an algebra slip, not a chemical discovery.

The errors that cost the most marks in this chapter

Error	Where it bites	The fix
Stoichiometric coefficients not carried into the powers	$K_c$ and $K_p$ expressions, and every ICE substitution	Each coefficient becomes an exponent. A 2 in front of a species means that term is squared, in the expression and in the change row alike
Pure solids and pure liquids left in the expression	Heterogeneous equilibria, $K_{sp}$, ester hydrolysis in dilute aqueous solution	Only gases and dissolved species appear. A pure solid or pure liquid has activity 1 and is absorbed into $K$
$\Delta n$ counted over all species	$K_p$ to $K_c$ conversion	$\Delta n$ counts gaseous moles only, products minus reactants. Solids and liquids contribute nothing
The approximation used without testing it	Weak acid and weak base pH, ICE tables with small $K$	Drop $x$ only if it comes out below about five per cent of the initial concentration. Otherwise solve the quadratic
pOH written down as the answer to a pH question	Strong bases, weak bases, salts of weak acids	Any route that ends in $[\mathrm{OH^-}]$ needs one more line: $\mathrm{pH} = 14 - \mathrm{pOH}$ at 298 K
Dilution ignored when two solutions are mixed	Precipitation prediction, buffers made by partial neutralisation	Recompute every concentration in the total volume first, then compare the ionic product with $K_{sp}$
Inert gas assumed to shift the equilibrium	Le Chatelier questions	Added at constant volume it changes nothing at all. Only at constant pressure does it act as a dilution

[JEE/NEET] The single most productive habit in this chapter is writing $\Delta n$ and the ICE row before touching the calculator.

Question 1: Writing the equilibrium constant expression

Write the expression for $K_c$ for each of the following.

(i) $2\mathrm{NOCl(g)} \rightleftharpoons 2\mathrm{NO(g)} + \mathrm{Cl_2(g)}$ (ii) $2\mathrm{Cu(NO_3)_2(s)} \rightleftharpoons 2\mathrm{CuO(s)} + 4\mathrm{NO_2(g)} + \mathrm{O_2(g)}$ (iii) $\mathrm{CH_3COOC_2H_5(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{CH_3COOH(aq)} + \mathrm{C_2H_5OH(aq)}$ (iv) $\mathrm{Fe^{3+}(aq)} + 3\mathrm{OH^-(aq)} \rightleftharpoons \mathrm{Fe(OH)_3(s)}$ (v) $\mathrm{I_2(s)} + 5\mathrm{F_2} \rightleftharpoons 2\mathrm{IF_5}$

Answer:

The rule is the same every time. Products over reactants, each concentration raised to its coefficient, and every pure solid or pure liquid dropped.

(i) All three are gases, nothing drops out.

$$K_c = \frac{[\mathrm{NO}]^2[\mathrm{Cl_2}]}{[\mathrm{NOCl}]^2}$$

(ii) Both $\mathrm{Cu(NO_3)_2}$ and $\mathrm{CuO}$ are pure solids. Only the two gases survive, and the denominator becomes 1.

$$K_c = [\mathrm{NO_2}]^4[\mathrm{O_2}]$$

(iii) Water here is the solvent, present in vast excess, so its concentration is effectively constant and it is absorbed into $K_c$.

$$K_c = \frac{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}{[\mathrm{CH_3COOC_2H_5}]}$$

(iv) $\mathrm{Fe(OH)_3}$ is a solid, so the numerator is 1.

$$K_c = \frac{1}{[\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3}$$

(v) Solid iodine drops out.

$$K_c = \frac{[\mathrm{IF_5}]^2}{[\mathrm{F_2}]^5}$$

Two things are worth noticing across the set. The reaction quotient $Q_c$ has exactly the same algebraic form in every case — the only difference is that the concentrations put into $Q_c$ need not be equilibrium values. And the units follow from the powers: in (i) the numerator carries three concentration terms against two in the denominator, so $K_c$ has units of $\mathrm{mol\,L^{-1}}$, while in (ii), with nothing in the denominator, the units are $\mathrm{mol^5\,L^{-5}}$. Only when the total power above and below match does $K_c$ come out dimensionless.

Ans: As written above in (i) to (v) Watch out: In (iii) water is omitted only because it is the solvent. If the same esterification is run in the liquid phase with no solvent, water is a genuine reagent and must appear in the expression.

---

Question 2: $K_c$ from equilibrium concentrations

(a) Find $K_c$ for $2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)}$ when the equilibrium concentrations are $[\mathrm{SO_2}] = 0.60\ \mathrm{M}$, $[\mathrm{O_2}] = 0.82\ \mathrm{M}$ and $[\mathrm{SO_3}] = 1.90\ \mathrm{M}$.

(b) At 500 K the equilibrium concentrations for ammonia synthesis are $[\mathrm{N_2}] = 1.5 \times 10^{-2}\ \mathrm{M}$, $[\mathrm{H_2}] = 3.0 \times 10^{-2}\ \mathrm{M}$ and $[\mathrm{NH_3}] = 1.2 \times 10^{-2}\ \mathrm{M}$. Find $K_c$.

Answer:

(a) The equation is already balanced, so I write the expression and substitute.

$$K_c = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]} = \frac{(1.90)^2}{(0.60)^2(0.82)} = \frac{3.61}{0.2952} = 12.23$$

The units are $\mathrm{mol^{-1}\,L}$, since $\Delta n$ for the concentration expression is $-1$.

(b) For $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$,

$$K_c = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} = \frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2})(3.0 \times 10^{-2})^3}$$

$$= \frac{1.44 \times 10^{-4}}{(1.5 \times 10^{-2})(2.7 \times 10^{-5})} = \frac{1.44 \times 10^{-4}}{4.05 \times 10^{-7}} = 3.56 \times 10^{2}$$

Both constants are comfortably above 1, so in both systems the products dominate the equilibrium mixture. That is all the magnitude of $K$ tells you. It says nothing about how fast either mixture got there — sulphur dioxide oxidation needs a vanadium catalyst and ammonia synthesis needs iron at 700 K, and neither fact appears anywhere in $K_c$.

Ans: (a) $K_c = 12.23$ (b) $K_c = 3.56 \times 10^{2}$ Watch out: In (b) the hydrogen term is cubed. Cubing $3.0 \times 10^{-2}$ gives $2.7 \times 10^{-5}$, not $9.0 \times 10^{-4}$; using the square instead returns $10.7$, an answer smaller than $K_c$ by a factor of 33.

---

Question 3: Two more constants, and one reversal

(a) At 800 K a sealed vessel holds $[\mathrm{N_2}] = 3.0 \times 10^{-3}\ \mathrm{M}$, $[\mathrm{O_2}] = 4.2 \times 10^{-3}\ \mathrm{M}$ and $[\mathrm{NO}] = 2.8 \times 10^{-3}\ \mathrm{M}$ at equilibrium. Find $K_c$ for $\mathrm{N_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{NO(g)}$, and for the reverse reaction.

(b) At 500 K, $[\mathrm{PCl_3}] = 1.59\ \mathrm{M}$, $[\mathrm{Cl_2}] = 1.59\ \mathrm{M}$ and $[\mathrm{PCl_5}] = 1.41\ \mathrm{M}$. Find $K_c$ for $\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$.

Answer:

(a)

$$K_c = \frac{[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]} = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})} = \frac{7.84 \times 10^{-6}}{1.26 \times 10^{-5}} = 0.622$$

Reversing the equation inverts the constant.

$$K_c' = \frac{1}{0.622} = 1.61$$

(b)

$$K_c = \frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} = \frac{1.59 \times 1.59}{1.41} = \frac{2.528}{1.41} = 1.79\ \mathrm{mol\,L^{-1}}$$

The equal concentrations of $\mathrm{PCl_3}$ and $\mathrm{Cl_2}$ in (b) are not a coincidence. They are produced in a $1:1$ ratio from a vessel that started with $\mathrm{PCl_5}$ alone, so they must stay equal all the way to equilibrium. Spotting that saves a line of algebra in the ICE problems later on, and it is also a quick check that the data in a question are self-consistent.

Ans: (a) $K_c = 0.622$; reverse $K_c' = 1.61$ (b) $K_c = 1.79\ \mathrm{mol\,L^{-1}}$ Watch out: $K_c$ in (a) is close to 1, which says reactants and products are present in comparable amounts — it does not mean the reaction has barely happened.

Question 4: Manipulating an equilibrium constant

(a) For $\mathrm{NO(g)} + \mathrm{O_3(g)} \rightleftharpoons \mathrm{NO_2(g)} + \mathrm{O_2(g)}$, $K_c = 6.3 \times 10^{14}$ at 1000 K. What is $K_c$ for the reverse reaction?

(b) For $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$, $K_c = 3.56 \times 10^{2}$ at 500 K. Find $K_c$ for $\tfrac{1}{2}\mathrm{N_2(g)} + \tfrac{3}{2}\mathrm{H_2(g)} \rightleftharpoons \mathrm{NH_3(g)}$ and for $2\mathrm{NH_3(g)} \rightleftharpoons \mathrm{N_2(g)} + 3\mathrm{H_2(g)}$.

(c) Given $\mathrm{A} \rightleftharpoons \mathrm{B}$ with $K_1 = 4.0$ and $\mathrm{B} \rightleftharpoons \mathrm{C}$ with $K_2 = 2.5$, find $K$ for $\mathrm{A} \rightleftharpoons \mathrm{C}$.

Answer:

Three rules cover all of this. Reversing an equation inverts $K$. Multiplying an equation by $n$ raises $K$ to the power $n$. Adding two equations multiplies their constants.

(a) Both directions are elementary bimolecular steps, but that does not matter for the arithmetic.

$$K_c' = \frac{1}{6.3 \times 10^{14}} = 1.6 \times 10^{-15}$$

(b) Halving the equation means raising $K_c$ to the power $\tfrac{1}{2}$.

$$K_c' = (3.56 \times 10^{2})^{1/2} = 18.9$$

Reversing the original equation inverts it.

$$K_c'' = \frac{1}{3.56 \times 10^{2}} = 2.81 \times 10^{-3}$$

(c) Adding the two steps gives $\mathrm{A} \rightleftharpoons \mathrm{C}$, so the constants multiply.

$$K = K_1 \times K_2 = 4.0 \times 2.5 = 10$$

The reason the constants multiply is visible in the expressions themselves. $K_1 = [\mathrm{B}]/[\mathrm{A}]$ and $K_2 = [\mathrm{C}]/[\mathrm{B}]$, so their product is $[\mathrm{C}]/[\mathrm{A}]$, which is exactly $K$ for the overall change. The same cancellation is what makes $K_a \times K_b = K_w$ work for a conjugate pair later in the chapter, and what lets a coupled biochemical reaction with an unfavourable $K$ be dragged forward by a second step with a large one.

Ans: (a) $1.6 \times 10^{-15}$ (b) $18.9$ and $2.81 \times 10^{-3}$ (c) $K = 10$ Watch out: Halving the equation takes the square root, not half the value. Writing $3.56 \times 10^{2}/2 = 178$ is the commonest slip here.

---

Question 5: $K_p$ from $K_c$

For $2\mathrm{NOCl(g)} \rightleftharpoons 2\mathrm{NO(g)} + \mathrm{Cl_2(g)}$, $K_c = 3.75 \times 10^{-6}$ at 1069 K. Calculate $K_p$.

Answer:

I use $K_p = K_c(RT)^{\Delta n}$ with $R = 0.0831\ \mathrm{L\,bar\,K^{-1}\,mol^{-1}}$, so $K_p$ comes out in bar.

Gaseous moles: products $2 + 1 = 3$, reactants $2$. So $\Delta n = 3 - 2 = 1$.

$$RT = 0.0831 \times 1069 = 88.83\ \mathrm{L\,bar\,mol^{-1}}$$

$$K_p = (3.75 \times 10^{-6})(88.83)^1 = 3.33 \times 10^{-4}\ \mathrm{bar}$$

A word on the gas constant. Using $R = 0.0831\ \mathrm{L\,bar\,K^{-1}\,mol^{-1}}$ delivers $K_p$ in bar, which is the standard state for pressure. If a question gives pressures in atmospheres, switch to $R = 0.0821\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}}$ and the answer comes out in atm. The two differ by less than 2%, so the numbers look similar, but mixing them within one calculation is untidy and the units in the final answer will be wrong.

Ans: $K_p = 3.33 \times 10^{-4}\ \mathrm{bar}$ Watch out: $\Delta n$ here is $+1$, so $K_p > K_c$. Using $\Delta n = -1$ by counting reactants minus products gives $4.2 \times 10^{-8}$, wrong by four orders of magnitude.

---

Question 6: $K_c$ from $K_p$, including a negative $\Delta n$

Find $K_c$ for each equilibrium from the given $K_p$. Take $R = 0.0831\ \mathrm{L\,bar\,K^{-1}\,mol^{-1}}$.

(i) $2\mathrm{NOCl(g)} \rightleftharpoons 2\mathrm{NO(g)} + \mathrm{Cl_2(g)}$, $K_p = 1.8 \times 10^{-2}$ at 500 K (ii) $\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}$, $K_p = 167\ \mathrm{kPa}$ at 1073 K (iii) $2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)}$, $K_p = 2.0 \times 10^{10}\ \mathrm{bar^{-1}}$ at 450 K

Answer:

Rearranging, $K_c = K_p(RT)^{-\Delta n}$.

(i) $\Delta n = 3 - 2 = 1$, $RT = 0.0831 \times 500 = 41.55$.

$$K_c = \frac{1.8 \times 10^{-2}}{41.55} = 4.33 \times 10^{-4}$$

(ii) The two solids contribute nothing to $\Delta n$. Gaseous products 1, gaseous reactants 0, so $\Delta n = 1$ and $RT = 0.0831 \times 1073 = 89.17$. The constant is supplied in kilopascals, so it goes into bar first to match $R$: $K_p = 167\ \mathrm{kPa} = 1.67\ \mathrm{bar}$, which is the equilibrium pressure of $\mathrm{CO_2}$ over the solid and sits sensibly below the $2\ \mathrm{bar}$ measured near 1100 K.

$$K_c = \frac{1.67}{89.17} = 1.87 \times 10^{-2}\ \mathrm{mol\,L^{-1}}$$

Dropping the unit and feeding a bare $167$ into the same division returns $1.87\ \mathrm{mol\,L^{-1}}$, which is the figure printed wherever the constant is read as bar. Writing the unit down is what keeps the two versions from looking like a contradiction.

(iii) Gaseous products 2, gaseous reactants 3, so $\Delta n = -1$ and $RT = 0.0831 \times 450 = 37.40$.

$$K_c = K_p(RT)^{+1} = (2.0 \times 10^{10})(37.40) = 7.48 \times 10^{11}$$

The sign of $\Delta n$ decides everything here, so it is worth reading off the pattern. When $\Delta n > 0$, $K_p$ exceeds $K_c$; when $\Delta n < 0$, $K_p$ is the smaller of the two; and when $\Delta n = 0$ the two are numerically equal and both dimensionless. That last case covers $\mathrm{H_2} + \mathrm{I_2} \rightleftharpoons 2\mathrm{HI}$ and the water gas shift reaction, which is why those two appear so often in questions that give a single constant without saying which it is.

Ans: (i) $4.33 \times 10^{-4}$ (ii) $1.87 \times 10^{-2}\ \mathrm{mol\,L^{-1}}$ (iii) $7.48 \times 10^{11}$ Watch out: In (ii) the temptation is to set $\Delta n = 0$ because there is one species on each side. Only gases are counted, and calcium carbonate and calcium oxide are solids, so $\Delta n = +1$.

Question 7: $K_p$ from percentage composition by volume

At a certain temperature and a total pressure of $10^5\ \mathrm{Pa}$, iodine vapour contains 40% by volume of I atoms. Calculate $K_p$ for $\mathrm{I_2(g)} \rightleftharpoons 2\mathrm{I(g)}$.

Answer:

For a gas mixture, percentage by volume is percentage by mole, so the mole fractions are $0.40$ for $\mathrm{I}$ and $0.60$ for $\mathrm{I_2}$.

The total pressure is $10^5\ \mathrm{Pa} = 1\ \mathrm{bar}$. Partial pressure is mole fraction times total pressure:

$$p_{\mathrm{I}} = 0.40 \times 1 = 0.40\ \mathrm{bar} \qquad p_{\mathrm{I_2}} = 0.60 \times 1 = 0.60\ \mathrm{bar}$$

$$K_p = \frac{(p_{\mathrm{I}})^2}{p_{\mathrm{I_2}}} = \frac{(0.40)^2}{0.60} = \frac{0.16}{0.60} = 0.267\ \mathrm{bar}$$

The same data give the degree of dissociation. If one mole of $\mathrm{I_2}$ dissociates to the extent $\alpha$, the mixture contains $(1-\alpha)$ of $\mathrm{I_2}$ and $2\alpha$ of $\mathrm{I}$, a total of $(1+\alpha)$. Setting the $\mathrm{I}$ mole fraction equal to $0.40$:

$$\frac{2\alpha}{1+\alpha} = 0.40 \quad\Rightarrow\quad 2\alpha = 0.40 + 0.40\alpha \quad\Rightarrow\quad \alpha = 0.25$$

So a quarter of the iodine molecules have split. This is the same bridge between composition and $\alpha$ that vapour density questions use.

Ans: $K_p = 0.267\ \mathrm{bar}$ (equivalently $2.67 \times 10^{4}\ \mathrm{Pa}$); $\alpha = 0.25$ Watch out: The 40% belongs to the atoms $\mathrm{I}$, not to $\mathrm{I_2}$. Swapping them gives $0.90\ \mathrm{bar}$.

---

Question 8: $K_p$ from an initial and an equilibrium pressure

A sample of $\mathrm{HI(g)}$ is placed in a flask at a pressure of $0.2\ \mathrm{atm}$. At equilibrium the partial pressure of $\mathrm{HI(g)}$ is $0.04\ \mathrm{atm}$. Find $K_p$ for $2\mathrm{HI(g)} \rightleftharpoons \mathrm{H_2(g)} + \mathrm{I_2(g)}$.

Answer:

I set up the pressure ICE table. Let $2p$ be the fall in $p_{\mathrm{HI}}$.

	$2\mathrm{HI}$	$\mathrm{H_2}$	$\mathrm{I_2}$
Initial (atm)	0.20	0	0
Change (atm)	$-2p$	$+p$	$+p$
Equilibrium (atm)	$0.20 - 2p$	$p$	$p$

The equilibrium pressure of $\mathrm{HI}$ is given as $0.04\ \mathrm{atm}$, so $0.20 - 2p = 0.04$, giving $2p = 0.16$ and $p = 0.08\ \mathrm{atm}$.

$$K_p = \frac{p_{\mathrm{H_2}} \cdot p_{\mathrm{I_2}}}{(p_{\mathrm{HI}})^2} = \frac{(0.08)(0.08)}{(0.04)^2} = \frac{6.4 \times 10^{-3}}{1.6 \times 10^{-3}} = 4.0$$

$K_p$ is dimensionless here because $\Delta n = 0$, which also means $K_c = K_p = 4.0$ without any conversion.

A check on the total pressure. At equilibrium it is $0.04 + 0.08 + 0.08 = 0.20\ \mathrm{atm}$, exactly what was there at the start — as it must be, since two moles of gas become two moles of gas. If the arithmetic had produced any other total, something in the change row would be wrong.

Ans: $K_p = K_c = 4.0$ Watch out: The decrease in $\mathrm{HI}$ pressure is $0.16\ \mathrm{atm}$, but only half of it, $0.08\ \mathrm{atm}$, appears as $\mathrm{H_2}$. Setting $p_{\mathrm{H_2}} = 0.16$ gives $K_p = 16$.

---

Question 9: A heterogeneous equilibrium from a mass percentage

At 1127 K and 1 atm, a gaseous mixture of $\mathrm{CO}$ and $\mathrm{CO_2}$ in equilibrium with solid carbon contains 90.55% $\mathrm{CO}$ by mass. Calculate $K_c$ for $\mathrm{C(s)} + \mathrm{CO_2(g)} \rightleftharpoons 2\mathrm{CO(g)}$.

Answer:

Mass percentages have to become mole fractions before they can become partial pressures. I take 100 g of the gas mixture.

$$n_{\mathrm{CO}} = \frac{90.55}{28} = 3.234\ \mathrm{mol} \qquad n_{\mathrm{CO_2}} = \frac{9.45}{44} = 0.2148\ \mathrm{mol}$$

Total $= 3.449\ \mathrm{mol}$, so the mole fractions are $0.9377$ and $0.0623$.

At a total pressure of 1 atm, $p_{\mathrm{CO}} = 0.9377\ \mathrm{atm}$ and $p_{\mathrm{CO_2}} = 0.0623\ \mathrm{atm}$. Solid carbon does not appear.

$$K_p = \frac{(p_{\mathrm{CO}})^2}{p_{\mathrm{CO_2}}} = \frac{(0.9377)^2}{0.0623} = \frac{0.8793}{0.0623} = 14.12\ \mathrm{atm}$$

Now to $K_c$, with $\Delta n = 2 - 1 = 1$ and $R = 0.0821\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}}$ to match the atm data:

$$K_c = \frac{K_p}{RT} = \frac{14.12}{0.0821 \times 1127} = \frac{14.12}{92.53} = 0.153\ \mathrm{mol\,L^{-1}}$$

The result makes physical sense. At 1127 K the Boudouard equilibrium lies well over towards carbon monoxide, which is why a hot bed of coke reduces carbon dioxide so effectively — the basis of the blast furnace.

Ans: $K_p = 14.1\ \mathrm{atm}$, $K_c = 0.153\ \mathrm{mol\,L^{-1}}$ Watch out: Treating 90.55% as a mole percentage skips the division by molar mass and gives $p_{\mathrm{CO}} = 0.9055$, $K_p = 8.68$ and $K_c = 0.094$.

Question 10: Reading the balanced equation off the expression

The equilibrium constant expression for a gas reaction is

$$K_c = \frac{[\mathrm{NH_3}]^4[\mathrm{O_2}]^5}{[\mathrm{NO}]^4[\mathrm{H_2O}]^6}$$

Write the balanced chemical equation.

Answer:

Everything in the numerator is a product, everything in the denominator a reactant, and every power is a coefficient. Reading straight off:

$$4\mathrm{NO(g)} + 6\mathrm{H_2O(g)} \rightleftharpoons 4\mathrm{NH_3(g)} + 5\mathrm{O_2(g)}$$

Checking the atom balance. Nitrogen: 4 on each side. Oxygen: $4 + 6 = 10$ on the left, $5 \times 2 = 10$ on the right. Hydrogen: $6 \times 2 = 12$ on the left, $4 \times 3 = 12$ on the right. It balances.

The corresponding pressure form follows the same reading:

$$K_p = \frac{(p_{\mathrm{NH_3}})^4(p_{\mathrm{O_2}})^5}{(p_{\mathrm{NO}})^4(p_{\mathrm{H_2O}})^6}$$

with $\Delta n = 9 - 10 = -1$.

Ans: $4\mathrm{NO(g)} + 6\mathrm{H_2O(g)} \rightleftharpoons 4\mathrm{NH_3(g)} + 5\mathrm{O_2(g)}$ Watch out: Writing the ammonia oxidation the usual way round, with $\mathrm{NH_3}$ and $\mathrm{O_2}$ as reactants, gives the reciprocal of the constant asked for.

---

Question 11: An ICE table that solves as a perfect square

$K_c = 4.24$ at 800 K for $\mathrm{CO(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO_2(g)} + \mathrm{H_2(g)}$. If only $\mathrm{CO}$ and $\mathrm{H_2O}$ are present initially, each at $0.10\ \mathrm{M}$, find all four equilibrium concentrations.

Answer:

Let $x$ be the concentration of $\mathrm{CO_2}$ formed. The stoichiometry is one-to-one throughout.

	$\mathrm{CO}$	$\mathrm{H_2O}$	$\mathrm{CO_2}$	$\mathrm{H_2}$
Initial (M)	0.10	0.10	0	0
Change (M)	$-x$	$-x$	$+x$	$+x$
Equilibrium (M)	$0.10-x$	$0.10-x$	$x$	$x$

$$K_c = \frac{x^2}{(0.10-x)^2} = 4.24$$

Both sides are perfect squares, so I take the square root rather than expanding.

$$\frac{x}{0.10-x} = \sqrt{4.24} = 2.059$$

$$x = 0.2059 - 2.059x \quad\Rightarrow\quad 3.059x = 0.2059 \quad\Rightarrow\quad x = 0.067$$

The last step of any ICE problem is to substitute back:

$$\frac{(0.067)^2}{(0.033)^2} = \frac{4.49 \times 10^{-3}}{1.09 \times 10^{-3}} = 4.12$$

which recovers $4.24$ to within rounding. That check costs one line and catches almost every arithmetic slip.

Ans: $[\mathrm{CO_2}] = [\mathrm{H_2}] = 0.067\ \mathrm{M}$; $[\mathrm{CO}] = [\mathrm{H_2O}] = 0.033\ \mathrm{M}$ Watch out: Expanding into $3.24x^2 - 0.848x + 0.0424 = 0$ gives roots $0.067$ and $0.194$. The second is impossible — it would require more $\mathrm{CO}$ to react than was present — so it is rejected. Taking the square root avoids the trap entirely.

---

Question 12: $K_c$ from a percentage reacted

One mole of $\mathrm{H_2O}$ and one mole of $\mathrm{CO}$ are taken in a 10 L vessel and heated to 725 K. At equilibrium 40% of the water by mass has reacted according to $\mathrm{H_2O(g)} + \mathrm{CO(g)} \rightleftharpoons \mathrm{H_2(g)} + \mathrm{CO_2(g)}$. Calculate the equilibrium constant.

Answer:

Moles first become concentrations. One mole in 10 L is $0.10\ \mathrm{M}$, for both reactants.

40% of the water reacts, so $x = 0.40 \times 0.10 = 0.04\ \mathrm{M}$.

	$\mathrm{H_2O}$	$\mathrm{CO}$	$\mathrm{H_2}$	$\mathrm{CO_2}$
Equilibrium (M)	0.06	0.06	0.04	0.04

$$K_c = \frac{[\mathrm{H_2}][\mathrm{CO_2}]}{[\mathrm{H_2O}][\mathrm{CO}]} = \frac{(0.04)(0.04)}{(0.06)(0.06)} = \frac{1.6 \times 10^{-3}}{3.6 \times 10^{-3}} = 0.44$$

$\Delta n = 0$, so $K_p = K_c = 0.44$ as well.

Two details in the wording deserve attention. "40% of water by mass" is the same as 40% by moles, because all the water in the vessel has the same molar mass — percentages by mass and by mole only diverge when different substances are being compared. And the vessel is 10 L, not 1 L, so the moles had to be converted before the ICE table was built, even though the volume happens to cancel in this particular expression.

Ans: $K_c = K_p = 0.44$ Watch out: Because $\Delta n = 0$ the volume cancels, so using moles instead of concentrations happens to give the same answer here. That coincidence does not survive to any reaction with $\Delta n \neq 0$, so build the habit of dividing by the volume.

Question 13: An ICE table needing the quadratic formula

$3.00\ \mathrm{mol}$ of $\mathrm{PCl_5}$ kept in a 1 L closed vessel is allowed to reach equilibrium at 380 K, where $K_c = 1.80$. Calculate the composition of the mixture at equilibrium.

Answer:

The volume is 1 L, so moles and molarities are numerically equal. Let $x$ be the molarity of $\mathrm{PCl_5}$ that dissociates.

	$\mathrm{PCl_5}$	$\mathrm{PCl_3}$	$\mathrm{Cl_2}$
Initial (M)	3.00	0	0
Equilibrium (M)	$3.00-x$	$x$	$x$

$$K_c = \frac{x^2}{3.00-x} = 1.80$$

$K_c$ is not small compared with the initial concentration, so the approximation is not available and I solve properly.

$$x^2 + 1.80x - 5.40 = 0$$

$$x = \frac{-1.80 \pm \sqrt{(1.80)^2 + 4(5.40)}}{2} = \frac{-1.80 \pm \sqrt{24.84}}{2} = \frac{-1.80 \pm 4.98}{2}$$

The negative root is rejected, leaving $x = 1.59$.

Substituting back, $(1.59)^2/1.41 = 2.528/1.41 = 1.79$, which is $K_c$ to two decimal places. The degree of dissociation follows immediately as $\alpha = 1.59/3.00 = 0.53$, so slightly over half the phosphorus pentachloride has broken up at 380 K.

Ans: $[\mathrm{PCl_5}] = 1.41\ \mathrm{M}$; $[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = 1.59\ \mathrm{M}$; $\alpha = 0.53$ Watch out: Dropping $x$ against $3.00$ here gives $x = \sqrt{5.40} = 2.32\ \mathrm{M}$, which is 77% of the initial concentration. The approximation is only safe when the answer it produces is small — always test it afterwards.

---

Question 14: Equilibrium concentrations for a $2\mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C}$ system

What is the equilibrium concentration of each substance when the initial concentration of $\mathrm{ICl}$ was $0.78\ \mathrm{M}$? $2\mathrm{ICl(g)} \rightleftharpoons \mathrm{I_2(g)} + \mathrm{Cl_2(g)}$, $K_c = 0.14$.

Answer:

Let $x$ be the concentration of $\mathrm{I_2}$ formed. Two $\mathrm{ICl}$ are consumed for each $\mathrm{I_2}$, so $\mathrm{ICl}$ falls by $2x$.

	$2\mathrm{ICl}$	$\mathrm{I_2}$	$\mathrm{Cl_2}$
Initial (M)	0.78	0	0
Equilibrium (M)	$0.78-2x$	$x$	$x$

$$K_c = \frac{x^2}{(0.78-2x)^2} = 0.14$$

Perfect squares again, so I take the root.

$$\frac{x}{0.78-2x} = \sqrt{0.14} = 0.374$$

$$x = 0.2918 - 0.748x \quad\Rightarrow\quad 1.748x = 0.2918 \quad\Rightarrow\quad x = 0.167$$

Then $[\mathrm{ICl}] = 0.78 - 2(0.167) = 0.78 - 0.334 = 0.446\ \mathrm{M}$.

Substituting back gives $(0.167)^2/(0.446)^2 = 0.0279/0.199 = 0.140$, which is $K_c$ exactly. A second check is available too: iodine and chlorine atoms must be conserved, and $0.446 + 2(0.167) = 0.78$ accounts for all the $\mathrm{ICl}$ originally present.

Ans: $[\mathrm{ICl}] = 0.446\ \mathrm{M}$; $[\mathrm{I_2}] = [\mathrm{Cl_2}] = 0.167\ \mathrm{M}$ Watch out: The change row for $\mathrm{ICl}$ is $-2x$, not $-x$. Using $-x$ gives $x = 0.212$ and an $\mathrm{ICl}$ concentration of $0.568\ \mathrm{M}$.

---

Question 15: A heterogeneous equilibrium with a quadratic

$K_p = 3.0$ at 1000 K for $\mathrm{CO_2(g)} + \mathrm{C(s)} \rightleftharpoons 2\mathrm{CO(g)}$. If initially $p_{\mathrm{CO_2}} = 0.48\ \mathrm{bar}$, $p_{\mathrm{CO}} = 0$, and pure graphite is present, calculate the equilibrium partial pressures.

Answer:

Graphite is a pure solid and stays out of the expression. Let $x$ be the fall in $p_{\mathrm{CO_2}}$; then $p_{\mathrm{CO}}$ rises by $2x$.

	$\mathrm{CO_2}$	$\mathrm{CO}$
Initial (bar)	0.48	0
Equilibrium (bar)	$0.48-x$	$2x$

$$K_p = \frac{(p_{\mathrm{CO}})^2}{p_{\mathrm{CO_2}}} = \frac{(2x)^2}{0.48-x} = 3.0$$

$$4x^2 = 1.44 - 3x \quad\Rightarrow\quad 4x^2 + 3x - 1.44 = 0$$

$$x = \frac{-3 \pm \sqrt{9 + 4(4)(1.44)}}{8} = \frac{-3 \pm \sqrt{32.04}}{8} = \frac{-3 + 5.66}{8} = 0.33$$

Checking: $(0.66)^2/0.15 = 0.4356/0.15 = 2.9$, close enough to 3.0 given the rounding of $x$. Notice also that the total pressure has risen from $0.48$ to $0.81\ \mathrm{bar}$, as it should when one mole of gas becomes two.

Ans: $p_{\mathrm{CO}} = 0.66\ \mathrm{bar}$; $p_{\mathrm{CO_2}} = 0.15\ \mathrm{bar}$ Watch out: Solid carbon must be present for the equilibrium to exist, but its amount never enters the calculation. Putting a concentration for graphite into the denominator is the standard heterogeneous-equilibrium error.

Question 16: When the small-$x$ approximation is justified

A mixture of $0.482\ \mathrm{mol}\ \mathrm{N_2}$ and $0.933\ \mathrm{mol}\ \mathrm{O_2}$ is placed in a 10 L vessel and allowed to form $\mathrm{N_2O}$ at a temperature where $K_c = 2.0 \times 10^{-37}$. Determine the composition of the equilibrium mixture.

$$2\mathrm{N_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{N_2O(g)}$$

Answer:

Initial concentrations: $[\mathrm{N_2}] = 0.482/10 = 0.0482\ \mathrm{M}$ and $[\mathrm{O_2}] = 0.933/10 = 0.0933\ \mathrm{M}$.

$K_c$ is $10^{-37}$, so almost nothing reacts. Let $2x$ be the $\mathrm{N_2O}$ formed; $x$ will be utterly negligible beside the initial concentrations, so I take them as unchanged.

$$K_c = \frac{[\mathrm{N_2O}]^2}{[\mathrm{N_2}]^2[\mathrm{O_2}]} = 2.0 \times 10^{-37}$$

$$[\mathrm{N_2O}]^2 = (2.0 \times 10^{-37})(0.0482)^2(0.0933) = (2.0 \times 10^{-37})(2.168 \times 10^{-4}) = 4.34 \times 10^{-41}$$

$$[\mathrm{N_2O}] = 6.6 \times 10^{-21}\ \mathrm{mol\,L^{-1}}$$

The check is immediate: $6.6 \times 10^{-21}$ against $0.0482$ is about $10^{-17}$ per cent, so treating the reactants as unchanged was more than safe.

A constant of this size is what makes the atmosphere possible. Nitrogen and oxygen sit together in air at room temperature without reacting to any measurable extent, and $K_c = 2.0 \times 10^{-37}$ says the equilibrium position itself forbids it, quite apart from how slow the reaction is. Only the enormous temperatures inside an engine cylinder or a lightning channel raise $K$ enough for nitrogen oxides to form in quantity.

Ans: $[\mathrm{N_2}] = 0.0482\ \mathrm{M}$, $[\mathrm{O_2}] = 0.0933\ \mathrm{M}$, $[\mathrm{N_2O}] = 6.6 \times 10^{-21}\ \mathrm{M}$ Watch out: The nitrogen term is squared. Leaving it as $0.0482$ instead of $(0.0482)^2$ gives $3.0 \times 10^{-20}\ \mathrm{M}$.

---

Question 17: Which way will the mixture move

(a) A mixture of $1.57\ \mathrm{mol}\ \mathrm{N_2}$, $1.92\ \mathrm{mol}\ \mathrm{H_2}$ and $8.13\ \mathrm{mol}\ \mathrm{NH_3}$ is put into a 20 L vessel at 500 K, where $K_c = 1.7 \times 10^{2}$ for $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightleftharpoons 2\mathrm{NH_3(g)}$. Is the mixture at equilibrium? If not, which way does it go?

(b) At 500 K, $K_c = 0.061$ for the same reaction. A mixture contains $3.0\ \mathrm{M}\ \mathrm{N_2}$, $2.0\ \mathrm{M}\ \mathrm{H_2}$ and $0.5\ \mathrm{M}\ \mathrm{NH_3}$. Same question.

Answer:

The tool is the reaction quotient $Q_c$, built exactly like $K_c$ but from whatever concentrations happen to be present.

(a) Concentrations: $[\mathrm{N_2}] = 1.57/20 = 0.0785\ \mathrm{M}$, $[\mathrm{H_2}] = 1.92/20 = 0.096\ \mathrm{M}$, $[\mathrm{NH_3}] = 8.13/20 = 0.4065\ \mathrm{M}$.

$$Q_c = \frac{(0.4065)^2}{(0.0785)(0.096)^3} = \frac{0.1652}{(0.0785)(8.85 \times 10^{-4})} = \frac{0.1652}{6.95 \times 10^{-5}} = 2.38 \times 10^{3}$$

$Q_c = 2.38 \times 10^3$ is far larger than $K_c = 1.7 \times 10^2$. Too much product is present, so the net reaction runs backwards, decomposing ammonia into nitrogen and hydrogen.

(b)

$$Q_c = \frac{(0.5)^2}{(3.0)(2.0)^3} = \frac{0.25}{24} = 0.0104$$

$Q_c = 0.0104$ is smaller than $K_c = 0.061$, so the net reaction runs forwards, making more ammonia.

The whole comparison can be done in pressures instead, using $Q_p$ against $K_p$, and the verdict is identical. What must never be done is compare a $Q_c$ with a $K_p$, or a $Q$ built from moles with a $K$ built from concentrations — the numbers are then in different currencies and the direction they suggest is meaningless.

Ans: (a) Not at equilibrium; $Q_c > K_c$, so it goes in the reverse direction (b) Not at equilibrium; $Q_c < K_c$, so it goes forward Watch out: Moles must be divided by the 20 L before $Q_c$ is built. Feeding the raw moles into (a) gives $Q_c = (8.13)^2/[(1.57)(1.92)^3] = 5.95$, which would suggest the forward direction — the exact opposite of the truth.

---

Question 18: $\Delta G^\circ$ and the equilibrium constant

Calculate (a) $\Delta_r G^\circ$ and (b) the equilibrium constant for the formation of $\mathrm{NO_2}$ from $\mathrm{NO}$ and $\mathrm{O_2}$ at 298 K.

$$\mathrm{NO(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightleftharpoons \mathrm{NO_2(g)}$$

Given $\Delta_f G^\circ(\mathrm{NO_2}) = 52.0\ \mathrm{kJ\,mol^{-1}}$, $\Delta_f G^\circ(\mathrm{NO}) = 87.0\ \mathrm{kJ\,mol^{-1}}$, $\Delta_f G^\circ(\mathrm{O_2}) = 0$.

Answer:

(a) Standard Gibbs energy of reaction is products minus reactants, each weighted by its coefficient. Oxygen is an element in its standard state, so it contributes nothing.

$$\Delta_r G^\circ = 52.0 - \left(87.0 + \tfrac{1}{2}(0)\right) = -35.0\ \mathrm{kJ\,mol^{-1}}$$

(b) At equilibrium $\Delta G = 0$ and $Q = K$, so $\Delta_r G^\circ = -RT\ln K$.

$$\ln K = -\frac{\Delta_r G^\circ}{RT} = \frac{35\,000}{8.314 \times 298} = \frac{35\,000}{2477.6} = 14.13$$

$$K = e^{14.13} = 1.4 \times 10^{6}$$

A negative $\Delta_r G^\circ$ and a $K$ far above 1 say the same thing: at 298 K the equilibrium lies well over towards $\mathrm{NO_2}$.

The distinction between $\Delta G$ and $\Delta G^\circ$ is worth holding on to. $\Delta G^\circ$ is a fixed property of the reaction at a given temperature and is tied to $K$ alone. $\Delta G$ describes the mixture actually in front of you, through $\Delta G = \Delta G^\circ + RT\ln Q$, and it changes continuously as the reaction proceeds. At equilibrium $Q = K$, the two terms cancel, and $\Delta G$ becomes zero — the reaction has no remaining driving force in either direction, even though molecules are still turning over.

Ans: (a) $\Delta_r G^\circ = -35.0\ \mathrm{kJ\,mol^{-1}}$ (b) $K = 1.4 \times 10^{6}$ Watch out: $\Delta_r G^\circ$ must be in joules before dividing by $R$ in $\mathrm{J\,K^{-1}\,mol^{-1}}$. Using $35$ instead of $35\,000$ gives $\ln K = 0.0141$ and $K = 1.01$.