Corrected exercises: chemical equilibrium (202-SN2-RE)

$K_{c}=\frac{[HI]^{2}}{[H_{2}][I_{2}]}$. For limestone: $K_{c}=[CO_{2}]$ only: both solids are left out. A pure solid has a \"concentration\" (molar density) fixed by its nature, not by the amount present: it is constant and is absorbed into $K$. This is why the pressure of $CO_{2}$ above heated limestone does not depend on the size of the pile of rock.

$K$ measures the ratio of products to reactants at the end. $K=4\times 10^{31}$: the reaction is almost complete, only traces of reactants remain. $K=10^{-30}$: the reaction practically does not occur. $K=54.3$: an interesting intermediate case, with products clearly favoured but reactants still present in measurable amounts: this is where ICE tables have work to do.

$Q=\frac{(0.10)^{2}}{(0.10)(0.10)}=1.0$. Since $Q=1.0<K_{c}=54.3$, the numerator has to grow: the system moves in the FORWARD direction, making $HI$ by consuming $H_{2}$ and $I_{2}$, until $Q$ reaches 54.3.

No: equilibrium is DYNAMIC. When $Q=K_{c}$, both reactions carry on at the molecular scale, but at the SAME rate: every $HI$ formed somewhere is matched by an $HI$ dissociating somewhere else. The macroscopic concentrations are frozen, while the molecular traffic is intense in both directions: the stillness is only an average.

$K_{c}=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}=\frac{(0.18)^{2}}{(0.30)(0.90)^{3}}=\frac{0.0324}{0.219}\approx 0.148$.

Reversing the equation inverts the ratio: $K'=\frac{1}{K_{c}}=\frac{1}{0.148}\approx 6.75$.

Dividing all the coefficients by 2 raises the constant to the power $\frac{1}{2}$: $K''=\sqrt{K_{c}}=\sqrt{0.148}\approx 0.385$.

No: one chemistry, three ways of writing it. The equilibrium mixture (0.30, 0.90, 0.18) satisfies all three expressions at once: you can check by substituting. The constant is not a property of nature alone, but of the PAIR formed by nature and the way it is written: hence the golden rule of these exercises, always attach a value of $K$ to the exact equation that defines it, coefficients and direction included.

Initial: 0.100 and 0. Change: $-x$ and $+2x$. Equilibrium: $0.100-x$ and $2x$. Hence $K_{c}=\frac{(2x)^{2}}{0.100-x}=\frac{4x^{2}}{0.100-x}=4.6\times 10^{-3}$.

If $x\ll 0.100$: $4x^{2}\approx 4.6\times 10^{-4}$, so $x\approx\sqrt{1.15\times 10^{-4}}\approx 0.0107$ mol/L. Test: $\frac{0.0107}{0.100}=10.7$% $>5$%: the approximation FAILS. The constant is not small enough compared with the initial concentration: the quadratic is needed. Trying the approximation was not a mistake, it was a controlled experiment: the 5% test is there to settle the matter.

$4x^{2}+4.6\times 10^{-3}x-4.6\times 10^{-4}=0$. Discriminant: $(4.6\times 10^{-3})^{2}+4(4)(4.6\times 10^{-4})=7.38\times 10^{-3}$, square root $0.0859$. $x=\frac{-4.6\times 10^{-3}+0.0859}{8}\approx 0.0102$ mol/L (the negative root is rejected). Equilibrium: $[NO_{2}]=2x\approx 0.0203$ mol/L and $[N_{2}O_{4}]=0.100-x\approx 0.0898$ mol/L. The approximation overestimated $x$ by 5%: visible, as the test announced.

$\frac{(0.0203)^{2}}{0.0898}=4.6\times 10^{-3}$: we land back on $K_{c}$, up to rounding. This reflex transforms an exam: it catches for sure a sign error in the table, a badly chosen root or a calculator slip, for the price of one division. An equilibrium result that has not been checked is a result that is not finished.

Adding $N_{2}$ swells the denominator of $Q=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$: $Q$ drops below $K_{c}$, and the system responds in the forward direction: more ammonia. Removing $NH_{3}$ crushes the numerator: same conclusion, $Q<K_{c}$, and production is restarted continuously. This is the industrial strategy: draw off the product (liquefied by cooling) so that equilibrium keeps running after it.

On the left: $1+3=4$ moles of gas; on the right: 2. Compressing pushes the system to reduce the number of moles of gas: a shift to the RIGHT, toward ammonia. Hence the 200 atmospheres of the process: pressure is a straightforward ally here.

The reaction is exothermic: heat is a PRODUCT. Heating means adding product: equilibrium retreats toward the reactants, and the yield of $NH_{3}$ falls. And unlike additions of matter or compressions, temperature changes the VALUE of $K_{c}$ itself (it decreases here): temperature is the only disturbance that redefines the target instead of moving the system relative to it.

When cold, equilibrium is favourable... but the KINETICS are hopeless: the triple bond of $N_{2}$ makes the reaction infinitely slow at 25 °C: an excellent yield that is never reached produces nothing. So we heat to 450 °C, giving up some equilibrium yield in exchange for speed, and make up for it with pressure (b) and withdrawal of the product (a). The iron catalyst does not improve the equilibrium one bit: it lowers the activation energy so that this less favourable but accessible equilibrium is REACHED in an industrial time. The whole Haber process is a trade-off between thermodynamics and kinetics: the quintessential exam question.

For an ideal gas, $P=\frac{n}{V}RT=[\,]RT$: a partial pressure is a concentration multiplied by $RT$. Replacing each concentration by $\frac{P}{RT}$ in $K_{c}$, each species contributes a factor $(RT)^{-1}$ raised to its coefficient, and what is left is $K_{p}=K_{c}(RT)^{\Delta n}$ with $\Delta n=$ (moles of GAS among the products) $-$ (moles of gas among the reactants). Only gaseous species count in $\Delta n$.

$\Delta n=2-1=+1$. So $K_{p}=K_{c}(RT)^{1}=(4.63\times 10^{-3})(0.08206)(298)=(4.63\times 10^{-3})(24.45)=0.113$.

When $\Delta n=0$: the same number of moles of gas on each side, so the factor $(RT)^{0}=1$. Example from the chapter: $H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)$, with $1+1=2$ on the left and 2 on the right. For this reaction, $K_{p}$ and $K_{c}$ are numerically identical at every temperature.

$K_{c}=[CO_{2}]$ and $K_{p}=P_{CO_{2}}$: a single term. A pure solid has a \"concentration\" fixed by its density and molar mass, independent of the amount present: grinding the limestone or putting in ten times more changes nothing. This constant value is absorbed into the equilibrium constant (formally, the activity of a pure solid is 1). A concrete and counterintuitive consequence: in a closed furnace at a given temperature, the equilibrium pressure of $CO_{2}$ does NOT depend on the amount of limestone, as long as some remains.

The student is off by a factor of $\frac{8.314}{0.08206}\approx 101.3$: exactly the number of kilopascals in an atmosphere, which is no accident, since the two values of $R$ differ only in the units of pressure (kPa versus atm) and of volume. The rule: $R$ must speak the same language as the quantities used. Pressures in atm and concentrations in mol/L require $R=0.08206$ L·atm/(mol·K); $R=8.314$ J/(mol·K) is the $R$ of energies (Arrhenius, thermodynamics), not that of gas equilibria expressed in atmospheres.

$[CO]_{0}=[H_{2}O]_{0}=\frac{1.00}{10.0}=0.100$ mol/L. ICE: at equilibrium $[CO]=[H_{2}O]=0.100-x$ and $[CO_{2}]=[H_{2}]=x$. Hence $K_{c}=\frac{x^{2}}{(0.100-x)^{2}}=4.0$.

Both sides are squares: $\left(\frac{x}{0.100-x}\right)^{2}=4.0$, so $\frac{x}{0.100-x}=2.0$ (the negative root has no physical meaning). Then $x=0.200-2x$, that is $3x=0.200$ and $x=0.0667$ mol/L. Spotting the perfect square spares you the quadratic: a reflex that pays off whenever the initial concentrations of the reactants are equal and the coefficients are symmetric.

$[CO_{2}]=[H_{2}]=0.0667$ mol/L; $[CO]=[H_{2}O]=0.0333$ mol/L. Conversion of $CO$: $\frac{0.0667}{0.100}\approx 67$%: two thirds of the monoxide has reacted.

$\frac{(0.0667)^{2}}{(0.0333)^{2}}=(2.0)^{2}=4.0$: this is indeed $K_{c}$.

Adding steam makes $Q$ drop below $K_{c}$: the system sets off again in the forward direction and consumes extra $CO$: the conversion of $CO$ INCREASES (at the price of excess steam that, for its part, is converted proportionally less). This is the classic industrial lever: to push the conversion of the EXPENSIVE reactant ($CO$ from coal or methane), you drown the equilibrium in the CHEAP reactant (water). Le Chatelier does the rest: the excess of one buys the conversion of the other.

$\Delta H>0$: the forward reaction is ENDOTHERMIC. Reasoning: heating favours the direction that absorbs heat; and heating increases $K_{c}$, so it favours the forward direction; the forward direction is therefore the one that consumes heat. (Actual value: about $+57$ kJ/mol, energy is needed to break the N-N bond of the dimer.)

In hot water, the bulb TURNS BROWN: equilibrium shifts toward brown $NO_{2}$. In ice, it fades: back toward colourless $N_{2}O_{4}$. The colour acts as a real-time indicator of the position of equilibrium: this is the favourite demonstration of the chapter, a Le Chatelier principle you can see with the naked eye, reversible at will.

$\frac{1.3}{4.6\times 10^{-3}}\approx 280$: fifty degrees multiply the constant by almost three hundred. The temperature dependence of $K$ is exponential (it is governed by $e^{-\Delta H/RT}$): for a substantial $\Delta H$, temperature is a massive lever on the position of equilibrium, out of all proportion to any adjustment of concentration.

Temperature changes the VALUE of $K_{c}$ (from $4.6\times 10^{-3}$ to 1.3): the target itself moves. Compression, at fixed temperature, leaves $K_{c}$ intact and only shifts the CONCENTRATIONS so that their combination recovers the same value: here, compressing favours the $N_{2}O_{4}$ side (1 mole of gas against 2). The bulb would confirm it: compressed suddenly, it first darkens (everything is more concentrated, $NO_{2}$ included), then FADES within moments, in the time equilibrium takes to re-establish toward the dimer, with $K_{c}$ unchanged. Two disturbances, two mechanisms: changing the rules of the game, or replaying the same game on a smaller field.

$[N_{2}O_{4}]_{0}=\frac{1.00}{2.00}=0.500$ mol/L. $0.500\alpha=0.100$ mol/L dissociates. ICE: $N_{2}O_{4}$: $0.500\rightarrow 0.500-0.100=0.400$ mol/L; $NO_{2}$: $0\rightarrow 2(0.100)=0.200$ mol/L (the factor 2 comes from the coefficient).

$K_{c}=\frac{[NO_{2}]^{2}}{[N_{2}O_{4}]}=\frac{(0.200)^{2}}{0.400}=\frac{0.0400}{0.400}=0.100$.

In 4.00 L: $[N_{2}O_{4}]_{0}=0.250$ mol/L. With $x$ dissociated: $\frac{(2x)^{2}}{0.250-x}=0.100$, that is $4x^{2}+0.100x-0.0250=0$. $x=\frac{-0.100+\sqrt{0.0100+0.400}}{8}=\frac{-0.100+0.6403}{8}=0.0675$ mol/L. So $\alpha=\frac{0.0675}{0.250}=27.0$%: the dissociation has INCREASED.

No contradiction: $K_{c}$ has not moved (0.100 in both cases, the temperature has not changed), and that is the only thing the constant promises. The degree of dissociation is not a constant: it is a composition, and it depends on dilution. Le Chatelier says it directly: diluting means lowering all the concentrations, and the system responds by shifting toward the side with the MOST moles of gas, here the 2 $NO_{2}$ against 1 $N_{2}O_{4}$, to make more of them. Confusing \"equilibrium constant\" with \"constant composition\" is the basic error of the chapter.

Prediction: compressing goes the opposite way from diluting, so equilibrium shifts toward the side with the FEWEST moles of gas, that is toward $N_{2}O_{4}$: $\alpha$ must DECREASE. Check with $[N_{2}O_{4}]_{0}=1.00$ mol/L: $4x^{2}+0.100x-0.100=0$ gives $x=\frac{-0.100+\sqrt{0.0100+1.60}}{8}=\frac{-0.100+1.2689}{8}=0.1461$, that is $\alpha=14.6$%. The prediction is confirmed: 27.0% at 4 L, 20.0% at 2 L, 14.6% at 1 L. The brown gas $NO_{2}$ fades when you compress it, which you can see with the naked eye in the demonstration syringe.

$K_{c}=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}=\frac{(0.10)^{2}}{(0.20)(0.30)^{3}}=\frac{0.010}{(0.20)(0.027)}=\frac{0.010}{0.0054}=1.85$.

Nothing happens. At constant volume, adding argon alters NONE of the three concentrations: $\frac{n}{V}$ is unchanged for $N_{2}$, $H_{2}$ and $NH_{3}$. So $Q$ keeps exactly the value $K_{c}$ and the system stays where it is. The TOTAL pressure rises, but equilibrium knows only the PARTIAL pressures of the species that take part: argon is a spectator.

This time equilibrium shifts to the LEFT, toward $N_{2}+H_{2}$. At constant pressure, adding argon forces the piston out: the volume increases, so all the concentrations of the reacting species DECREASE. It is a dilution in disguise, and the system responds by going to the side with the most moles of gas (4 on the left, 2 on the right). The whole difference from b) comes from what is held fixed: at constant $V$ argon has no effect, at constant $P$ it acts as a diluent.

Doubling $V$ divides each concentration by 2: $[N_{2}]=0.10$, $[H_{2}]=0.15$, $[NH_{3}]=0.050$. $Q=\frac{(0.050)^{2}}{(0.10)(0.15)^{3}}=\frac{0.0025}{3.375\times 10^{-4}}=7.41$. Since $Q>K_{c}$ (7.41 against 1.85), the system consumes products: a shift to the LEFT, consistent with c). Check: $\Delta n=2-4=-2$, and dividing the concentrations by 2 multiplies $Q$ by $2^{-\Delta n}=2^{2}=4$. And $\frac{7.41}{1.85}=4.00$ exactly. The reaction quotient gives the direction without anything to memorize.

No, it is higher than before the addition, somewhere between 0.20 mol/L and the value just after injection. Le Chatelier's principle says that the system PARTLY opposes the disturbance, not that it cancels it: the shift to the right consumes part of the added $N_{2}$, never all of it, otherwise nothing would have changed and equilibrium would be indifferent to what is done to it. It is a moderation, not a restoration. Incidentally, this shift increases the yield of $NH_{3}$ while LOWERING the percent conversion of $N_{2}$, which is precisely why an excess of the cheaper reactant is injected instead.

$\Delta H^{\circ}>0$: the reaction is ENDOTHERMIC. The constant increases with temperature, so heating favours the products. Now Le Chatelier requires the system to absorb the added heat: it can only do so by shifting in the direction that CONSUMES heat, that is, the endothermic direction. If that direction is the one toward the products, the forward reaction is endothermic.

$\ln\frac{0.50}{0.040}=\ln 12.5=2.526$. $\frac{1}{298}-\frac{1}{350}=3.3557\times 10^{-3}-2.8571\times 10^{-3}=4.986\times 10^{-4}$. So $\Delta H^{\circ}=\frac{R\ln 12.5}{4.986\times 10^{-4}}=\frac{(8.314)(2.526)}{4.986\times 10^{-4}}=4.21\times 10^{4}$ J/mol, that is $\Delta H^{\circ}\approx +42.1$ kJ/mol. The positive sign confirms a).

From 350 K to 400 K: $\ln\frac{K_{400}}{0.50}=\frac{42{,}118}{8.314}\left(\frac{1}{350}-\frac{1}{400}\right)=(5066)(3.571\times 10^{-4})=1.809$, so $K_{400}=0.50\,e^{1.809}\approx 3.1$. The constant has been multiplied by more than 75 between 298 K and 400 K: endothermic equilibria are very sensitive to temperature.

Adding two reactions MULTIPLIES their constants (whereas enthalpy, for its part, adds: $K$ is exponential in $\Delta G^{\circ}$). $A\rightleftharpoons C$ is the sum of the two steps: $K=K_{1}K_{2}=(2.0)(3.0)=6.0$. Multiplying an equation by 2 SQUARES the constant: $2A\rightleftharpoons 2C$ gives $K=6.0^{2}=36$. Reversing an equation takes the INVERSE: $C\rightleftharpoons A$ gives $K=\frac{1}{6.0}=0.167$.

It assumes $\Delta H^{\circ}$ is CONSTANT over the whole temperature interval considered. This is an approximation: the heat capacities of the reactants and products differ, so $\Delta H^{\circ}$ drifts slowly with $T$. Over 50 K the error stays modest; extrapolating from 350 K to 400 K, beyond the range where the two measurements were made, accumulates this drift and the uncertainty on $\Delta H^{\circ}$ itself. Hence \"estimate\": the calculation is right, it is the model that is only approximately valid, and less so the further you move from the measured points.