Homogeneous Equilibria and the Relation Between Kp and Kc

One Phase, One Set of Rules

An equilibrium is homogeneous when every reactant and every product sits in the same phase. Nothing is a solid while something else is a gas; nothing is a liquid while something else is dissolved. One phase, throughout.

Key Point (Definition): In a homogeneous equilibrium all the reacting species — reactants and products alike — are present in a single phase, either all gaseous or all in the same solution.

Two families of homogeneous equilibria turn up again and again.

All species gaseous

The synthesis of ammonia is the standard example:

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

Nitrogen, hydrogen and ammonia are all gases at the temperature of the reaction, so the mixture is a single gas phase filling the vessel uniformly. The same is true of

$$\mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons 2\mathrm{HI}(g)$$

$$2\mathrm{SO_2}(g) + \mathrm{O_2}(g) \rightleftharpoons 2\mathrm{SO_3}(g)$$

$$\mathrm{PCl_5}(g) \rightleftharpoons \mathrm{PCl_3}(g) + \mathrm{Cl_2}(g)$$

$$\mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO_2}(g)$$

Each of these can be handled with concentrations, and each can equally well be handled with pressures. That second option is what this section is about.

All species in one solution

Homogeneous does not mean gaseous. A reaction in which everything is dissolved in the same solvent is equally homogeneous. The hydrolysis of ethyl acetate is written

$$\mathrm{CH_3COOC_2H_5}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{CH_3COOH}(aq) + \mathrm{C_2H_5OH}(aq)$$

and the formation of the blood-red thiocyanato complex of iron is

$$\mathrm{Fe^{3+}}(aq) + \mathrm{SCN^-}(aq) \rightleftharpoons \mathrm{[Fe(SCN)]^{2+}}(aq)$$

Every species here is in the aqueous solution phase. For such equilibria only $K_c$ makes sense, because dissolved species have concentrations but no partial pressures of their own. Nothing in this section about $K_p$ applies to them.

What homogeneous buys you

The reason the distinction matters is a practical one. In a homogeneous equilibrium every species that appears in the balanced equation also appears in the equilibrium constant expression. Nothing gets left out. The moment a pure solid or a pure liquid enters the equation the system becomes heterogeneous, and those species drop out of the expression — a separate rule dealt with in the next section.

So for

$$a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$$

with all four species in one phase,

$$K_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$$

and no term is missing.

[NEET] A one-line test for homogeneous: read the state symbols. If they are all $(g)$, or all $(aq)$, the equilibrium is homogeneous. A single $(s)$, or an $(l)$ present as a separate pure phase, makes it heterogeneous — with one exception, the solvent. The $\mathrm{H_2O}(l)$ written in the ethyl acetate hydrolysis above is the medium the reaction is run in and not a second phase, so that equilibrium is homogeneous even though an $(l)$ appears in the equation.

Why Gases Prefer Pressure

A gas mixture at equilibrium can be described in two equivalent ways. You can quote the molar concentration of each gas, $[\mathrm{A}] = n_\mathrm{A}/V$ in $\mathrm{mol\ L^{-1}}$. Or you can quote the partial pressure of each gas.

Key Point (Definition): The partial pressure of a gas in a mixture is the pressure that gas alone would exert if it occupied the whole container at the same temperature. The total pressure is the sum of the partial pressures (Dalton's law), and $p_i = x_i P_{\text{total}}$ where $x_i$ is the mole fraction of that gas.

Pressure wins on convenience for three reasons.

It is what the instrument reads. A pressure gauge on a reaction vessel gives the total pressure directly. Getting concentrations instead would mean withdrawing a sample, analysing it and dividing by the volume — slower and less accurate.

Composition follows from pressure without extra measurement. If a gas mixture is 40% by volume of one component, that component's mole fraction is 0.40, and its partial pressure is $0.40 \times P_{\text{total}}$. Volume percentage, mole percentage and pressure percentage are the same number for ideal gases.

Standard states for gases are defined in pressure. The standard state of a gas is a pressure of exactly 1 bar. Thermodynamic quantities such as $\Delta G^{\circ}$ are tabulated against that reference, so an equilibrium constant already written in pressures links straight to thermodynamics.

Writing $K_p$

The recipe is identical to the one for $K_c$; only the quantity substituted changes. Products over reactants, each raised to its stoichiometric coefficient, with partial pressures in place of concentrations.

For $\mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons 2\mathrm{HI}(g)$:

$$K_p = \frac{(p_{\mathrm{HI}})^2}{(p_{\mathrm{H_2}})(p_{\mathrm{I_2}})}$$

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

$$K_p = \frac{(p_{\mathrm{NH_3}})^2}{(p_{\mathrm{N_2}})(p_{\mathrm{H_2}})^3}$$

For $2\mathrm{SO_2}(g) + \mathrm{O_2}(g) \rightleftharpoons 2\mathrm{SO_3}(g)$:

$$K_p = \frac{(p_{\mathrm{SO_3}})^2}{(p_{\mathrm{SO_2}})^2(p_{\mathrm{O_2}})}$$

Key Point: $K_p$ is the equilibrium constant written with equilibrium partial pressures of the gaseous species, each raised to its stoichiometric coefficient, products in the numerator.

Both constants describe the same equilibrium at the same temperature, so they must be connected. The connection is the ideal gas equation.

Deriving $K_p = K_c(RT)^{\Delta n}$

The whole derivation rests on one substitution. Start from the ideal gas equation for a single gas in the mixture:

$$pV = nRT$$

Divide both sides by $V$:

$$p = \frac{n}{V}RT$$

The ratio $n/V$ is moles per unit volume, which is exactly molar concentration. Writing $c$ for it,

$$p = cRT$$

or, for a species $\mathrm{A}$ in the mixture,

$$p_{\mathrm{A}} = [\mathrm{A}]RT$$

At a fixed temperature $RT$ is a constant, so the partial pressure of each gas is directly proportional to its molar concentration: $p \propto [\text{gas}]$.

The general case, step by step

Take the general homogeneous gaseous equilibrium

$$a\mathrm{A}(g) + b\mathrm{B}(g) \rightleftharpoons c\mathrm{C}(g) + d\mathrm{D}(g)$$

Step 1. Write $K_p$ from the definition:

$$K_p = \frac{(p_{\mathrm{C}})^c(p_{\mathrm{D}})^d}{(p_{\mathrm{A}})^a(p_{\mathrm{B}})^b}$$

Step 2. Replace every partial pressure by $p_i = [i]RT$:

$$K_p = \frac{([\mathrm{C}]RT)^c([\mathrm{D}]RT)^d}{([\mathrm{A}]RT)^a([\mathrm{B}]RT)^b}$$

Step 3. Expand each bracket. A product raised to a power is the product of the powers, so $([\mathrm{C}]RT)^c = [\mathrm{C}]^c(RT)^c$:

$$K_p = \frac{[\mathrm{C}]^c(RT)^c \cdot [\mathrm{D}]^d(RT)^d}{[\mathrm{A}]^a(RT)^a \cdot [\mathrm{B}]^b(RT)^b}$$

Step 4. Collect the concentration terms in one fraction and the $RT$ terms in another:

$$K_p = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \times \frac{(RT)^{c+d}}{(RT)^{a+b}}$$

Step 5. The first fraction is $K_c$. The second collapses by the law of indices, $x^m/x^n = x^{m-n}$:

$$K_p = K_c \times (RT)^{(c+d)-(a+b)}$$

Step 6. Name the exponent. Writing $\Delta n$ for $(c+d) - (a+b)$,

$$\boxed{\,K_p = K_c(RT)^{\Delta n}\,}$$

Key Point: $K_p = K_c(RT)^{\Delta n}$, where $\Delta n =$ (total number of moles of gaseous products) $-$ (total number of moles of gaseous reactants), read off the balanced equation.

A worked check on the formula

Apply it to $\mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons 2\mathrm{HI}(g)$ from scratch instead of quoting the result:

$$K_p = \frac{(p_{\mathrm{HI}})^2}{(p_{\mathrm{H_2}})(p_{\mathrm{I_2}})} = \frac{([\mathrm{HI}]RT)^2}{([\mathrm{H_2}]RT)([\mathrm{I_2}]RT)} = \frac{[\mathrm{HI}]^2(RT)^2}{[\mathrm{H_2}][\mathrm{I_2}](RT)^2}$$

The $(RT)^2$ cancels top and bottom:

$$K_p = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} = K_c$$

Consistent with $\Delta n = 2 - 2 = 0$ and $(RT)^0 = 1$.

Now the ammonia synthesis, where the cancellation is incomplete:

$$K_p = \frac{(p_{\mathrm{NH_3}})^2}{(p_{\mathrm{N_2}})(p_{\mathrm{H_2}})^3} = \frac{[\mathrm{NH_3}]^2(RT)^2}{[\mathrm{N_2}](RT) \cdot [\mathrm{H_2}]^3(RT)^3} = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}(RT)^{2-4}$$

$$K_p = K_c(RT)^{-2}$$

which matches $\Delta n = 2 - (1+3) = -2$.

The Three Cases of $\Delta n$

Everything the relation can do is decided by the sign of one small integer.

Case 1: $\Delta n > 0$ — more gas moles on the right

$$\mathrm{PCl_5}(g) \rightleftharpoons \mathrm{PCl_3}(g) + \mathrm{Cl_2}(g), \qquad \Delta n = 2 - 1 = +1$$

$$K_p = K_c(RT)^{+1} = K_c(RT)$$

At any ordinary temperature $RT$ is a large number — at 500 K it is $0.0831 \times 500 = 41.6$ — so $K_p$ comes out larger than $K_c$, here by a factor of about 42.

Another with $\Delta n = +1$: $\ \mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO_2}(g)$, since $2 - 1 = +1$.

Case 2: $\Delta n < 0$ — more gas moles on the left

$$2\mathrm{SO_2}(g) + \mathrm{O_2}(g) \rightleftharpoons 2\mathrm{SO_3}(g), \qquad \Delta n = 2 - 3 = -1$$

$$K_p = K_c(RT)^{-1} = \frac{K_c}{RT}$$

Now $K_p$ is smaller than $K_c$, divided by that same large factor.

The strongest common example is ammonia synthesis, $\mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightleftharpoons 2\mathrm{NH_3}(g)$, with $\Delta n = 2 - 4 = -2$ and

$$K_p = \frac{K_c}{(RT)^2}$$

At 500 K that divides by $41.6^2 \approx 1730$.

Case 3: $\Delta n = 0$ — the numbers are equal

$$\mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons 2\mathrm{HI}(g), \qquad \Delta n = 2 - 2 = 0$$

$$K_p = K_c(RT)^0 = K_c \times 1 = K_c$$

Key Point: $K_p = K_c$ exactly when $\Delta n = 0$, and only then. This holds at every temperature, because $(RT)^0 = 1$ whatever $T$ may be.

Other equilibria with $\Delta n = 0$:

$$\mathrm{CO}(g) + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{CO_2}(g) + \mathrm{H_2}(g), \qquad \Delta n = 2 - 2 = 0$$

$$\mathrm{N_2}(g) + \mathrm{O_2}(g) \rightleftharpoons 2\mathrm{NO}(g), \qquad \Delta n = 2 - 2 = 0$$

$$2\mathrm{HI}(g) \rightleftharpoons \mathrm{H_2}(g) + \mathrm{I_2}(g), \qquad \Delta n = 2 - 2 = 0$$

The three cases side by side

$\Delta n$	Meaning	Relation	Size comparison	Named example
$+1$	one extra mole of gas formed	$K_p = K_c(RT)$	$K_p > K_c$	$\mathrm{PCl_5} \rightleftharpoons \mathrm{PCl_3} + \mathrm{Cl_2}$
$+2$	two extra moles of gas formed	$K_p = K_c(RT)^2$	$K_p \gg K_c$	$\mathrm{NH_4HS}(s) \rightleftharpoons \mathrm{NH_3}(g) + \mathrm{H_2S}(g)$
$0$	gas moles unchanged	$K_p = K_c$	equal	$\mathrm{H_2} + \mathrm{I_2} \rightleftharpoons 2\mathrm{HI}$
$-1$	one mole of gas consumed	$K_p = K_c/(RT)$	$K_p < K_c$	$2\mathrm{SO_2} + \mathrm{O_2} \rightleftharpoons 2\mathrm{SO_3}$
$-2$	two moles of gas consumed	$K_p = K_c/(RT)^2$	$K_p \ll K_c$	$\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$

[JEE Main] A question that gives $K_c$ and asks for $K_p$ without giving $T$ is answerable only if $\Delta n = 0$. Check $\Delta n$ before hunting for a missing temperature.

$\Delta n$ Counts Gases and Nothing Else

This is where the marks go.

Key Point: $\Delta n$ in $K_p = K_c(RT)^{\Delta n}$ is the change in the number of moles of gaseous species only. Solids, pure liquids and dissolved species contribute zero, on both sides.

The reason follows from the derivation. The factors of $RT$ appeared because each gaseous partial pressure was replaced by $[i]RT$. A solid or a pure liquid never entered the equilibrium expression at all, so it never contributed an $RT$, so it cannot contribute to the exponent.

The error to avoid

The commonest mistake is to count every species in the balanced equation — the total change in moles — instead of only the gaseous ones. Three reactions expose the difference immediately.

$$\mathrm{C}(s) + \mathrm{CO_2}(g) \rightleftharpoons 2\mathrm{CO}(g)$$

Total moles: $2$ on the right, $2$ on the left, so a total change of $0$. Gaseous moles: $2$ on the right, $1$ on the left, so $\Delta n = 2 - 1 = +1$. The correct relation is $K_p = K_c(RT)$, and a student who used the total change would have written $K_p = K_c$ and lost the whole answer.

$$\mathrm{NH_4HS}(s) \rightleftharpoons \mathrm{NH_3}(g) + \mathrm{H_2S}(g)$$

Total change: $2 - 1 = +1$. Gaseous change: $2 - 0 = +2$. So $K_p = K_c(RT)^2$, not $K_c(RT)$.

$$\mathrm{Fe_2O_3}(s) + 3\mathrm{CO}(g) \rightleftharpoons 2\mathrm{Fe}(s) + 3\mathrm{CO_2}(g)$$

Total change: $5 - 4 = +1$. Gaseous change: $3 - 3 = 0$. So $K_p = K_c$ exactly, at every temperature, even though the equation looks lopsided.

The habit that prevents it

Before touching the formula, rewrite the equation with the non-gaseous species struck out of the count:

1. List the coefficients of the $(g)$ species on the product side and add them.
2. List the coefficients of the $(g)$ species on the reactant side and add them.
3. Subtract: products minus reactants. That number, and only that number, is $\Delta n$.

$(aq)$ species are counted the same way as solids — that is, not at all. A species dissolved in water has no partial pressure, so it contributes no $RT$.

[JEE/NEET] In a heterogeneous equilibrium, the solid and liquid species are absent from the $K_p$ expression and absent from $\Delta n$. Both omissions are the same omission.

Units, $R$ and the Standard-State Convention

Choosing $R$ to match the pressure unit

The value of $R$ substituted into $(RT)^{\Delta n}$ must carry the same pressure unit that $K_p$ is expressed in. Two combinations are in common use.

Pressure unit for $K_p$	Value of $R$ to use	Units of $R$
bar	$0.0831$	$\mathrm{L\ bar\ K^{-1}\ mol^{-1}}$
atm	$0.0821$	$\mathrm{L\ atm\ K^{-1}\ mol^{-1}}$

Concentrations are in $\mathrm{mol\ L^{-1}}$ in both cases, and $T$ is always in kelvin.

This section uses $R = 0.0831\ \mathrm{L\ bar\ K^{-1}\ mol^{-1}}$ and expresses every $K_p$ in bar, because the standard state for pressure is defined as exactly 1 bar. Where a problem quotes $K_p$ in atm, $R = 0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$ is used instead so that the units agree. Mixing them — a $K_p$ in atm with $R = 0.0831$ — produces a wrong number by a factor of about 1.2% per power of $\Delta n$, which is enough to miss a printed option.

Useful conversions: $1\ \mathrm{bar} = 10^5\ \mathrm{Pa}$, and $1\ \mathrm{atm} = 1.013\ \mathrm{bar} = 1.013 \times 10^5\ \mathrm{Pa}$.

Units of $K_c$ and $K_p$

Substituting concentrations in $\mathrm{mol\ L^{-1}}$ and pressures in bar leaves the constant with units unless the powers on top and bottom cancel:

• units of $K_c$ are $(\mathrm{mol\ L^{-1}})^{\Delta n}$
• units of $K_p$ are $(\mathrm{bar})^{\Delta n}$

Worked through for three reactions:

Reaction	$\Delta n$	Units of $K_c$	Units of $K_p$
$\mathrm{H_2} + \mathrm{I_2} \rightleftharpoons 2\mathrm{HI}$	$0$	none	none
$\mathrm{N_2O_4} \rightleftharpoons 2\mathrm{NO_2}$	$+1$	$\mathrm{mol\ L^{-1}}$	$\mathrm{bar}$
$\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$	$-2$	$\mathrm{mol^{-2}\ L^{2}}$	$\mathrm{bar^{-2}}$

The same $\Delta n$ that sets the power of $RT$ sets the units — one fact doing two jobs.

Why the constants are strictly dimensionless

Each concentration is properly divided by the standard concentration $c^{\circ} = 1\ \mathrm{mol\ L^{-1}}$, and each pressure by the standard pressure $p^{\circ} = 1\ \mathrm{bar}$. A pressure of 4 bar becomes $4\ \mathrm{bar}/1\ \mathrm{bar} = 4$, a pure number. Every term entering the expression is then a ratio, and $K_p$ and $K_c$ come out dimensionless.

The numerical values are still different whenever $\Delta n \neq 0$, because the two standard states are different. Being dimensionless does not make them equal.

Key Point: Written against standard states, both $K_p$ and $K_c$ are dimensionless. In routine numerical work the units $(\mathrm{bar})^{\Delta n}$ and $(\mathrm{mol\ L^{-1}})^{\Delta n}$ are still quoted, and quoting them is accepted.

[Board] State the temperature and the pressure unit whenever a $K_p$ value is written down. A bare number with neither is incomplete.

Worked Questions

Question 1: Reading $\Delta n$ off four equations

State $\Delta n$ for each: (i) $2\mathrm{NOCl}(g) \rightleftharpoons 2\mathrm{NO}(g) + \mathrm{Cl_2}(g)$, (ii) $\mathrm{CaCO_3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO_2}(g)$, (iii) $\mathrm{C}(s) + \mathrm{CO_2}(g) \rightleftharpoons 2\mathrm{CO}(g)$, (iv) $\mathrm{Fe_2O_3}(s) + 3\mathrm{CO}(g) \rightleftharpoons 2\mathrm{Fe}(s) + 3\mathrm{CO_2}(g)$.

Answer:

I count only the species carrying $(g)$.

(i) Gaseous products $2 + 1 = 3$; gaseous reactants $2$. So $\Delta n = 3 - 2 = +1$.

(ii) Gaseous products $1$ (the $\mathrm{CO_2}$); gaseous reactants $0$, since $\mathrm{CaCO_3}$ is a solid. So $\Delta n = 1 - 0 = +1$.

(iii) Gaseous products $2$; gaseous reactants $1$, because the graphite does not count. So $\Delta n = 2 - 1 = +1$.

(iv) Gaseous products $3$; gaseous reactants $3$. So $\Delta n = 3 - 3 = 0$, and $K_p = K_c$ for this one.

Ans: (i) $+1$, (ii) $+1$, (iii) $+1$, (iv) $0$ Watch out: In (iii) the total number of moles does not change at all, and in (iv) the total goes up by one. Both totals are irrelevant.

Question 2: $K_c$ to $K_p$ for nitrosyl chloride

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:

First I find $\Delta n$. Gaseous products $= 2 + 1 = 3$, gaseous reactants $= 2$, so $\Delta n = +1$.

The relation is $K_p = K_c(RT)^{\Delta n} = K_c(RT)^1$.

With $R = 0.0831\ \mathrm{L\ bar\ K^{-1}\ mol^{-1}}$ and $T = 1069\ \mathrm{K}$:

$$RT = 0.0831 \times 1069 = 88.83$$

$$K_p = 3.75 \times 10^{-6} \times 88.83 = 3.33 \times 10^{-4}$$

Ans: $K_p = 3.33 \times 10^{-4}\ \mathrm{bar}$ Watch out: $\Delta n$ is $+1$, so $RT$ multiplies. Dividing by 88.83 instead gives $4.2 \times 10^{-8}$, which is the answer to the reverse conversion.

Question 3: $K_p$ to $K_c$ for the contact process

At 450 K, $K_p = 2.0 \times 10^{10}\ \mathrm{bar^{-1}}$ for $2\mathrm{SO_2}(g) + \mathrm{O_2}(g) \rightleftharpoons 2\mathrm{SO_3}(g)$. Find $K_c$.

Answer:

Gaseous products $= 2$, gaseous reactants $= 2 + 1 = 3$, so $\Delta n = 2 - 3 = -1$.

$$K_p = K_c(RT)^{-1} \quad \Rightarrow \quad K_c = K_p \times RT$$

$$RT = 0.0831 \times 450 = 37.40$$

$$K_c = 2.0 \times 10^{10} \times 37.40 = 7.48 \times 10^{11}$$

Ans: $K_c = 7.48 \times 10^{11}\ \mathrm{L\ mol^{-1}}$ Watch out: The unit $\mathrm{bar^{-1}}$ printed with $K_p$ is itself a check: $(\mathrm{bar})^{\Delta n}$ with $\Delta n = -1$. If your $\Delta n$ disagrees with the printed unit, one of them is wrong.

Question 4: A carbonate with a solid on each side

$K_p = 167\ \mathrm{kPa}$ at 1073 K for $\mathrm{CaCO_3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO_2}(g)$. Calculate $K_c$.

Answer:

The unit on $K_p$ decides everything here. Both solids drop out, so $K_p$ is simply $p_{\mathrm{CO_2}}$, and the figure 167 is in kilopascals — that is $1.67\ \mathrm{bar}$, sitting sensibly beside the $2.0 \times 10^{5}\ \mathrm{Pa}$ measured for this equilibrium near 1100 K. Read as 167 bar it would be absurd. So I take $R$ in the matching unit, $R = 8.31\ \mathrm{L\ kPa\ K^{-1}\ mol^{-1}}$.

Only $\mathrm{CO_2}$ is a gas, so $\Delta n = 1 - 0 = +1$.

$$RT = 8.31 \times 1073 = 8916$$

$$K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{167}{8916} = 1.87 \times 10^{-2}$$

Ans: $K_c = 1.87 \times 10^{-2}\ \mathrm{mol\ L^{-1}}$ Watch out: Both solids are absent from the $K_p$ expression and absent from $\Delta n$. The other trap is the unit: pairing a $K_p$ of 167 kPa with $R = 0.0831\ \mathrm{L\ bar\ K^{-1}\ mol^{-1}}$ gives $1.87$, a hundred times too large, because $R$ and the pressure unit no longer agree.

Question 5: Phosphorus pentachloride, $K_c$ to $K_p$

For $\mathrm{PCl_5}(g) \rightleftharpoons \mathrm{PCl_3}(g) + \mathrm{Cl_2}(g)$, $K_c = 8.3 \times 10^{-3}$ at 473 K. Find $K_p$.

Answer:

$\Delta n = 2 - 1 = +1$.

$$RT = 0.0831 \times 473 = 39.31$$

$$K_p = K_c(RT) = 8.3 \times 10^{-3} \times 39.31 = 0.326$$

Ans: $K_p = 0.326\ \mathrm{bar}$

Question 6: Iodine vapour partly dissociated

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

Answer:

A total pressure of $10^5\ \mathrm{Pa}$ is $1\ \mathrm{bar}$.

For ideal gases, percentage by volume equals mole percentage, so the mole fraction of I atoms is $0.40$ and that of $\mathrm{I_2}$ is $0.60$.

Using $p_i = x_i P_{\text{total}}$:

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

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

Ans: $K_p = 0.267\ \mathrm{bar}$ Watch out: The coefficient 2 in $2\mathrm{I}$ becomes a square, not a multiplier. Writing $2 \times 0.40/0.60$ gives $1.33$, a wrong answer produced by a very common slip.

Question 7: Hydrogen iodide decomposing, and why $R$ never appears

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

Answer:

The pressure of HI falls from 0.2 atm to 0.04 atm, so 0.16 atm of HI has reacted.

From the stoichiometry, 2 mol HI give 1 mol $\mathrm{H_2}$ and 1 mol $\mathrm{I_2}$. So the pressure of each product formed is $0.16/2 = 0.08\ \mathrm{atm}$.

$$K_p = \frac{p_{\mathrm{H_2}} \cdot p_{\mathrm{I_2}}}{(p_{\mathrm{HI}})^2} = \frac{0.08 \times 0.08}{(0.04)^2} = \frac{0.0064}{0.0016} = 4.0$$

Here $\Delta n = 2 - 2 = 0$, so $K_p = K_c(RT)^0 = K_c$. No value of $R$ and no temperature are needed.

Ans: $K_p = 4.0$ and $K_c = 4.0$, both dimensionless Watch out: The 0.16 atm consumed is halved to get each product pressure. Using 0.16 for $\mathrm{H_2}$ and $\mathrm{I_2}$ gives $16$, four times too large.

Question 8: Working in atmospheres

$K_p = 0.04\ \mathrm{atm}$ at 899 K for $\mathrm{C_2H_6}(g) \rightleftharpoons \mathrm{C_2H_4}(g) + \mathrm{H_2}(g)$. Calculate $K_c$ at this temperature.

Answer:

$K_p$ is quoted in atm, so I use $R = 0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$ to keep the units consistent.

$\Delta n = 2 - 1 = +1$.

$$K_c = \frac{K_p}{RT} = \frac{0.04}{0.0821 \times 899}$$

$$RT = 0.0821 \times 899 = 73.81$$

$$K_c = \frac{0.04}{73.81} = 5.42 \times 10^{-4}$$

Ans: $K_c = 5.42 \times 10^{-4}\ \mathrm{mol\ L^{-1}}$ Watch out: Using $R = 0.0831$ with a $K_p$ in atm gives $5.35 \times 10^{-4}$ — close enough to look right and wrong enough to miss the printed option.

Question 9: The ratio $K_p/K_c$ for ammonia synthesis

For $\mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightleftharpoons 2\mathrm{NH_3}(g)$ at 500 K, find the ratio $K_p/K_c$.

Answer:

$\Delta n = 2 - 4 = -2$, so

$$\frac{K_p}{K_c} = (RT)^{\Delta n} = (RT)^{-2} = \frac{1}{(RT)^2}$$

$$RT = 0.0831 \times 500 = 41.55$$

$$(RT)^2 = 1726.4$$

$$\frac{K_p}{K_c} = \frac{1}{1726.4} = 5.79 \times 10^{-4}$$

Ans: $K_p/K_c = 5.79 \times 10^{-4}$, so $K_p$ is about 1700 times smaller than $K_c$ Watch out: $\Delta n$ is $-2$, not $-1$. Using $-1$ gives $2.41 \times 10^{-2}$.

Question 10: When the answer needs no temperature

For which of these can $K_p$ be found from $K_c$ without knowing $T$? (i) $\mathrm{CO}(g) + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{CO_2}(g) + \mathrm{H_2}(g)$, (ii) $2\mathrm{NO_2}(g) \rightleftharpoons \mathrm{N_2O_4}(g)$, (iii) $\mathrm{Fe_2O_3}(s) + 3\mathrm{CO}(g) \rightleftharpoons 2\mathrm{Fe}(s) + 3\mathrm{CO_2}(g)$.

Answer:

I check $\Delta n$ in each case.

(i) $2 - 2 = 0$, so $K_p = K_c$. No temperature needed.

(ii) $1 - 2 = -1$, so $K_p = K_c/(RT)$. Temperature needed.

(iii) Gaseous only: $3 - 3 = 0$, so $K_p = K_c$. No temperature needed, even though solids appear in the equation.

Ans: (i) and (iii) Watch out: In (iii), counting all species gives a total change of $+1$ and the false conclusion that $T$ is required.

Question 11: Building $K_p$ from a starting pressure

A vessel at $400\ {}^{\circ}\mathrm{C}$ is charged with an equimolar mixture of CO and steam at $p_{\mathrm{CO}} = p_{\mathrm{H_2O}} = 4.0\ \mathrm{bar}$. For $\mathrm{CO}(g) + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{CO_2}(g) + \mathrm{H_2}(g)$, $K_p = 10.1$. Find the equilibrium partial pressure of $\mathrm{H_2}$.

Answer:

Let $x$ bar of CO react. Then $x$ bar of steam also reacts, and $x$ bar each of $\mathrm{CO_2}$ and $\mathrm{H_2}$ form.

At equilibrium: $p_{\mathrm{CO}} = p_{\mathrm{H_2O}} = (4.0 - x)$, and $p_{\mathrm{CO_2}} = p_{\mathrm{H_2}} = x$.

$$K_p = \frac{x \cdot x}{(4.0-x)(4.0-x)} = \frac{x^2}{(4.0-x)^2} = 10.1$$

Both sides are perfect squares, so I take the square root:

$$\frac{x}{4.0-x} = \sqrt{10.1} = 3.178$$

$$x = 3.178(4.0 - x) = 12.71 - 3.178x$$

$$4.178x = 12.71 \quad \Rightarrow \quad x = 3.04$$

Ans: $p_{\mathrm{H_2}} = 3.04\ \mathrm{bar}$ Watch out: $\Delta n = 0$ here, so no conversion between $K_p$ and $K_c$ is involved at any stage. Taking the square root turns a quadratic into one line.

Question 12: Same reaction, opposite direction

$K_p$ for $\mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO_2}(g)$ is $0.115\ \mathrm{bar}$ at 298 K. Find $K_c$ for this reaction, and then $K_c$ for $2\mathrm{NO_2}(g) \rightleftharpoons \mathrm{N_2O_4}(g)$.

Answer:

For the forward equation as written, $\Delta n = 2 - 1 = +1$.

$$RT = 0.0831 \times 298 = 24.76$$

$$K_c = \frac{K_p}{RT} = \frac{0.115}{24.76} = 4.64 \times 10^{-3}$$

Reversing an equation inverts its equilibrium constant:

$$K_c' = \frac{1}{4.64 \times 10^{-3}} = 215$$

Ans: $K_c = 4.64 \times 10^{-3}\ \mathrm{mol\ L^{-1}}$ for the dissociation, the value quoted for this equilibrium at 298 K throughout the chapter; $K_c' = 215\ \mathrm{L\ mol^{-1}}$ for the dimerisation Watch out: Convert first, invert second — or invert first and convert with $\Delta n = -1$. Both routes agree; mixing the sign of $\Delta n$ with the inversion does not.

Fixing the Section in Place

The chain of reasoning in six lines

1. All species in one phase $\Rightarrow$ homogeneous equilibrium, nothing dropped from the expression.
2. For gases, partial pressure is the natural variable: it is measured directly and defines the standard state.
3. $pV = nRT$ gives $p = (n/V)RT = cRT$ for each gas separately.
4. Substituting $p_i = [i]RT$ into $K_p$ pulls out a factor $(RT)$ for every mole of gas in the equation.
5. The factors collect into $(RT)^{\Delta n}$, giving $K_p = K_c(RT)^{\Delta n}$.
6. $\Delta n$ = gaseous products $-$ gaseous reactants. Nothing else is counted.

Four errors that cost marks

Counting non-gaseous species in $\Delta n$. $\mathrm{C}(s) + \mathrm{CO_2}(g) \rightleftharpoons 2\mathrm{CO}(g)$ has $\Delta n = +1$, not 0. Strike out the solids and liquids before counting.

Getting the sign backwards. $K_p = K_c(RT)^{\Delta n}$ means multiply by $(RT)^{\Delta n}$ when going from $K_c$ to $K_p$, and divide by it when going the other way. A quick sanity check: for $\Delta n > 0$, $K_p$ must be the bigger number.

Mismatching $R$ with the pressure unit. Bar goes with $0.0831$, atm goes with $0.0821$. Choose $R$ after looking at the unit printed on $K_p$.

Forgetting the stoichiometric powers inside the pressure expression. $K_p$ for $\mathrm{I_2} \rightleftharpoons 2\mathrm{I}$ is $(p_{\mathrm{I}})^2/p_{\mathrm{I_2}}$, never $2p_{\mathrm{I}}/p_{\mathrm{I_2}}$.

Formula card

$$p_i = [i]RT \qquad p_i = x_i P_{\text{total}} \qquad P_{\text{total}} = \sum p_i$$

$$K_p = K_c(RT)^{\Delta n}, \qquad \Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$$

$$R = 0.0831\ \mathrm{L\ bar\ K^{-1}\ mol^{-1}} \text{ with } K_p \text{ in bar}; \qquad R = 0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}} \text{ with } K_p \text{ in atm}$$

$$\text{Units: } K_c \text{ in } (\mathrm{mol\ L^{-1}})^{\Delta n}, \quad K_p \text{ in } (\mathrm{bar})^{\Delta n}$$

$$\Delta n = 0 \iff K_p = K_c$$