Kc and Kp Calculations

Spec Mapping — OCR H432 Module 5.1.2 — How far? Equilibrium constant calculations using ICE (Initial, Change, Equilibrium) tables for both $K_c$Kc​ (concentrations) and $K_p$Kp​ (partial pressures); the relationship $K_p = K_c (RT)^{\Delta n_{\text{gas}}}$Kp​=Kc​(RT)Δngas​; problems in which initial moles, equilibrium amounts, equilibrium constants or compositions must be deduced from each other using stoichiometry (refer to the official OCR H432 specification document for exact wording).

This lesson is the operational backbone of OCR Module 5.1.2 — it turns the definitions of $K_c$Kc​ and $K_p$Kp​ (from earlier lessons) into a recipe for solving any equilibrium-composition problem. The central tool is the ICE table (Initial, Change, Equilibrium), a three-row bookkeeping device that tracks how many moles of each species are present at each stage. With the ICE table built correctly, applying stoichiometry to the Change row is automatic; converting to concentrations or partial pressures is mechanical; and substituting into the $K_c$Kc​ or $K_p$Kp​ expression is the easy final step. We work through five complete examples — $K_c$Kc​ for $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$H2​+I2​⇌2HI, $K_c$Kc​ for $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$N2​O4​⇌2NO2​ (units example), $K_p$Kp​ for the Contact process via ICE plus Dalton, $K_c$Kc​ for $\text{PCl}_5$PCl5​ dissociation using an $x$x variable, and inverse $K_c$Kc​ problems where the equilibrium constant is given and a concentration must be solved by quadratic algebra. The relationship $K_p = K_c (RT)^{\Delta n_{\text{gas}}}$Kp​=Kc​(RT)Δngas​ closes the loop between the two formulations.

Key Equation: the ICE-table workflow $\boxed{n_{\text{eq}} = n_{\text{initial}} + \Delta n_{\text{stoichiometric}}}$neq​=ninitial​+Δnstoichiometric​​ Then for $K_c$Kc​: divide $n_{\text{eq}}$neq​ by volume $V$V to get concentrations; for $K_p$Kp​: divide $n_{\text{eq}}$neq​ by $n_{\text{total, eq}}$ntotal, eq​ to get mole fractions, then multiply by $P_{\text{total}}$Ptotal​ to get partial pressures. The link is $K_p = K_c (RT)^{\Delta n_{\text{gas}}}$Kp​=Kc​(RT)Δngas​ (only valid with $R$R in matching units; OCR uses this qualitatively — see Going Further).

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The ICE table method

An ICE table is a three-row table that keeps track of moles (or concentrations) at each stage of an equilibrium calculation:

• I — Initial: amounts before any reaction.
• C — Change: how much reacted (negative for reactants, positive for products), with stoichiometric ratios applied.
• E — Equilibrium: $I + C$I+C, the amounts at equilibrium.

Example — simple 1:1

$\text{A(g)} \rightleftharpoons \text{B(g)}$A(g)⇌B(g) with 1.00 mol A initially and no B. At equilibrium, 0.60 mol A remains.

	A	B
Initial	1.00	0
Change	$-0.40$−0.40	$+0.40$+0.40
Equilibrium	0.60	0.40

The change $-0.40$−0.40 in A is matched by $+0.40$+0.40 in B because the stoichiometry is 1:1.

Example — 1:2 stoichiometry

$\text{A(g)} \rightleftharpoons 2\text{B(g)}$A(g)⇌2B(g), starting with 1.00 mol A and 0 mol B. At equilibrium, 0.70 mol A.

	A	2B
Initial	1.00	0
Change	$-0.30$−0.30	$+0.60$+0.60
Equilibrium	0.70	0.60

Note the 1:2 ratio: 0.30 mol of A dissociates to give 0.60 mol of B (twice as many moles).

Worked Example 1 — $K_c$Kc​ for $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$H2​+I2​⇌2HI

2.0 mol H$_2$2​ and 2.0 mol I$_2$2​ are mixed in a 2.0 dm$^3$3 flask. At equilibrium, 3.2 mol HI is present. Calculate $K_c$Kc​.

Step 1 — ICE in moles

Change in HI $= +3.2$=+3.2. Stoichiometry $2$2 HI ↔ $1$1 H$_2$2​ $+ 1$+1 I$_2$2​ means H$_2$2​ and I$_2$2​ each decrease by $3.2/2 = 1.6$3.2/2=1.6.

	H$_2$2​	I$_2$2​	2HI
Initial	2.0	2.0	0
Change	$-1.6$−1.6	$-1.6$−1.6	$+3.2$+3.2
Equilibrium	0.4	0.4	3.2

Step 2 — convert moles to concentrations (divide by $V = 2.0\,\text{dm}^{3}$V=2.0dm3)

• $[\text{H}_2] = 0.4/2.0 = 0.20\,\text{mol dm}^{-3}$[H2​]=0.4/2.0=0.20mol dm−3
• $[\text{I}_2] = 0.4/2.0 = 0.20\,\text{mol dm}^{-3}$[I2​]=0.4/2.0=0.20mol dm−3
• $[\text{HI}] = 3.2/2.0 = 1.6\,\text{mol dm}^{-3}$[HI]=3.2/2.0=1.6mol dm−3

Step 3 — substitute into $K_c$Kc​

$$\begin{aligned} K_c &= \frac{[\text{HI}]^{2}}{[\text{H}_2][\text{I}_2]} \\ &= \frac{(1.6)^{2}}{0.20 \times 0.20} \\ &= \frac{2.56}{0.040} \\ &= 64 \end{aligned}$$Kc​​=[H2​][I2​][HI]2​=0.20×0.20(1.6)2​=0.0402.56​=64​

Step 4 — units

$\Delta n = 2 - 2 = 0$Δn=2−2=0, so $K_c$Kc​ is dimensionless. $K_c = \mathbf{64}$Kc​=64.

Worked Example 2 — $K_c$Kc​ for $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$N2​O4​⇌2NO2​ with units

$\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$N2​O4​(g)⇌2NO2​(g). A 2.0 dm$^3$3 flask initially contains 1.0 mol $\text{N}_2\text{O}_4$N2​O4​ and no $\text{NO}_2$NO2​. At equilibrium, 0.20 mol $\text{N}_2\text{O}_4$N2​O4​ is present. Calculate $K_c$Kc​.

ICE table (moles)

Change in $\text{N}_2\text{O}_4 = -0.80$N2​O4​=−0.80 (decreased from 1.0 to 0.20). Therefore change in $\text{NO}_2 = +1.60$NO2​=+1.60 (stoichiometric factor 2).

	$\text{N}_2\text{O}_4$N2​O4​	$2\text{NO}_2$2NO2​
Initial	1.0	0
Change	$-0.80$−0.80	$+1.60$+1.60
Equilibrium	0.20	1.60

Concentrations

• $[\text{N}_2\text{O}_4] = 0.20/2.0 = 0.10\,\text{mol dm}^{-3}$[N2​O4​]=0.20/2.0=0.10mol dm−3
• $[\text{NO}_2] = 1.60/2.0 = 0.80\,\text{mol dm}^{-3}$[NO2​]=1.60/2.0=0.80mol dm−3

$K_c$Kc​

$$\begin{aligned} K_c &= \frac{[\text{NO}_2]^{2}}{[\text{N}_2\text{O}_4]} \\ &= \frac{(0.80)^{2}}{0.10} \\ &= \frac{0.64}{0.10} \\ &= 6.4\,\text{mol dm}^{-3} \end{aligned}$$Kc​​=[N2​O4​][NO2​]2​=0.10(0.80)2​=0.100.64​=6.4mol dm−3​

$\Delta n = 2 - 1 = +1$Δn=2−1=+1, so units = $\text{mol dm}^{-3}$mol dm−3. $K_c = \mathbf{6.4\,\text{mol dm}^{-3}}$Kc​=6.4mol dm−3.

Worked Example 3 — $K_p$Kp​ with ICE for the Contact process

For $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$2SO2​(g)+O2​(g)⇌2SO3​(g), a flask initially contains 2.0 mol $\text{SO}_2$SO2​, 1.0 mol $\text{O}_2$O2​ and no $\text{SO}_3$SO3​. Total pressure at equilibrium $= 150\,\text{kPa}$=150kPa. At equilibrium, 1.6 mol $\text{SO}_3$SO3​ is present.

Step 1 — ICE table

Change in $\text{SO}_3 = +1.60$SO3​=+1.60. By stoichiometry, $\text{SO}_2$SO2​ decreases by 1.60 (2:2 ratio) and $\text{O}_2$O2​ decreases by $0.80$0.80 (2:1 ratio).

	$2\text{SO}_2$2SO2​	$\text{O}_2$O2​	$2\text{SO}_3$2SO3​
Initial	2.0	1.0	0
Change	$-1.6$−1.6	$-0.80$−0.80	$+1.6$+1.6
Equilibrium	0.40	0.20	1.60

Step 2 — mole fractions

$n_{\text{total}} = 0.40 + 0.20 + 1.60 = 2.20\,\text{mol}$ntotal​=0.40+0.20+1.60=2.20mol.

• $x_{\text{SO}_2} = 0.40/2.20 = 0.1818$xSO2​​=0.40/2.20=0.1818
• $x_{\text{O}_2} = 0.20/2.20 = 0.0909$xO2​​=0.20/2.20=0.0909
• $x_{\text{SO}_3} = 1.60/2.20 = 0.7273$xSO3​​=1.60/2.20=0.7273

Check: $0.1818 + 0.0909 + 0.7273 = 1.0000$0.1818+0.0909+0.7273=1.0000. ✓

Step 3 — partial pressures

• $p_{\text{SO}_2} = 0.1818 \times 150 = 27.3\,\text{kPa}$pSO2​​=0.1818×150=27.3kPa
• $p_{\text{O}_2} = 0.0909 \times 150 = 13.6\,\text{kPa}$pO2​​=0.0909×150=13.6kPa
• $p_{\text{SO}_3} = 0.7273 \times 150 = 109.1\,\text{kPa}$pSO3​​=0.7273×150=109.1kPa

Step 4 — $K_p$Kp​

$$\begin{aligned} K_p &= \frac{p_{\text{SO}_3}^{2}}{p_{\text{SO}_2}^{2} \cdot p_{\text{O}_2}} \\ &= \frac{(109.1)^{2}}{(27.3)^{2} \times 13.6} \\ &= \frac{11\,903}{745.3 \times 13.6} \\ &= \frac{11\,903}{10\,136} \\ &= 1.17 \end{aligned}$$Kp​​=pSO2​2​⋅pO2​​pSO3​2​​=(27.3)2×13.6(109.1)2​=745.3×13.611903​=1013611903​=1.17​

$\Delta n_{\text{gas}} = 2 - (2 + 1) = -1$Δngas​=2−(2+1)=−1. Units = $\text{kPa}^{-1}$kPa−1. $K_p = \mathbf{1.17\,\text{kPa}^{-1}}$Kp​=1.17kPa−1.

Worked Example 4 — using $x$x for the extent of dissociation ($\text{PCl}_5$PCl5​)

$\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$PCl5​(g)⇌PCl3​(g)+Cl2​(g). 1.00 mol PCl$_5$5​ is heated in a 2.0 dm$^3$3 flask. At equilibrium, 20% of the PCl$_5$5​ has dissociated. Calculate $K_c$Kc​.

Step 1 — define the change

Let $x =$x= moles of PCl$_5$5​ dissociated at equilibrium. $x = 0.20 \times 1.00 = 0.20\,\text{mol}$x=0.20×1.00=0.20mol.

	$\text{PCl}_5$PCl5​	$\text{PCl}_3$PCl3​	$\text{Cl}_2$Cl2​
Initial	1.00	0	0
Change	$-0.20$−0.20	$+0.20$+0.20	$+0.20$+0.20
Equilibrium	0.80	0.20	0.20

Step 2 — concentrations

• $[\text{PCl}_5] = 0.80/2.0 = 0.40\,\text{mol dm}^{-3}$[PCl5​]=0.80/2.0=0.40mol dm−3
• $[\text{PCl}_3] = 0.20/2.0 = 0.10\,\text{mol dm}^{-3}$[PCl3​]=0.20/2.0=0.10mol dm−3
• $[\text{Cl}_2] = 0.20/2.0 = 0.10\,\text{mol dm}^{-3}$[Cl2​]=0.20/2.0=0.10mol dm−3

Step 3 — $K_c$Kc​

$$\begin{aligned} K_c &= \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \\ &= \frac{0.10 \times 0.10}{0.40} \\ &= \frac{0.010}{0.40} \\ &= 0.025\,\text{mol dm}^{-3} \end{aligned}$$Kc​​=[PCl5​][PCl3​][Cl2​]​=0.400.10×0.10​=0.400.010​=0.025mol dm−3​

$\Delta n = 2 - 1 = +1$Δn=2−1=+1, units = $\text{mol dm}^{-3}$mol dm−3.

Worked Example 5 — inverse problem: $K_c$Kc​ given, find concentration

For $\text{A} \rightleftharpoons \text{B}$A⇌B, $K_c = 4.0$Kc​=4.0 (dimensionless). Starting $[\text{A}]_0 = 1.0\,\text{mol dm}^{-3}$[A]0​=1.0mol dm−3, $[\text{B}]_0 = 0$[B]0​=0. Find the equilibrium concentrations.

Let $x =$x= amount of A reacted (mol dm$^{-3}$−3).

	A	B
Initial	1.0	0
Change	$-x$−x	$+x$+x
Equilibrium	$1.0 - x$1.0−x	$x$x

$$\begin{aligned} K_c &= \frac{[\text{B}]}{[\text{A}]} = \frac{x}{1.0 - x} = 4.0 \\ x &= 4.0(1.0 - x) \\ x &= 4.0 - 4.0x \\ 5.0x &= 4.0 \\ x &= 0.80 \end{aligned}$$Kc​xx5.0xx​=[A][B]​=1.0−xx​=4.0=4.0(1.0−x)=4.0−4.0x=4.0=0.80​

Therefore $[\text{A}]_{\text{eq}} = 1.0 - 0.80 = \mathbf{0.20\,\text{mol dm}^{-3}}$[A]eq​=1.0−0.80=0.20mol dm−3, $[\text{B}]_{\text{eq}} = \mathbf{0.80\,\text{mol dm}^{-3}}$[B]eq​=0.80mol dm−3.

Check: $K_c = 0.80/0.20 = 4.0$Kc​=0.80/0.20=4.0. ✓

For more complex stoichiometries (e.g. $\text{A} \rightleftharpoons 2\text{B}$A⇌2B), the equilibrium expression in $x$x becomes a quadratic; solve using the quadratic formula or, for $K_c \ll [\text{A}]_0$Kc​≪[A]0​, the small-$x$x approximation $[\text{A}]_{\text{eq}} \approx [\text{A}]_0$[A]eq​≈[A]0​.

Worked Example 6 — ICE table for the Haber process at industrial conditions

A reactor operating at industrial Haber conditions (700 K, 200 atm) is charged with 1.0 mol $\text{N}_2$N2​ and 3.0 mol $\text{H}_2$H2​ (the stoichiometric feed ratio) per dm$^3$3 of reactor volume. At equilibrium, the mole fraction of $\text{NH}_3$NH3​ in the gas mixture is measured to be $x_{\text{NH}_3} = 0.150$xNH3​​=0.150. Calculate $K_p$Kp​ for $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$N2​(g)+3H2​(g)⇌2NH3​(g) at this temperature, with units.

Step 1 — define the extent of reaction $\xi$ξ. Let $\xi$ξ be the amount (mol) of N$_2$2​ that has reacted at equilibrium. By stoichiometry, 3$\xi$ξ mol H$_2$2​ has reacted and 2$\xi$ξ mol NH$_3$3​ has formed. ICE table in moles:

	N$_2$2​	3H$_2$2​	2NH$_3$3​
Initial	1.00	3.00	0
Change	$-\xi$−ξ	$-3\xi$−3ξ	$+2\xi$+2ξ
Equilibrium	$1.00 - \xi$1.00−ξ	$3.00 - 3\xi$3.00−3ξ	$2\xi$2ξ

Step 2 — total moles and mole-fraction constraint. $n_{\text{total, eq}} = (1.00 - \xi) + (3.00 - 3\xi) + 2\xi = 4.00 - 2\xi$ntotal, eq​=(1.00−ξ)+(3.00−3ξ)+2ξ=4.00−2ξ. The given $x_{\text{NH}_3} = 0.150$xNH3​​=0.150 provides one equation in $\xi$ξ:

$x_{\text{NH}_3} = \frac{2\xi}{4.00 - 2\xi} = 0.150$xNH3​​=4.00−2ξ2ξ​=0.150

Solving: $2\xi = 0.150(4.00 - 2\xi) = 0.600 - 0.300\xi$2ξ=0.150(4.00−2ξ)=0.600−0.300ξ, so $2.300\xi = 0.600$2.300ξ=0.600, giving $\xi = 0.261\,\text{mol}$ξ=0.261mol.

Step 3 — equilibrium moles and mole fractions.

• $n_{\text{N}_2} = 1.00 - 0.261 = 0.739\,\text{mol}$nN2​​=1.00−0.261=0.739mol
• $n_{\text{H}_2} = 3.00 - 3(0.261) = 3.00 - 0.783 = 2.217\,\text{mol}$nH2​​=3.00−3(0.261)=3.00−0.783=2.217mol
• $n_{\text{NH}_3} = 2(0.261) = 0.522\,\text{mol}$nNH3​​=2(0.261)=0.522mol
• $n_{\text{total}} = 0.739 + 2.217 + 0.522 = 3.478\,\text{mol}$ntotal​=0.739+2.217+0.522=3.478mol (check: $4.00 - 2(0.261) = 3.478$4.00−2(0.261)=3.478 ✓)

Mole fractions: $x_{\text{N}_2} = 0.739/3.478 = 0.213$xN2​​=0.739/3.478=0.213; $x_{\text{H}_2} = 2.217/3.478 = 0.637$xH2​​=2.217/3.478=0.637; $x_{\text{NH}_3} = 0.522/3.478 = 0.150$xNH3​​=0.522/3.478=0.150 (matches the given value, sanity-check passed).

Step 4 — partial pressures at $P_{\text{total}} = 200\,\text{atm}$Ptotal​=200atm.

• $p_{\text{N}_2} = 0.213 \times 200 = 42.6\,\text{atm}$pN2​​=0.213×200=42.6atm
• $p_{\text{H}_2} = 0.637 \times 200 = 127.4\,\text{atm}$pH2​​=0.637×200=127.4atm
• $p_{\text{NH}_3} = 0.150 \times 200 = 30.0\,\text{atm}$pNH3​​=0.150×200=30.0atm

(Check: $42.6 + 127.4 + 30.0 = 200.0\,\text{atm}$42.6+127.4+30.0=200.0atm ✓)

Step 5 — $K_p$Kp​ and units.

$K_p = \frac{p_{\text{NH}_3}^{2}}{p_{\text{N}_2} \cdot p_{\text{H}_2}^{3}} = \frac{(30.0)^{2}}{42.6 \times (127.4)^{3}} = \frac{900}{42.6 \times 2.068 \times 10^{6}} = \frac{900}{8.81 \times 10^{7}} = 1.02 \times 10^{-5}$Kp​=pN2​​⋅pH2​3​pNH3​2​​=42.6×(127.4)3(30.0)2​=42.6×2.068×106900​=8.81×107900​=1.02×10−5

$\Delta n_{\text{gas}} = 2 - 4 = -2$Δngas​=2−4=−2, units = $\text{atm}^{-2}$atm−2. $K_p = 1.02 \times 10^{-5}\,\text{atm}^{-2}$Kp​=1.02×10−5atm−2.

Comment on the answer. A $K_p$Kp​ of order $10^{-5}\,\text{atm}^{-2}$10−5atm−2 at 700 K (a typical industrial Haber operating temperature) reflects an equilibrium that lies substantially toward the reactants — only 15% of the gas at equilibrium is NH$_3$3​. This is exactly why industrial Haber plants run at high pressure (200 atm) — Le Chatelier predicts that compression shifts equilibrium toward the side with fewer moles of gas (the NH$_3$3​ side, with $\Delta n_{\text{gas}} = -2$Δngas​=−2), partially compensating for the temperature compromise. Even higher pressure would shift equilibrium further but at prohibitive capital cost; lower temperature would raise $K_p$Kp​ but slow the kinetics to industrially uneconomic rates. The chosen 700 K/200 atm operating point is the engineering compromise between yield and rate that BASF first determined in 1909–1913. The Worked Example 6 calculation reproduces, with modern data, exactly the kind of equilibrium-yield analysis that informed that original industrial decision.

The $\Delta G^{\ominus} = -RT\ln K$ΔG⊖=−RTlnK connection — bridging to undergraduate thermodynamics

The numerical value of $K_p$Kp​ (or $K_c$Kc​) is not a free parameter — it is determined by the Gibbs free energy change of the reaction under standard conditions. The bridging relation, which is the cornerstone of undergraduate physical chemistry, is:

$\Delta G^{\ominus} = -RT \ln K^{\ominus}$ΔG⊖=−RTlnK⊖

where $K^{\ominus}$K⊖ is the dimensionless thermodynamic equilibrium constant (each species' activity referenced to standard state; for gases, $p_A/p^{\ominus}$pA​/p⊖ with $p^{\ominus} = 1\,\text{bar}$p⊖=1bar). At A-Level, the practical interpretation is that the sign of $\Delta G^{\ominus}$ΔG⊖ predicts whether $K$K is "large" ($K > 1$K>1, $\Delta G^{\ominus} < 0$ΔG⊖<0, equilibrium lies to products) or "small" ($K < 1$K<1, $\Delta G^{\ominus} > 0$ΔG⊖>0, equilibrium lies to reactants).

For the Haber equilibrium at 700 K with $K_p \approx 10^{-5}\,\text{atm}^{-2}$Kp​≈10−5atm−2 (Worked Example 6), the corresponding standard Gibbs energy change is approximately:

$\Delta G^{\ominus} \approx -RT \ln(10^{-5}) = -(8.314)(700)(-11.5) \approx +67\,000\,\text{J mol}^{-1} = +67\,\text{kJ mol}^{-1}$ΔG⊖≈−RTln(10−5)=−(8.314)(700)(−11.5)≈+67000J mol−1=+67kJ mol−1

(treating $K_p$Kp​ as the dimensionless $K^{\ominus}$K⊖ for the sake of the estimate). A positive $\Delta G^{\ominus}$ΔG⊖ of +67 kJ mol$^{-1}$−1 at 700 K confirms that the thermodynamic drive at this temperature actually opposes NH$_3$3​ formation — the reaction is endergonic at industrial $T$T. This is consistent with the exothermicity of the reaction at low $T$T being overwhelmed by the unfavourable entropy change at high $T$T: $\Delta H^{\ominus} \approx -92\,\text{kJ mol}^{-1}$ΔH⊖≈−92kJ mol−1 but $\Delta S^{\ominus} \approx -198\,\text{J K}^{-1}\,\text{mol}^{-1}$ΔS⊖≈−198J K−1mol−1, so $T\Delta S^{\ominus} \approx -139\,\text{kJ mol}^{-1}$TΔS⊖≈−139kJ mol−1 at 700 K and $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} \approx -92 - (-139) = +47\,\text{kJ mol}^{-1}$ΔG⊖=ΔH⊖−TΔS⊖≈−92−(−139)=+47kJ mol−1 — order-of-magnitude agreement with the value extracted from $K_p$Kp​. The connection between kinetic practicality (large enough $k$k) and thermodynamic feasibility (sufficient $K$K) is what governs every industrial process decision in chemical engineering.

OCR introduces $\Delta G^{\ominus}$ΔG⊖ in Module 5.2.3 — Acids, Bases and Buffers (specifically, the energetics of acid-base equilibria) and revisits it in Module 5.2.4 — Redox and Electrode Potentials (with $\Delta G^{\ominus} = -nFE^{\ominus}_{\text{cell}}$ΔG⊖=−nFEcell⊖​). Knowing in advance that $K_p$Kp​ and $K_c$Kc​ have a direct thermodynamic interpretation as $-\Delta G^{\ominus}/RT$−ΔG⊖/RT exponentials prepares you for both modules.

Mermaid — $K_c$Kc​/$K_p$Kp​ workflow