Solvent selection — water, supercritical CO₂, ionic liquids: Worked examples

This page is the drill hall for the solvent-selection topic. The parent note explained what the three green solvents are. Here we hit every kind of question the topic can throw at you — every sign of the "is it supercritical?" test, the degenerate/boundary cases, the limiting behaviour of density and compressibility, a real-world word problem, and an exam twist that tries to trap you.

Before we start, we re-earn every symbol so a reader who skipped the parent can still follow.

Definition The symbols we will use (each anchored to a picture)

$T$ = temperature of the substance, in kelvin (K). Think "how hot the molecules jiggle." $0 ° C = 273.15 K$ .
$P$ = pressure, the push per area the fluid exerts, in bar or MPa. $1 bar = 0.1 MPa$ .
$V$ = volume, the amount of space the fluid occupies, in m³. Squeeze harder (raise $P$ ) and $V$ shrinks — that shrinking is the whole story of Figure 2.
$T_{c}$ (critical temperature) and $P_{c}$ (critical pressure) = the single point on the map where the liquid–gas boundary line ends (the butter-yellow dot in Figure 1). For CO₂: $T_{c} = 304.25 K$ ( $31.1 ° C$ ), $P_{c} = 7.38 MPa$ ( $73.8 bar$ ).
$ρ$ (Greek "rho") = density, mass packed into a volume, in kg/m³. Higher $ρ$ = molecules closer = better at dissolving.
$κ_{T}$ (Greek "kappa") = isothermal compressibility, how much the volume $V$ shrinks when you squeeze at fixed $T$ . See Phase Diagrams & Critical Point.
$E_{a}$ (activation energy) = the energy "hill" reactant molecules must climb before they can turn into product. Lower $E_{a}$ = faster reaction. (Used only in Example 7.)
LCA = Life Cycle Assessment, the accounting of a substance's total environmental cost from raw material through synthesis, use, and disposal — not just its behaviour in the flask. See Life Cycle Assessment (LCA).

The scenario matrix

Every question about "is this fluid supercritical, and how strong a solvent is it?" reduces to the two-part test $T > T_{c}$ and $P > P_{c}$ . That gives a truth table with four sign-combinations, plus the boundary (equalities), plus limiting behaviour, plus real-world/exam cases. Here is the complete map — each row is a cell we must cover.

Cell	Case class	Concrete question
A	$T > T_{c}$ and $P > P_{c}$ (both pass)	Genuinely supercritical?
B	$T > T_{c}$ , $P < P_{c}$ (only temp passes)	Gas, not supercritical
C	$T < T_{c}$ , $P > P_{c}$ (only pressure passes)	Liquid, not supercritical
D	$T < T_{c}$ and $P < P_{c}$ (both fail)	Ordinary gas region
E	Boundary / degenerate ( $T = T_{c}$ or $P = P_{c}$ exactly, all three sub-cases)	Sitting on a critical line
F	Limiting behaviour ( $κ_{T} \to \infty$ near $T_{c}$ )	Why a tiny $Δ P$ swings solvent power
G	Real-world word problem	Decaf, unit conversions, co-solvent
H	Selection logic (which of 3 solvents?)	Polar vs non-polar, recovery
I	Exam trap	"Supercritical = super hot?" + LCA twist

The examples below are labelled with the cell(s) they hit. Together they fill every row.

The two figures at a glance

Figure 1 is the pressure–temperature map of CO₂. The horizontal axis is temperature $T$ , the vertical axis is pressure $P$ . A black curve — the liquid–gas boundary — separates liquid from gas, and it stops at the butter-yellow critical-point dot. A dashed lavender vertical line marks $T_{c}$ ; a dashed coral horizontal line marks $P_{c}$ . The top-right box (past both lines, where the black curve no longer exists) is shaded mint: that is the supercritical region. Every worked point (Examples 1–4) is plotted on this map so you can see which region it lands in.

Figure 2 is the density-versus-pressure curve at fixed temperature just above $T_{c}$ . Horizontal axis is pressure $P$ , vertical axis is density $ρ$ (which stands in for solvent power). The curve rises steeply just past $P_{c}$ (coral dot) and then flattens at high pressure (mint dot). That steep-then-flat shape is the visual heart of Example 5.

Worked examples

Worked example Example 1 — Cell A (both pass)

A vessel holds CO₂ at $T = 320 K$ and $P = 9.0 MPa$ . Is it supercritical?

Forecast: Guess yes/no before reading — jot down whether both conditions clear the bar.

Compare temperature: $T = 320 K$ vs $T_{c} = 304.25 K$ . Since $320 > 304.25$ , temperature passes. Why this step? The liquid–gas line only vanishes once you are hotter than $T_{c}$ ; below it, boiling still exists.
Compare pressure: $P = 9.0 MPa$ vs $P_{c} = 7.38 MPa$ . Since $9.0 > 7.38$ , pressure passes. Why this step? Above $T_{c}$ you still need enough squeeze to reach the dense, liquid-like state that dissolves things. Low pressure above $T_{c}$ is just hot gas.
Both pass ⇒ supercritical. Why do two strict passes guarantee a single supercritical phase? Past $T_{c}$ the liquid–gas boundary line no longer exists (it ended at the critical point), so there is no interface for the substance to split across — clearing both bars lands you in the region beyond the line's endpoint, where only one indistinguishable fluid phase can exist. In Figure 1 this is the green mint dot sitting in the top-right "supercritical" box, past where the black boundary curve stops.

Verify: Margins are $320 - 304.25 = 15.75 K$ and $9.0 - 7.38 = 1.62 MPa$ . Both are strictly positive ( $> 0$ ), so both strict " $>$ " tests pass ⇒ answer is consistent.

Figure — Solvent selection — water, supercritical CO₂, ionic liquids

Worked example Example 2 — Cell B (only temperature passes)

CO₂ at $T = 340 K$ but $P = 5.0 MPa$ . Supercritical?

Forecast: It's hot — tempting to say yes. Guess first.

Temperature: $340 > 304.25$ ⇒ passes.
Pressure: $5.0 < 7.38$ ⇒ fails. Why this step? Without enough pressure, the molecules stay far apart; density stays gas-like and it dissolves almost nothing.
One fail ⇒ not supercritical. Why does a failed pressure test alone sink it? Solvent power tracks density, and below $P_{c}$ the fluid stays low-density gas — so even though it is hot, it lacks the liquid-like packing that gives a supercritical fluid its dissolving power. Missing either requirement drops you out of the useful region. In Figure 1 this is the coral triangle sitting above the lavender $T_{c}$ line but below the coral $P_{c}$ line — the "gas" strip.

Verify: Pressure margin $5.0 - 7.38 = - 2.38 MPa$ is strictly negative ( $< 0$ ) ⇒ the strict " $P > P_{c}$ " test fails ⇒ "not supercritical." This is exactly the trap the parent's mistake box warns about ("supercritical = super hot") — heat alone is not enough.

Worked example Example 3 — Cell C (only pressure passes)

CO₂ at $T = 290 K$ ( $16.85 ° C$ ) and $P = 10.0 MPa$ . State?

Forecast: High pressure, cool temperature — liquid, gas, or supercritical?

Temperature: $290 < 304.25$ ⇒ fails.
Pressure: $10.0 > 7.38$ ⇒ passes.
Below $T_{c}$ but heavily squeezed ⇒ you have pushed CO₂ across the liquid–gas boundary into the liquid region, not the supercritical one. Why this step? Below $T_{c}$ the phase boundary still exists; enough pressure just condenses gas to liquid. The special "no boundary" trick only lives past $T_{c}$ . In Figure 1 this is the lavender diamond: above the coral $P_{c}$ line but left of the lavender $T_{c}$ line — the "liquid" corner.

Verify: $290 - 304.25 = - 14.25 K$ (strictly negative) and $10.0 - 7.38 = 2.62 MPa$ (strictly positive). One strict test fails ⇒ not supercritical; it's liquid CO₂. Consistent.

Worked example Example 4 — Cell E (all three boundary sub-cases)

The boundary is not one point — it is every place where $T = T_{c}$ or $P = P_{c}$ exactly. Test all three: (a) $T = 304.25 K$ (= $T_{c}$ ), $P = 9.0 MPa$ (> $P_{c}$ ). (b) $T = 320 K$ (> $T_{c}$ ), $P = 7.38 MPa$ (= $P_{c}$ ). (c) $T = 304.25 K$ (= $T_{c}$ ), $P = 7.38 MPa$ (= $P_{c}$ , the point itself).

Forecast: Which of these three counts as supercritical under the strict " $>$ " test?

(a) Temperature is equal, not greater: $T = T_{c}$ makes " $T > T_{c}$ " false. Pressure passes. One strict test fails ⇒ not supercritical — you are sitting on the lavender $T_{c}$ line in Figure 1, at its right (high-pressure) end. Why this step? The definition demands strict "greater than." Equality means you are exactly on the wall, not through it.
(b) Pressure is equal: " $P > P_{c}$ " is false; temperature passes. Again one fail ⇒ not supercritical — you are on the coral $P_{c}$ line, at its right end.
(c) Both equal ⇒ both strict tests false ⇒ this is the critical point itself (butter-yellow dot). The single degenerate spot where liquid and gas become indistinguishable. In practice you step just past both to guarantee one clean fluid phase. Why treat all three? A boundary is a whole line, not a dot; an exam can put you on either arm. In every sub-case, " $=$ " fails the strict " $>$ ", so none is counted supercritical.

Verify: (a) margins $0 K$ (fails " $> 0$ ") and $+ 1.62 MPa$ ⇒ not supercritical. (b) margins $+ 15.75 K$ and $0 MPa$ (fails) ⇒ not supercritical. (c) both margins $0$ ⇒ not supercritical. All consistent: any " $=$ " kills the strict test.

Worked example Example 6 — Cell F (limiting behaviour of

$κ_{T}$ ) Explain, with a limit, why moving from $8.0 MPa$ to $8.5 MPa$ just above $T_{c}$ changes the solvent power far more than the same $0.5 MPa$ jump at high pressure ( $20 \to 20.5 MPa$ ).

Forecast: Guess: does the small squeeze do more near the critical point, or far from it?

Recall compressibility $κ_{T} = \frac{1}{ρ} (\frac{\partial ρ}{\partial P})_{T}$ , equivalently $κ_{T} = - \frac{1}{V} (\frac{\partial V}{\partial P})_{T}$ where $V$ is the volume. In words: it measures how much density $ρ$ (or volume $V$ ) responds to a pressure nudge. Why this tool and not another? Solvent power tracks density; the sensitivity of density to pressure is exactly $κ_{T}$ . So $κ_{T}$ is literally "how big a solvent-strength swing per bar."
At the critical point $(\frac{\partial P}{\partial V})_{T} = 0$ . Since $κ_{T} = - \frac{1}{V} (\frac{\partial V}{\partial P})_{T} = - \frac{1}{V} / (\frac{\partial P}{\partial V})_{T}$ , dividing by something heading to $0$ sends $κ_{T} \to \infty$ . Why this step? A flat slope $\partial P / \partial V = 0$ means the substance offers no resistance to compression — squeeze a hair, volume (and hence density) leaps.
Limiting statement: $(T, P) \to (T_{c}, P_{c}) lim κ_{T} = + \infty$ . So near $T_{c}$ a small $Δ P$ = large $Δ ρ$ = large $Δ$ (solvent power). Far above $P_{c}$ the fluid is already dense and stiff, $κ_{T}$ is small, so the same $Δ P$ barely moves $ρ$ . See the steep-then-flat density curve in Figure 2: the coral dot near $P_{c}$ sits on the steep part, the mint dot near $20 MPa$ on the flat part.

Verify: Use a toy model where density rises steeply then saturates: $ρ (P) = 900 \cdot \frac{P}{P + 2}$ (kg/m³, $P$ in MPa). Slope $\frac{d ρ}{d P} = \frac{1800}{( P + 2 ) ^{2}}$ . At $P = 8$ : slope $= 18.0$ ; at $P = 20$ : slope $\approx 3.72$ . The near-critical slope is $\approx 4.8 \times$ larger — a small $Δ P$ does much more there. Consistent.

Worked example Example 7 — Cell G (real-world word problem, unit conversion)

A decaffeination plant runs scCO₂ at $35 ° C$ and $80 bar$ . (i) Convert to K and MPa. (ii) Confirm it is supercritical. (iii) Caffeine is somewhat polar — what one tweak raises the CO₂'s pull for it, and why?

Forecast: Will $80 bar$ clear the pressure bar? Guess the margin.

Convert: $T = 35 + 273.15 = 308.15 K$ ; $P = 80 bar \times 0.1 = 8.0 MPa$ . Why this step? The critical values are stored in K and MPa; you must compare like with like.
Test: $308.15 > 304.25$ ✓ and $8.0 > 7.38$ ✓ ⇒ supercritical. Mild temperature ( $35 ° C$ ) protects flavour — the industrial reason (parent note).
Tweak: add a small amount of water or ethanol co-solvent. Why this step? Pure CO₂ is non-polar; caffeine is mildly polar. A polar co-solvent nudges the medium's polarity up so it "grips" caffeine better — same logic as parent Worked selection #1.

Verify: Margins $308.15 - 304.25 = 3.90 K$ and $8.0 - 7.38 = 0.62 MPa$ are both strictly positive ( $> 0$ ) ⇒ both strict " $>$ " tests pass ⇒ supercritical. Unit check: $80 bar = 8.0 MPa$ since $10 bar = 1 MPa$ . ✓

Worked example Example 8 — Cell H (selection logic across all three solvents)

Match the best green solvent to each job and justify: (a) Dissolve a non-polar natural oil, then recover it residue-free. (b) A Diels–Alder of barely-soluble non-polar dienes you want to run faster and cheaply. (c) Reuse an expensive Rh catalyst across many batches without contaminating product.

Forecast: Guess the three answers (scCO₂ / water / IL, in some order).

(a) → scCO₂. Why? Dissolves non-polar solutes; depressurize and it flashes to gas ⇒ zero residue. Easy recovery is its signature strength.
(b) → water. Why? The Hydrophobic Effect forces non-polar reactants together (higher effective concentration) and stabilises the more polar transition state, lowering the activation energy $E_{a}$ (the energy hill defined in the symbol list) — greener and faster. No pressure kit needed = cheapest.
(c) → ionic liquid. Why? Run biphasic catalysis: catalyst stays in the IL layer, product in a second layer; pour off product, keep catalyst. The IL's negligible vapour pressure means it never contaminates the product stream.

Verify (cross-check against the trade-off table): (a) matches scCO₂'s "easy recovery + dissolves non-polar." (b) matches water's "cheap + hydrophobic effect." (c) matches IL's "reusable + negligible VOC." Each job hits the solvent's distinct strength — no two answers overlap. Consistent.

Worked example Example 9 — Cell I (the exam trap)

True or false, with one-line justification each: (i) "scCO₂ at $400 K$ and $2 MPa$ is supercritical because it is very hot." (ii) "Ionic liquids are automatically green because they don't evaporate." (iii) "Using scCO₂ worsens global warming because CO₂ is a greenhouse gas."

Forecast: How many of the three are false?

(i) FALSE. Temperature passes ( $400 > 304.25$ ) but pressure fails ( $2 < 7.38$ ) ⇒ hot gas, not supercritical. Supercritical needs both $T > T_{c}$ and $P > P_{c}$ ; "super" is about phase, not heat.
(ii) FALSE. Low volatility removes only air pollution (VOCs). Many ILs are aquatic-toxic, poorly biodegradable, and energy-intensive to make — judge over the whole Life Cycle Assessment (LCA) (defined in the symbol list; see Life Cycle Assessment (LCA)).
(iii) FALSE. The CO₂ is captured and recycled in a closed loop, not newly emitted; it is contained, then reused.

Verify: For (i), pressure margin $2 - 7.38 = - 5.38 MPa < 0$ ⇒ strict test fails ⇒ statement false, numerically confirmed. (ii) and (iii) are the parent's two mistake boxes verbatim ⇒ consistent. All three false ⇒ count $= 3$ .

Recall

Recall The two-part test (fill the blanks)

A pure substance is supercritical only when == $T > T_{c}$ and $P > P_{c}$ == — both must pass. If only temperature passes, the state is ::: gas. If only pressure passes (below $T_{c}$ ), the state is ::: liquid. Near the critical point, a tiny change in pressure swings solvent power a lot because ::: $κ_{T} \to \infty$ , i.e. density is hugely pressure-sensitive there.