Conversion, Selectivity & Yield: Three Numbers That Aren’t the Same

Conversion is the number everyone quotes. “We hit 95% conversion” sounds like a win — and sometimes it’s a disaster. Because conversion only tells you how much reactant disappeared. It says nothing about where it went. And in any real process with side reactions, where it went is the entire game.

This is the article that untangles the three numbers students blur together: conversion, selectivity, and yield. For a single clean reaction they’re practically the same fact stated three ways. The moment a second reaction shows up, they split apart — and optimizing the wrong one is how you end up with a reactor full of waste.

The three definitions, made concrete

Let’s pin them down for a reactant $A$ that can form a desired product $D$ (and maybe some junk you don’t want). Every one of these is just a ratio — the trick is being ruthless about what’s on top and what’s on the bottom.

Conversion — of the $A$ you fed, how much reacted at all:

Selectivity — of the $A$ that reacted, how much became what you wanted. (This is the fractional definition; there’s also a ratio version, more on that below.)

Yield — of the $A$ you fed, how much ended up as $D$:

Stack those three fractions next to each other and a clean relationship falls out — the numerator of yield is the numerator of selectivity, and the denominator of yield is the denominator of conversion:

(One bookkeeping note: I’m writing everything “per mole of $A$.” If the stoichiometry isn’t 1:1 — say two $A$ make one $D$ — you scale the desired-product terms by the stoichiometric ratio. The logic of the three ratios never changes; only the conversion factor does.)

Why they’re the same number — until they’re not

For a single reaction $A \to D$, every mole of $A$ that reacts becomes $D$. There’s nowhere else for it to go, so selectivity is pinned at $S = 1$, and yield equals conversion exactly. That’s why in intro problems the three words feel interchangeable — in that world, they are.

Reality has side reactions. And they come in two flavors that behave completely differently:

Series: your product is a middle step that keeps reacting. Parallel: your product competes with a side reaction for the same feed. Different traps, different fixes.

The series-reaction trap

Take the series case $A \to B \to C$, where $B$ is the money product and $C$ is over-reacted waste. Here’s the cruel part: $B$ doesn’t just form, it also disappears. Early on there’s barely any $B$ around, so it mostly builds up. But as $B$ accumulates, the second reaction speeds up and starts eating it. There’s a moment where formation and destruction balance — and that’s the peak. Run past it and you’re actively converting your product into garbage.

$B$ has a maximum. Before it, you’re still making product; after it, you’re destroying it. The optimal reactor size or run time is the one that stops right at the peak.

This is why “maximize conversion” is dangerous advice for series reactions. Higher conversion means more time, and more time means more of your $B$ turns into $C$. The optimum isn’t maximum conversion — it’s the conversion that maximizes yield of B, and that sits at the peak of that middle curve.

Worked example: finding the peak

Let both steps be first order: $A \to B$ with rate constant $k_1 = 0.2\ \text{min}^{-1}$, and $B \to C$ with $k_2 = 0.1\ \text{min}^{-1}$. In a batch (or plug-flow) reactor, the time that maximizes $B$ is:

And the best yield of $B$ you can possibly get — the height of that peak — has a clean closed form:

So the ceiling on $B$ is 50%, hit at about 6.9 minutes. Notice what that means: even with a reactant twice as eager to form $B$ as to destroy it, you can never turn more than half your feed into $B$ in one pass. No amount of “pushing conversion” beats that — past the peak, yield drops. The move in practice is to stop early, separate the $B$, and recycle the unreacted $A$. Conversion per pass stays modest on purpose; overall yield is what you actually bank.

Selectivity is a concentration knob — which means the reactor sets it

Now the parallel case, and this is where Part 1 comes roaring back. Suppose $A$ can go two ways, and the two pathways have different orders in $A$:

The instantaneous selectivity — the ratio of the rates at a given instant — is:

Look at what that says: selectivity is proportional to concentration. Run at high $C_A$ and you favor $D$; run at low $C_A$ and you feed the junk pathway. Concentration isn’t just something that falls out of the reactor — it’s a dial you set by choosing the reactor.

And you already know from Part 1 how each reactor sets that dial:

• A PFR or batch spends time at high inlet concentrations — great when the desired reaction is the higher-order one, like here.
• A CSTR pins everything at the low exit concentration — which would be the worst choice for this system, but the right choice if the junk reaction were the higher-order one and you wanted to starve it.

Put numbers on it. With $k_D/k_U = 2\ \text{L/mol}$, feed at $C_{A0} = 2$ vs. an exit at $C_A = 0.4$:

That’s a 5× swing in selectivity from concentration alone — same chemistry, same rate constants, just a different place on the concentration axis. Choose the reactor that keeps you where you want to be.

Instantaneous vs. overall — the distinction that trips people up

One last subtlety, because it’s a classic exam gotcha. The selectivity above is instantaneous — it’s the value at one specific concentration. But a PFR sweeps through a whole range of concentrations from inlet to outlet, so its overall selectivity is an average of the instantaneous value over that range. A CSTR, because it sits at a single exit concentration, has its overall selectivity equal to the instantaneous value at that exit — no averaging needed.

That’s the same structural split you saw in the design equations last time: the PFR integrates over a range, the CSTR evaluates at a point. Selectivity, yield, reactor sizing — it’s the same two personalities showing up everywhere in reaction engineering.

Quantity	Asks	Single reaction
Conversion $X_A$	How much $A$ reacted?	= yield
Selectivity $S$	Of that, how much became $D$?	= 1
Yield $Y$	How much fed $A$ became $D$?	= conversion

For a single reaction the right-hand column shows why nobody bothers distinguishing them. For everything else — which is to say, every real process — keep them straight, and remember the one equation that ties them together: $Y = X_A \, S$.