Reaction Extent and Coefficient Ratios

Introduction

Balanced equations give mole ratios, but a reaction need not go to completion. Reaction extent ξ provides a single variable for how far a specified equation has proceeded. Each reactant decreases and each product increases by its coefficient times ξ. This makes limiting-reagent, equilibrium and gas-mixture calculations more consistent than separate guesses for every species.

Core explanation

For aA + bB → cC, let ξ be measured in moles of reaction as written. Then nA = nA,0 − aξ, nB = nB,0 − bξ and nC = nC,0 + cξ. If ξ = 1 mol, the equation consumes a mol A and b mol B and creates c mol C. The coefficient ratio describes a change per extent, not necessarily the starting or final amount of each species. If some C was initially present, nC,0 must be included.

For N₂ + 3H₂ → 2NH₃, an extent of 0.60 mol consumes 0.60 mol N₂ and 1.80 mol H₂ while producing 1.20 mol NH₃. Suppose initial amounts are 2.00 mol N₂, 3.00 mol H₂ and zero NH₃. Final amounts at this extent are 1.40 mol N₂, 1.20 mol H₂ and 1.20 mol NH₃. The fact that NH₃ produced equals remaining H₂ numerically in this example is coincidental; they came from different expressions.

Feasibility bounds follow from nonnegative reactants. For the starting amounts above, N₂ allows ξ at most 2.00/1 = 2.00 mol, while H₂ allows ξ at most 3.00/3 = 1.00 mol. Therefore ξ cannot exceed 1.00 mol for a forward-only reaction before H₂ is exhausted. The chosen ξ = 0.60 mol is feasible but not complete. If a problem gives an equilibrium constant, an energy constraint or a measured conversion, it may determine a smaller extent.

Signed stoichiometric numbers compress the equations. Set νN₂ = −1, νH₂ = −3 and νNH₃ = +2, then ni = ni,0 + νi ξ for every species. The same pattern works for multiple products and supports mass balance because a correctly balanced equation makes the net element changes cancel. A stoichiometric coefficient is exact for the chosen equation; ξ inherits the amount unit of moles when ni is in moles.

Sometimes a problem uses concentration changes rather than mole amounts. At a fixed known volume, changes in concentration can be represented by a concentration-scale variable x with units mol L⁻¹. It is related to mole extent by x = ξ/V if volume stays fixed. Do not treat an ICE-table x in M as though it were automatically a mole amount. If volume changes, especially in gas reactions or mixing solutions, that simple relation requires reconsideration.

For a reversible reaction, ξ can move backward relative to an initial reference state; its allowable range is constrained by nonnegative amounts of all species in both directions. “Forward ξ ≥ 0” is appropriate only when the chosen starting point and problem describe forward progress from that state. Equilibrium calculations often use a signed change variable and then test physical roots.

Step-by-step reasoning

1. Balance the reaction and assign negative coefficients to reactants, positive to products. 2. List initial moles of every relevant species, including any initial products. 3. Write ni = ni,0 + νi ξ for each species. 4. Apply nonnegative bounds to determine the feasible interval for ξ. 5. Use measured conversion, equilibrium or complete-reaction assumptions to determine its actual value.

Visual explanation

Draw a table with rows N₂, H₂ and NH₃ and columns initial, change and final. Fill initial 2.00, 3.00, 0; change −ξ, −3ξ, +2ξ; final 2.00−ξ, 3.00−3ξ, 2ξ. At ξ = 0.60, fill final values 1.40, 1.20, 1.20 mol. Draw a feasible ξ number line from 0 to 1.00 mol.

Real-world analogy

A recipe consumes one unit of ingredient A and three of B to make two portions of C each time it is run. The number of recipe runs is like ξ. Starting stock limits how many runs are possible, while actually choosing fewer runs leaves ingredients unused. Chemical extent applies exact mole ratios, not kitchen volume assumptions.

Real-world example

An ammonia-process calculation may specify feed moles and a measured conversion rather than complete reaction. The extent method updates nitrogen, hydrogen and ammonia consistently. It also supplies final gas moles for pressure or composition calculations, showing why stoichiometry should be completed before applying Dalton's law.

Why?

Why is hydrogen's maximum extent 3.00/3 = 1.00 mol in the example? Every mole of reaction as written consumes three moles H₂. With only 3.00 mol initially, at most one mole of the written reaction can proceed before H₂ reaches zero.

Common misconception

“A coefficient ratio means final amounts always have that ratio.” Coefficients relate changes . Final amounts also include initial unreacted material and any initial product, so their ratio can differ.

Worked example

Initial N₂ = 2.00 mol, H₂ = 3.00 mol and NH₃ = 0 for N₂ + 3H₂ → 2NH₃. A measured nitrogen conversion is 30.0%, so N₂ consumed is 0.300 × 2.00 = 0.600 mol. Since N₂ coefficient is one, ξ = 0.600 mol. Final N₂ = 2.00 − 0.600 = 1.40 mol; H₂ = 3.00 − 3(0.600) = 1.20 mol; NH₃ = 2(0.600) = 1.20 mol. Total gas amount changes from 5.00 to 3.80 mol, relevant for pressure at fixed V and T.

Quick check

1. In N₂ + 3H₂ → 2NH₃, how much H₂ is consumed at ξ = 0.20 mol? Answer: 3 × 0.20 = 0.60 mol H₂; NH₃ formation would be 0.40 mol.

Exam focus

Use extent for changes, not final ratios. Show the initial–change–final table, check nonnegative amounts and keep mole extent separate from concentration-scale ICE variables. State whether complete reaction or a partial conversion is assumed.

Advanced insight

For several independent reactions, each has its own extent ξr and species amounts satisfy ni = ni,0 + Σr νir ξr. This matrix form is the basis of reaction-network models and makes element conservation visible through stoichiometric linear algebra.

Summary

Reaction extent is one common progress variable for a balanced equation. Every species amount equals its initial amount plus signed coefficient times ξ. Starting inventories bound feasible extent, while equilibrium or measured conversion determines how far the reaction actually goes.

Practice questions

1. For 2A + B → 3C at ξ = 0.50 mol, how much A is consumed? Answer: 2 × 0.50 = 1.00 mol A. 2. If initial A is 1.50 mol and B is 1.00 mol, what is the maximum forward ξ? Answer: min(1.50/2, 1.00/1) = 0.75 mol, limited by A. 3. Why can final C:A mole ratio differ from 3:2? Answer: Final amounts include initial inventories and unreacted A; 3:2 applies to changes, not necessarily totals.