Equations and particle diagrams say the same thing
The equation can be read as: every 2 H₂ molecules react with 1 O₂ molecule to form 2 H₂O molecules. The coefficients化学计量数 give the ratio of particles that react and form, which is also the mole ratio, not the mass ratio.
A particle diagram粒子图 of an equation must show:
- Conservation of atoms. Each kind of atom has the same count before and after. A chemical change only rearranges atoms; none are created or destroyed, which is why mass is conserved.
- Ratios matching the coefficients. The reactants used up and the products formed are in the same ratio as the coefficients.
- Matching states. Gas molecules are far apart; particles in liquids and solids are close together; ions in solution are spread among water molecules, and a precipitate collects at the bottom.
Leftover molecules aren’t in the equation
In real particle diagrams, reactants usually aren’t supplied in exactly the coefficient ratio, so some of the excess reactant is left over after the reaction. When reading a diagram:
- count the change for each substance, the difference between its count before and after;
- only the changes match the coefficients; leftover molecules didn’t react and aren’t part of the equation;
- reduce the changes to the lowest whole-number ratio.
Look at the diagrams, set a coefficient for each substance, and click “Check”:
| Substance | Before | After | Change |
|---|---|---|---|
| H₂ | 5 | 1 | −4 |
| O₂ | 4 | 2 | −2 |
| H₂O | 0 | 4 | +4 |
| Substance | Before | After | Change |
|---|---|---|---|
| N₂ | 3 | 1 | −2 |
| H₂ | 6 | 0 | −6 |
| NH₃ | 0 | 4 | +4 |
| Substance | Before | After | Change |
|---|---|---|---|
| CO | 4 | 0 | −4 |
| O₂ | 3 | 1 | −2 |
| CO₂ | 0 | 4 | +4 |
| Substance | Before | After | Change |
|---|---|---|---|
| H₂ | 3 | 1 | −2 |
| Cl₂ | 2 | 0 | −2 |
| HCl | 0 | 4 | +4 |
| Substance | Before | After | Change |
|---|---|---|---|
| CH₄ | 2 | 0 | −2 |
| O₂ | 5 | 1 | −4 |
| CO₂ | 0 | 2 | +2 |
| H₂O | 0 | 4 | +4 |
| Substance | Before | After | Change |
|---|---|---|---|
| NO | 6 | 2 | −4 |
| O₂ | 2 | 0 | −2 |
| NO₂ | 0 | 4 | +4 |
| Substance | Before | After | Change |
|---|---|---|---|
| H₂O₂ | 4 | 0 | −4 |
| H₂O | 0 | 4 | +4 |
| O₂ | 0 | 2 | +2 |
Gas molecules and pressure
At constant temperature:
- In a rigid container (constant volume), the pressure of a gas is proportional to the moles of gas: fewer gas molecules after the reaction means lower pressure, and more means higher pressure.
- In a container at constant pressure with a movable piston, the volume is proportional to the moles of gas: fewer gas molecules and the piston falls, more and it rises.
For example, in , 3 gas molecules become 2, so in a rigid container the pressure drops to 2/3 of its starting value. Count only gases: at 125 °C, forms water vapor, which counts; at room temperature the water forms as a liquid and doesn’t.
The pressure changes because the number of gas particles changes, not because product molecules are “smaller”, and not because of mass: the mass before and after a reaction is always the same.
Example 1: From particle diagrams to an equation
Before the reaction, a box holds 6 NO molecules and 4 O₂ molecules; afterward it holds 6 NO₂ molecules and 1 O₂ molecule. Write the equation for the reaction.
-
Count the change in each molecule. Coefficients give the ratio of particles that react, so use the changes, not the totals before or after:
-
Leave out the leftovers. The 1 molecule left over didn’t react ( is in excess), so it doesn’t count toward the coefficients.
-
Reduce to lowest whole numbers.
The equation is .
-
Check the atoms.
N and O balance on both sides.
Example 2: From an equation to a particle diagram
For the reaction , a box starts with 4 SO₂ molecules and 4 O₂ molecules. Describe what the box contains after the reaction is complete.
-
Find the reactant that runs out. By the equation, SO₂ and O₂ react 2 : 1. The O₂ needed for 4 SO₂:
There is enough O₂, so SO₂ runs out first (the limiting reactant).
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Products. SO₂ and SO₃ are 2 : 2 = 1 : 1, so 4 SO₂ molecules form 4 SO₃ molecules.
-
Leftover reactant.
-
Describe the box. It holds and no SO₂.
-
Check the atoms.
- N₂ goes from 3 to 1, H₂ from 6 to 0, and NH₃ from 0 to 4. The coefficient of N₂?
The changes are 2 : 6 : 4 = 1 : 3 : 2.
- Same reaction: the coefficient of H₂?
In 1 : 3 : 2, H₂ is 3.
- Same reaction: the coefficient of NH₃?
In 1 : 3 : 2, NH₃ is 2.
- CO goes from 4 to 0, O₂ from 3 to 1, and CO₂ from 0 to 4. The coefficient of O₂?
The changes are 4 : 2 : 4 = 2 : 1 : 2.
- Same reaction: the coefficient of CO₂?
In 2 : 1 : 2, CO₂ is 2.
- H₂ goes from 3 to 1, Cl₂ from 2 to 0, and HCl from 0 to 4. The coefficient of HCl?
The changes are 2 : 2 : 4 = 1 : 1 : 2.
- CH₄ goes from 2 to 0, O₂ from 5 to 1, and H₂O from 0 to 4. The coefficient of O₂?
The changes for CH₄ and O₂ are 2 : 4 = 1 : 2.
- Same reaction: the coefficient of H₂O?
The changes for CH₄ and H₂O are 2 : 4 = 1 : 2.
- SO₂ goes from 4 to 0, O₂ from 4 to 2, and SO₃ from 0 to 4. The coefficient of O₂?
The changes are 4 : 2 : 4 = 2 : 1 : 2; the 2 leftover O₂ molecules don't count.
- NO goes from 6 to 2, O₂ from 2 to 0, and NO₂ from 0 to 4. The coefficient of NO?
The changes are 4 : 2 : 4 = 2 : 1 : 2, not the 6 in the box.
Common mistakes
Mistake 1: “The coefficients are the numbers of each molecule in the boxes.” The boxes often contain excess molecules that didn’t react. The coefficients match the changes (the amounts used up and formed), reduced to the lowest whole numbers.
Mistake 2: “The coefficient ratio is the mass ratio.” Coefficients give the ratio of particles, which is also the mole ratio. In 2H₂ + O₂ → 2H₂O, the masses of H₂ and O₂ are in the ratio 4 : 32, not 2 : 1.
Mistake 3: “If it won’t balance, change a subscript.” Changing a subscript changes the substance: H₂O becomes H₂O₂, hydrogen peroxide. Balance only by changing coefficients.
Quick check
Before a reaction, a box holds 5 H₂ molecules and 4 O₂ molecules; afterward it holds 4 H₂O molecules, 1 H₂ molecule, and 2 O₂ molecules. Which equation correctly represents the reaction?
The changes: H₂ down 4, O₂ down 2, H₂O up 4, and 4 : 2 : 4 = 2 : 1 : 2. A includes the leftover molecules; C doesn't conserve atoms; D has the right ratio but isn't reduced.
For , a box starts with 2 N₂ molecules and 9 H₂ molecules. What does it contain after the reaction is complete?
2 N₂ molecules need 6 H₂ molecules and there are 9, so N₂ runs out first, forming 4 NH₃ and leaving 3 H₂. C is calculated from H₂; B forgets the leftover H₂.
Which statement about is correct?
Coefficients give the ratio of particles. A treats them as masses; B is wrong because 3 molecules become 2; D is wrong because atoms are conserved.
In a particle diagram, 6 A₂ molecules react completely to form 4 A₃ molecules. Which equation is correct?
6 : 4 = 3 : 2, and the atoms balance: 6 A on each side. B isn't reduced; C and D don't conserve atoms.
Which particle diagram best represents ?
(aq) means a solution, where NaCl exists as Na⁺ and Cl⁻ ions. A and B show no ionization; C is solid NaCl(s).
Balancing Fe + O₂ → Fe₂O₃, a student writes Fe + O₂ → FeO₂. What's wrong?
Balance only with coefficients, never subscripts. The correct equation is 4Fe + 3O₂ → 2Fe₂O₃.
A rigid container holds 4 CO molecules and 3 O₂ molecules, and the reaction 2CO + O₂ → 2CO₂ goes as far as possible. Which diagram best represents the particles in the container afterward?
4 CO molecules need 2 O₂ molecules, so CO runs out first, forming 4 CO₂ and leaving 1 O₂. A uses no O₂ and doesn't conserve atoms; B conserves atoms but the reaction hasn't gone as far as possible; D loses the leftover O₂.
A rigid container holds 4 H₂ molecules and 3 O₂ molecules at 125 °C. The reaction 2H₂ + O₂ → 2H₂O goes as far as possible, and the temperature stays at 125 °C. Which diagram best represents the particles in the container afterward?
4 H₂ molecules need 2 O₂ molecules, forming 4 H₂O and leaving 1 O₂. At 125 °C, above water's boiling point, H₂O is a gas and spreads through the whole container like the O₂, so C is wrong; A uses no O₂; D doesn't conserve atoms.
In a rigid container, 2 mol H₂ and 1 mol O₂ react completely at 125 °C to form water vapor, and the temperature stays at 125 °C. What fraction of the starting pressure is the final pressure, and why?
At constant volume and temperature, pressure is proportional to moles of gas. A: equal mass doesn't mean an equal number of gas particles; C: molecular size doesn't set an ideal gas's pressure; D: the temperature is held constant.
At constant temperature and pressure (with a freely moving piston), 2NH₃(g) → N₂(g) + 3H₂(g) goes to completion. By what factor does the gas volume change?
At constant pressure, volume is proportional to moles of gas: 2 mol becomes 1 + 3 = 4 mol, so the volume doubles and the piston rises.
A particle diagram shows 4 H₂O₂ molecules, which decompose completely by 2H₂O₂ → 2H₂O + O₂. What should the box contain afterward?
4 H₂O₂ molecules are twice the equation, forming 4 H₂O and 2 O₂. Check: 8 H and 8 O before and after. A gives the equation's coefficients, not the counts in this box.
A solution containing 4 Ag⁺ and 4 NO₃⁻ ions is mixed with one containing 3 Na⁺ and 3 Cl⁻ ions, and AgCl precipitates. Which ions remain in solution (not counting the precipitate)?
3 Cl⁻ and 3 Ag⁺ form 3 AgCl, leaving 1 Ag⁺ in excess; the spectators NO₃⁻ and Na⁺ stay in solution.
AP-style free response
Hydrogen reacts with chlorine to form hydrogen chloride: . Before the reaction, a box holds 5 H₂ molecules and 3 Cl₂ molecules.
(a) Draw (or describe) the particles in the box after the reaction is complete, giving the number of each kind of molecule.
(b) Use atom counts to show that your diagram is consistent with conservation of mass.
(c) Looking at the box after the reaction, a student writes . Identify two errors in this equation.
(d) If the box instead started with 4 H₂ and 4 Cl₂ molecules, what would it contain after the reaction? Is there an excess reactant?
Model answer and scoring
(a) 3 Cl₂ molecules need 3 H₂ molecules and there are 5, so Cl₂ runs out first. Afterward the box holds 6 HCl molecules and 2 H₂ molecules, with no Cl₂.
(b) Before: 5 × 2 = 10 H atoms and 3 × 2 = 6 Cl atoms. After: 6 + 2 × 2 = 10 H atoms and 6 Cl atoms. The number of each kind of atom is unchanged, so mass is conserved.
(c) First, the atoms don’t balance: 4 H and 6 Cl on the left, 6 H and 6 Cl on the right. Second, the coefficients should be the lowest whole-number ratio of the changes (H₂ : Cl₂ : HCl = 3 : 3 : 6 = 1 : 1 : 2), not counts copied from the box; the correct equation is .
(d) The 4 H₂ and 4 Cl₂ molecules react exactly, forming 8 HCl molecules; there is no excess reactant.
Scoring: (a) 6 HCl and 2 H₂; (b) both kinds of atoms checked; (c) atoms not conserved, and coefficients can’t be copied from the box; (d) 8 HCl, nothing left over.
Summary
- Coefficients give the particle ratio (and the mole ratio), not the mass ratio; balance only by changing coefficients.
- A particle diagram must conserve atoms, show ratios that match the coefficients, and show the right states.
- From diagrams to an equation: count the change for each substance, leave out leftovers, and reduce.
- From an equation to a diagram: find the reactant that runs out first, then work out the products and the leftover reactant.
AP tip: Particle-diagram questions often draw the excess reactant in the “after” box precisely to see whether you’ll put it in the equation. Make a quick list of each substance’s count before, after, and the change, then write the equation.