Combining Volumes

Gas Volumes Follow Coefficients

The law of combining volumes states that reacting gases, and gaseous products, have volumes in simple whole-number ratios when measured at the same temperature and pressure. This law is also known as Gay-Lussac's law of combining gas volumes.

When all gas volumes are measured at the same temperature and pressure, their ratios follow the coefficients in the balanced equation. Hydrogen and oxygen combine by volume in a $2:1$ ratio to form water.

This rule applies only to substances in the gas phase. Their volumes must also be compared at the same temperature and pressure.

Water Must Be in the Gas Phase

Look at the balanced equation for forming water vapor.

2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O(g)}

The coefficients $2:1:2$ mean $2$ volume parts of hydrogen react with $1$ volume part of oxygen to form $2$ volume parts of water vapor.

V_{\mathrm{H_2}} : V_{\mathrm{O_2}} : V_{\mathrm{H_2O}} = 2:1:2

So, if every gas is measured at the same temperature and pressure, $200\ \mathrm{mL}$ of hydrogen reacts with $100\ \mathrm{mL}$ of oxygen to produce $200\ \mathrm{mL}$ of water vapor. If the water has condensed into liquid, the liquid volume is no longer read directly from this gas ratio.

Comparing Gas Volumes by Tube

Choose a reaction below. Each tube represents one volume part of gas at the same temperature and pressure.

Gas Volume Ratio Model

The tube heights compare gas volumes at the same temperature and pressure, so their ratios follow the reaction coefficients.

Two equal-volume portions of $\mathrm{H_2}$ react with one equal-volume portion of $\mathrm{O_2}$ and form two equal-volume portions of $\mathrm{H_2O(g)}$ at the same temperature and pressure.

Volume ratio: $\mathrm{H_2:O_2:H_2O} = 2:1:2$ .
Scaled example: $200\ \mathrm{mL}$ of $\mathrm{H_2}$ reacts with $100\ \mathrm{mL}$ of $\mathrm{O_2}$ .

Each tube height represents a gas volume at the shared temperature and pressure. The number of equal volume parts therefore matches the coefficient in the balanced equation.

Ammonia Decomposition Example

Suppose $600\ \mathrm{mL}$ of ammonia gas decomposes completely at the same temperature and pressure. The balanced equation is:

2\mathrm{NH_3(g)} \rightarrow 3\mathrm{H_2(g)} + \mathrm{N_2(g)}

The coefficients give this volume ratio:

V_{\mathrm{NH_3}} : V_{\mathrm{H_2}} : V_{\mathrm{N_2}} = 2:3:1

Because $2$ parts of ammonia equal $600\ \mathrm{mL}$ , $1$ volume part equals $300\ \mathrm{mL}$ .

\begin{aligned} V_{\mathrm{H_2}} &= 3 \times 300 = 900\ \mathrm{mL} \\ V_{\mathrm{N_2}} &= 1 \times 300 = 300\ \mathrm{mL} \end{aligned}

So decomposing $600\ \mathrm{mL}$ of ammonia gas produces $900\ \mathrm{mL}$ of hydrogen gas and $300\ \mathrm{mL}$ of nitrogen gas.

Conditions for Using Volume Ratios

The reaction equation must already be balanced.
The substances being compared must be gases.
All gas volumes must be measured at the same temperature and pressure.
If a product turns into a liquid or solid, that product volume is not read directly from the gas coefficients.

If any of these conditions is missing, compute the mole amounts first and convert them to volumes.