Combining Volumes: Basic Laws of Chemistry - Chemistry (Grade 10)

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.

OpenStax Chemistry 2e explains that gas volume ratios in a reaction are given by the coefficients in the balanced equation when all gas volumes are measured at the same temperature and pressure in openstax.org. Britannica also summarizes Gay-Lussac's finding that hydrogen and oxygen combine by volume in a $2:1$ ratio to form water in britannica.com.

The important word is gas. Coefficients can be read directly as volume ratios only for substances in the gas phase and only when those volumes are compared at the same temperature and pressure.

Read Water as Vapor

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.

Gas Volume Meter

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

Gas Volume Meter

Compare tube heights to see reaction coefficients as gas volume ratios.

Two parts of $\mathrm{H_2}$ meet one part of $\mathrm{O_2}$ and form two parts of $\mathrm{H_2O(g)}$ .

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

The model is not comparing mass. It compares gas volume. Because the temperature and pressure are the same, each volume part follows a coefficient in the balanced reaction.

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.

Where the Shortcut Stops

This law is useful, but its conditions must stay visible.

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.

With those conditions in place, reaction coefficients do more than count moles. For gases under the same conditions, they also give a quick map of relative volumes.

Gas Volumes Follow Coefficients

The important word is gas. Coefficients can be read directly as volume ratios only for substances in the gas phase and only when those volumes are compared at the same temperature and pressure.

Read Water as Vapor

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

Gas Volume Meter

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

Gas Volume Meter

Compare tube heights to see reaction coefficients as gas volume ratios.

Two parts of $\mathrm{H_2}$ meet one part of $\mathrm{O_2}$ and form two parts of $\mathrm{H_2O(g)}$ .

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

The model is not comparing mass. It compares gas volume. Because the temperature and pressure are the same, each volume part follows a coefficient in the balanced reaction.

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.

Where the Shortcut Stops

This law is useful, but its conditions must stay visible.

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.

With those conditions in place, reaction coefficients do more than count moles. For gases under the same conditions, they also give a quick map of relative volumes.

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