Gas Evolution Equations: Videos & Practice Problems

Gas Evolution Reaction Example

In this example, we explore the reaction between sodium carbonate and hydrobromic acid, focusing on predicting whether a reaction occurs and writing the balanced molecular equation. The first step involves breaking down the reactants into their ionic forms. Sodium carbonate consists of sodium ions (Na⁺) and carbonate ions (CO₃^2-), while hydrobromic acid is made up of hydrogen ions (H⁺) and bromide ions (Br^-).

Next, we apply the principle of charge attraction to swap the ionic partners. The positive sodium ion will combine with the negative bromide ion to form sodium bromide (NaBr). For the combination of hydrogen ions and carbonate ions, we need to crisscross the charges, resulting in carbonic acid (H₂CO₃).

Following this, we refer to solubility rules. Sodium bromide is soluble in water due to sodium being a group 1A ion. Carbonic acid, however, is unstable and will decompose into water (H₂O) and carbon dioxide gas (CO₂). Thus, we replace carbonic acid in our equation with its decomposition products.

Finally, we balance the molecular equation. Initially, we have 2 sodium ions on the reactant side, so we place a coefficient of 2 in front of sodium bromide. The carbon and oxygen counts remain consistent, but we also need to balance the hydrogen and bromine atoms. Since we have 2 bromide ions from the sodium bromide, we adjust the coefficient in front of hydrobromic acid to 2, ensuring that both sides of the equation are balanced.

The final balanced equation for the reaction is:

2 HBr + Na₂CO₃ → 2 NaBr + H₂O + CO₂ (g)

Predict the products formed from the following gas evolution equation.
NH₄Cl (aq) + NaOH (aq) →

• A. NH₄Cl (aq) + NaOH (aq) → NH₄OH (aq) + NaCl (aq)
• B. NH₄Cl (aq) + NaOH (aq) → NH₃ (g) + H₂O (l) + NaCl (aq)
• C. NH₄Cl (aq) + NaOH (aq) → NH₄OH (g) + NaCl (s)
• D. No reaction

Answer: (B)

Here's what students ask on this topic:

What is a gas evolution equation in chemistry?

A gas evolution equation is a type of molecular equation that describes chemical reactions where specific gases are produced as final products. These gases commonly include ammonia ($N_{3}H$), carbon dioxide ($C_{2}O2$), sulfur dioxide ($S_{2}O2$), and hydrogen sulfide ($H_{2}S$). These gases form when intermediate or medium products lose a water molecule ($H2O$). The medium product is the compound formed initially by the combination of reactive ions before it converts into the final gaseous product. For example, ammonium ion (${\mathrm{NH}}_4+$) and hydroxide ion (${\mathrm{OH}}_{-}$) combine to form ammonium hydroxide (${\mathrm{NH}}_4\mathrm{OH}$), which upon losing water produces ammonia gas. Understanding these steps helps predict the gases evolved in chemical reactions.

How do intermediate products lead to the formation of gases like NH₃, CO₂, and SO₂ in gas evolution reactions?

In gas evolution reactions, intermediate or medium products form first by the combination of reactive ions. These intermediates then lose a water molecule ($H2O$) to produce the final gaseous products. For example, ammonium ion (${\mathrm{NH}}_4+$) reacts with hydroxide ion (${\mathrm{OH}}_{-}$) to form ammonium hydroxide (${\mathrm{NH}}_4\mathrm{OH}$), which loses water to yield ammonia gas (${\mathrm{NH}}_3$). Similarly, hydrogen ion ($H_{+}$) and bicarbonate ion (${\mathrm{HCO}}_{}$₃^-) form carbonic acid ($H_2\mathrm{CO}3$), which loses water to produce carbon dioxide (${\mathrm{CO}}_2$). Sulfurous acid ($H_2\mathrm{SO}3$) loses water to form sulfur dioxide (${\mathrm{SO}}_2$). This loss of water from intermediate products is key to gas formation in these reactions.

Why does hydrogen sulfide (H₂S) not lose water in gas evolution reactions?

Hydrogen sulfide ($H_{2}S$) is unique among common gases formed in gas evolution reactions because it is already in its final gaseous form and does not contain oxygen. Unlike other intermediate products such as ammonium hydroxide or carbonic acid, which lose a water molecule ($H2O$) to form gases, hydrogen sulfide does not have water to lose. It forms directly from the combination of hydrogen ions ($H_{+}$) and sulfide ions ($S_{2-}$) without an intermediate that contains water. Therefore, no dehydration step is needed, and $H_{2}S$ is produced directly as a gas in the reaction.

How do you write the gas evolution equation for the reaction between ammonium ion and hydroxide ion?

To write the gas evolution equation for ammonium ion (${\mathrm{NH}}_4+$) reacting with hydroxide ion (${\mathrm{OH}}_{-}$), first combine the ions to form the intermediate product ammonium hydroxide (${\mathrm{NH}}_4\mathrm{OH}$):

$$\mathrm{NH}{4}_{}++\; OH-\to \; NH$$₄OH

Next, ammonium hydroxide loses a water molecule ($H2O$) to form ammonia gas (${\mathrm{NH}}_3$):

$$NH$$₄OH → NH₃(g) + H₂O

Combining these steps, the overall gas evolution equation is:

$$NH$$₄⁺ + OH^- → NH₃(g) + H₂O

This equation shows the formation of ammonia gas through the loss of water from the intermediate ammonium hydroxide.

What role do charges and ion numbers play in forming intermediate products in gas evolution reactions?

Charges and ion numbers are crucial in determining how reactive ions combine to form intermediate or medium products in gas evolution reactions. When ions have charges of equal magnitude but opposite signs, they combine directly, and their charges cancel out. For example, ammonium ion (${\mathrm{NH}}_4+$) and hydroxide ion (${\mathrm{OH}}_{-}$) combine to form ammonium hydroxide (${\mathrm{NH}}_4\mathrm{OH}$). However, when the charges or numbers differ, the ions crisscross to balance the charges, forming compounds like carbonic acid ($H_2\mathrm{CO}3$) from hydrogen ion ($H_{+}$) and bicarbonate ion (${\mathrm{HCO}}_{}$₃^-). This balancing ensures the intermediate product is electrically neutral before it loses water to form the final gas.

How is carbon dioxide formed in gas evolution reactions involving bicarbonate ions?

Carbon dioxide (${\mathrm{CO}}_2$) forms in gas evolution reactions when hydrogen ions ($H_{+}$) react with bicarbonate ions (${\mathrm{HCO}}_{}$₃^-) to form carbonic acid ($H_2\mathrm{CO}3$) as an intermediate. The reaction is:

$$H++\; HCO$$₃^- → H₂CO₃

Carbonic acid is unstable and loses a water molecule ($H2O$) to produce carbon dioxide gas and water:

$$H$$₂CO₃ → CO₂(g) + H₂O

This process explains how carbon dioxide gas evolves in reactions involving bicarbonate ions and acids, a common occurrence in acid-base chemistry.