Problem 8 - Chemistry and Chemical Reactivity

Question

Write balanced chemical equations for the following reactions: (a) production of ammonia, \(\mathrm{NH}_{3}(\mathrm{g}),\) by combin\(\operatorname{ing} \mathrm{N}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (b) production of methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\ell)\) by combining \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) (c) production of sulfuric acid by combining sulfur, oxygen, and water

Step-by-Step Solution

Answer

(a) \( \mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3 \) (b) \( \mathrm{CO} + 2\mathrm{H}_2 \rightarrow \mathrm{CH}_3\mathrm{OH} \) (c) \( 2\mathrm{S} + 3\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{H}_2\mathrm{SO}_4 \)

Step 1: Write Down the Unbalanced Equation for Ammonia Production

The reaction for forming ammonia involves nitrogen gas and hydrogen gas. The unbalanced equation is: \( \mathrm{N}_2(\mathrm{g}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{NH}_3(\mathrm{g}) \).

Step 2: Balance the Ammonia Production Equation

We need to balance the number of atoms for each element on both sides. Nitrogen is balanced by having 1 molecule of \( \mathrm{N}_2 \). Hydrogen requires 3 molecules of \( \mathrm{H}_2 \) to balance the 6 hydrogen atoms needed for 2 ammonia molecules. So, the balanced equation is: \( \mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightarrow 2\mathrm{NH}_3(\mathrm{g}) \).

Step 3: Write Down the Unbalanced Equation for Methanol Production

The reaction for forming methanol involves hydrogen gas and carbon monoxide gas. The unbalanced equation is: \( \mathrm{CO}(\mathrm{g}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{CH}_3\mathrm{OH}(\ell) \).

Step 4: Balance the Methanol Production Equation

To balance the equation, you need 2 molecules of hydrogen to balance the 4 hydrogen atoms in methanol. No balancing is needed for carbon. The balanced equation is: \( \mathrm{CO}(\mathrm{g}) + 2\mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{CH}_3\mathrm{OH}(\ell) \).

Step 5: Write Down the Unbalanced Equation for Sulfuric Acid Production

Sulfuric acid is produced through a multi-step process involving sulfur, oxygen, and water. The unbalanced primary reactions are: \( \mathrm{S}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_2(\mathrm{g}) \), \( \mathrm{2SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{SO}_3(\mathrm{g}) \), followed by \( \mathrm{SO}_3(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\ell) \rightarrow \mathrm{H}_2\mathrm{SO}_4(\ell) \).

Step 6: Balance the Sulfuric Acid Production Equation

First, balance the sulfur and oxygen in each step. Combine steps: add 2 moles of \( \mathrm{SO}_2 \) with 1 mole of \( \mathrm{O}_2 \). The equation from \( \mathrm{SO}_3 \) to sulfuric acid is balanced as is, resulting in the overall process equation: \( 2\mathrm{S}(\mathrm{s}) + 3\mathrm{O}_2(\mathrm{g}) + 2\mathrm{H}_2\mathrm{O}(\ell) \rightarrow 2\mathrm{H}_2\mathrm{SO}_4(\ell) \).

Key Concepts

Ammonia Production

Ammonia production is a crucial industrial process, commonly known as the Haber-Bosch process. This method synthesizes ammonia from nitrogen and hydrogen gases. Ammonia is essential for agricultural fertilizers, making its production vital for global food supply.
In the chemical reaction, nitrogen and hydrogen gases are combined to form ammonia \( \text{NH}_3 \). The unbalanced chemical equation for this reaction is:\[ \text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3 \]
To balance this equation:

• Begin with the nitrogen atoms. One nitrogen molecule \( \text{N}_2 \) provides two nitrogen atoms, matching the two nitrogen atoms needed for two \( \text{NH}_3 \) molecules.
• For hydrogen, three hydrogen molecules \( \text{H}_2 \) are required to supply six hydrogen atoms, balancing the six hydrogen atoms present in two ammonia molecules.

Thus, the balanced chemical equation becomes:\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]
This chemical reaction is conducted under high pressure and temperature, utilizing an iron catalyst to increase efficiency.

Methanol Production

Methanol production is another critical industrial process vital for the chemical industry. Methanol is used as a solvent, antifreeze, fuel, and a precursor for manufacturing various chemicals like formaldehyde.
The process of forming methanol involves the reaction between carbon monoxide \( \text{CO} \) and hydrogen \( \text{H}_2 \) gases. The initial unbalanced chemical equation reads:\[ \text{CO} + \text{H}_2 \rightarrow \text{CH}_3\text{OH} \]
To balance the equation:

• The carbon atom in carbon monoxide matches the one carbon atom in methanol, so no changes are needed here.
• For hydrogen, the equation requires two hydrogen molecules \( 2\text{H}_2 \) to balance the four hydrogen atoms present in \( \text{CH}_3\text{OH} \).

The balanced chemical equation thus becomes:\[ \text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH} \]
This synthesis typically occurs at high temperatures and pressures, with a catalyst, such as a mixture of copper and zinc oxides, to facilitate the reaction.

Sulfuric Acid Production

Sulfuric acid is a fundamental chemical used in many industrial applications, including fertilizer production, mineral processing, and wastewater treatment. The process involves multiple steps, starting from elemental sulfur.

• Step 1: Sulfur \( \text{S} \) is burned in oxygen \( \text{O}_2 \) to produce sulfur dioxide \( \text{SO}_2 \): \[ \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \]
• Step 2: The \( \text{SO}_2 \) reacts with additional oxygen to produce sulfur trioxide \( \text{SO}_3 \):\[ 2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \]
• Step 3: Finally, sulfur trioxide reacts with water to create sulfuric acid \( \text{H}_2\text{SO}_4 \):\[ \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \]

The balanced overall equation for producing sulfuric acid is:\[ 2\text{S} + 3\text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4 \]
This process requires specific conditions such as controlled temperatures and catalysts like vanadium oxide \( \text{V}_2\text{O}_5 \) to drive the conversion efficiently.