Problem 64 - Chemistry: The Central Science

Question

Write a balanced equation for each of the following reactions: (a) hydrolysis of \(\mathrm{PCl}_{5}\) (b) dehydration of phosphoric acid (also called orthophosphoric acid) to form pyrophosphoric acid, (c) reaction of \(\mathrm{P}_{4} \mathrm{O}_{10}\) with water. Carbon, the Other Group 4A Elements, and Boron (Sections 22.9, 22.10, and 22.11)

Step-by-Step Solution

Answer

(a) \( \mathrm{PCl}_{5} + 4\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{H}_{3}\mathrm{PO}_{4} + 5\mathrm{HCl} \) (b) \( 2\mathrm{H}_{3}\mathrm{PO}_{4} \rightarrow \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7} + \mathrm{H}_{2}\mathrm{O} \) (c) \( \mathrm{P}_{4}\mathrm{O}_{10} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 4\mathrm{H}_{3}\mathrm{PO}_{4} \)

Step 1: Hydrolysis of PCl5

In the hydrolysis of PCl5, the reactants are PCl5 and water (H2O). The products are phosphoric acid (H3PO4) and hydrogen chloride (HCl). To balance the equation, we must ensure that the number of atoms of each element is equal on both sides of the equation. Reactants: PCl5 + H2O Products: H3PO4 + HCl

Step 2: Balance P and O atoms in the Quick phosphorus and oxygen.

P: There is one P atom on each side of the equation, so it is already balanced. O: There are 4 O atoms on the product side, and just 1 O atom on the reactant side. To balance, we need to multiply the number of water molecules by 4: Reactants: PCl5 + 4H2O Products: H3PO4 + HCl

Step 3: Balance Cl and H atoms

Cl: There are 5 Cl atoms in the reactant PCl5, and only 1 Cl atom in one HCl molecule in the product. We need to multiply the number of HCl molecules by 5: Reactants: PCl5 + 4H2O Products: H3PO4 + 5HCl H: Now there are 5 x 2 = 10 H atoms in 5 molecules of HCl, and 4 x 2 = 8 H atoms in 4 molecules of H2O (reactants). A total of 10 + 8 = 18 H atoms exist on the product side (3 x 6 = 18 H atoms in 1 molecule of H3PO4, and 5 x 1 = 5 H atoms in 5 molecules of HCl). So, the equation is balanced: \( \mathrm{PCl}_{5} + 4\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{H}_{3}\mathrm{PO}_{4} + 5\mathrm{HCl} \)

Step 4: Dehydration of phosphoric acid

Dehydration of phosphoric acid (H3PO4) forms pyrophosphoric acid (H4P2O7). The balanced equation can be written as follows: \( 2\mathrm{H}_{3}\mathrm{PO}_{4} \rightarrow \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7} + \mathrm{H}_{2}\mathrm{O} \)

Step 5: Reaction of P4O10 with water

In the reaction of P4O10 with water, the reactants are P4O10 and water (H2O). The product of this reaction is phosphoric acid (H3PO4). To balance the equation, we need to equalize the number of atoms of each element on both sides of the equation. Reactants: P4O10 + H2O Products: H3PO4

Step 6: Balance P, O, and H atoms

P: There are 4 P atoms on the reactant side, so we need to multiply the number of H3PO4 molecules by 4: Reactants: P4O10 + H2O Products: 4H3PO4 O: There are 10 O atoms in P4O10, and we have 4 x 4 = 16 O atoms in 4 molecules of H3PO4. To balance the O atoms, we need to add 6 O atoms to the reactants side. We can do this by multiplying the number of water molecules by 6: Reactants: P4O10 + 6H2O Products: 4H3PO4 H: With 6 molecules of H2O, we have 6 x 2 = 12 H atoms on the reactant side, and there are 4 x 3 = 12 H atoms in 4 molecules of H3PO4 on the product side, so the equation is now balanced: \( \mathrm{P}_{4}\mathrm{O}_{10} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 4\mathrm{H}_{3}\mathrm{PO}_{4} \)

Key Concepts

hydrolysis reaction

The term hydrolysis comes from two Greek words: "hydro" meaning water and "lysis" meaning breaking. A hydrolysis reaction involves the interaction of a chemical compound with water, resulting in the breakdown of the compound and formation of new substances. In the case of PCl extsubscript{5}, water molecules attack the phosphorus atom, replacing the chlorine atoms with hydroxyl groups.

• This results in the formation of phosphoric acid (H extsubscript{3}PO extsubscript{4}) and hydrochloric acid (HCl).
• The balanced chemical equation for this hydrolysis is: \[ \mathrm{PCl}_{5} + 4\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{H}_{3}\mathrm{PO}_{4} + 5\mathrm{HCl} \]
• Starting with the reactants and products, we ensure each element is balanced across both sides of the equation.

Balancing hydrolysis reactions can be tricky as it often results in multiple products, requiring careful manipulation and counting of atoms. This is crucial to ensuring the law of conservation of mass holds.

dehydration reaction

Dehydration reactions involve the removal of water from a molecule. Essentially, two molecules combine to form a larger molecule, with water as a side product. In the dehydration of phosphoric acid, two molecules of this acid (H extsubscript{3}PO extsubscript{4}) combine to form pyrophosphoric acid (H extsubscript{4}P extsubscript{2}O extsubscript{7}), releasing a water molecule in the process.

• The balanced equation for this process is:\[ 2\mathrm{H}_{3}\mathrm{PO}_{4} \rightarrow \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7} + \mathrm{H}_{2}\mathrm{O} \]
• This type of reaction is commonly found in organic chemistry during the synthesis of various compounds.
• The dehydration reaction is primarily driven by heating, which promotes the removal of water.

The key to understanding dehydration reactions is recognizing how molecules can link together, releasing water as a byproduct. Such reactions are fundamental in biochemical processes like the formation of complex carbohydrates.

reaction balance

Balancing chemical reactions involves ensuring the number of each type of atom is the same on both sides of the equation. This process reflects the principle of conservation of mass, as the mass of products must be equal to the mass of reactants.

• The process starts by writing down the unbalanced equation and identifying reactants and products.
• For reactions like the one between P extsubscript{4}O extsubscript{10} and water, balance is achieved by adjusting coefficients:\[ \mathrm{P}_{4}\mathrm{O}_{10} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 4\mathrm{H}_{3}\mathrm{PO}_{4} \]
• This involves complex steps, but an organized approach can simplify it.

Balancing is essential in chemistry as it translates to accurate experimental results and ensures chemical reactions proceed as intended. Understanding how to systematically balance reactions can demystify complex equations and enhance comprehension of chemical behavior.