Problem 1
Question
The reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g})\) is an example of (a) backward reaction (b) forward reaction (c) irreversible reaction (d) reversible reaction
Step-by-Step Solution
Verified Answer
(d) reversible reaction
1Step 1: Understanding the Reaction Equation
The given chemical equation is \( \mathrm{PCl}_{5} (\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3} (\mathrm{s}) + \mathrm{Cl}_{2} (\mathrm{g}) \). The double-headed arrow (\( \rightleftharpoons \)) indicates that the reaction can proceed in both directions, from left to right and from right to left.
2Step 2: Analyzing Terms Related to Reactions
- **Backward Reaction**: A reaction proceeding from products to reactants.
- **Forward Reaction**: A reaction proceeding from reactants to products.
- **Irreversible Reaction**: A reaction that cannot proceed in reverse, typically indicated by a single-headed arrow.
- **Reversible Reaction**: A reaction that can proceed in both directions, indicated by a double-headed arrow.
3Step 3: Identifying the Reaction Type
Since the reaction is indicated by a double-headed arrow (\( \rightleftharpoons \)), it suggests that the reaction can proceed in both forward and backward directions. Hence, the reaction is a reversible reaction.
Key Concepts
Chemical EquilibriumReaction TypesArrow Symbols in Chemical Equations
Chemical Equilibrium
Chemical equilibrium is a fascinating concept in chemistry where a reaction and its reverse occur simultaneously and at equal rates. This means that the concentrations of reactants and products remain constant over time. When a system reaches this state, it is said to be in dynamic equilibrium. It's dynamic because even though amounts don't change, molecules keep reacting with each other.
Here are some important features of chemical equilibrium:
Here are some important features of chemical equilibrium:
- The rates of the forward and backward reactions are equal.
- No net change occurs in the concentrations of reactants and products.
- Equilibrium is reached in a closed system where reactants and products can't escape.
Reaction Types
Chemical reactions can be grouped into various types based on different criteria. The directionality of the reaction is an essential aspect of its classification:
- Forward Reaction: This reaction proceeds from reactants to products. In our given reaction, this means the transformation starts with \( \text{PCl}_5 \) changing into \( \text{PCl}_3 \) and \( \text{Cl}_2 \).
- Backward Reaction: Also known as the reverse reaction, this goes from products back to reactants. Here, \( \text{PCl}_3 \) combines with \( \text{Cl}_2 \) to reform \( \text{PCl}_5 \).
- Irreversible Reaction: These reactions can proceed in one direction only. Once products are formed, they don't change back into reactants.
- Reversible Reaction: These reactions can move in both forward and backward directions. They are usually represented by a special arrow symbol, which we'll explore next.
Arrow Symbols in Chemical Equations
In chemistry, arrow symbols in equations define the nature and direction of reactions. Understanding these symbols helps in interpreting what the reaction can do:
Recognizing these arrow symbols is key to understanding complex reactions and predicting how they will behave under different conditions. They are essential tools in the chemist's toolkit for navigating the microscopic world of molecules.
- Single-headed Arrow: Indicated as (→), signals an irreversible reaction, pointing from reactants to products.
- Double-headed Arrow: Shown as (↔ or \( \rightleftharpoons \)), denotes a reversible reaction, where the reaction can proceed in both directions.
Recognizing these arrow symbols is key to understanding complex reactions and predicting how they will behave under different conditions. They are essential tools in the chemist's toolkit for navigating the microscopic world of molecules.
Other exercises in this chapter
Problem 2
For a reversible reaction, if the concentrations of the reactants are doubled, at constant temperature the equilibrium constant will be (a) one-fourth (b) halve
View solution Problem 3
The value of \(K_{p}\) in the reaction \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(\mat
View solution Problem 4
In the reaction \(\mathrm{BaO}_{2}(\mathrm{~s}) \longrightarrow \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}=\) tve. In equilibriu
View solution