Problem 26
Question
If for the reaction given below \(2 \mathrm{PQ} \rightleftharpoons{\mathrm{B}} \rightleftharpoons{\mathrm{P}_{2}}+\underset{\mathrm{g}}{\mathrm{Q}_{2}} \mathrm{~K}_{1}=2.5 \times 10^{5}\) \(\mathrm{PQ}+\frac{1}{2} \mathrm{R}_{2} \rightleftharpoons{\mathrm{PQR}} \mathrm{K}_{2}=5 \times 10^{-3}\) find \(\mathrm{K}_{3}\) for the reaction \(\frac{1}{2} \mathrm{P}_{2}+\frac{1}{2} \mathrm{Q}_{2}+\frac{1}{2} \mathrm{R}_{2} \rightleftharpoons \mathrm{PQR}\) (a) \(2.5 \times 10^{-3}\) (b) \(2.5 \times 10^{3}\) (c) \(1 \times 10^{-3}\) (d) \(5 \times 10^{-3}\)
Step-by-Step Solution
Verified Answer
The equilibrium constant \( K_3 \) is \( 1 \times 10^{-3} \) (option c).
1Step 1: Understand the Problem
We are given two reactions with their equilibrium constants, \( K_1 \) and \( K_2 \). We need to find the equilibrium constant \( K_3 \) for a third reaction \( \frac{1}{2} \mathrm{P}_2 + \frac{1}{2} \mathrm{Q}_2 + \frac{1}{2} \mathrm{R}_2 \rightleftharpoons \mathrm{PQR} \).
2Step 2: Relate Reactions with Known Equilibria
The first reaction \( 2 \mathrm{PQ} \rightleftharpoons \mathrm{P}_2 + \mathrm{Q}_2 \) has an equilibrium constant \( K_1 = 2.5 \times 10^5 \). The second reaction \( \mathrm{PQ} + \frac{1}{2} \mathrm{R}_2 \rightleftharpoons \mathrm{PQR} \) has an equilibrium constant \( K_2 = 5 \times 10^{-3} \).
3Step 3: Manipulate Reactions for Desired Equation
We want to derive the third reaction \( \frac{1}{2} \mathrm{P}_2 + \frac{1}{2} \mathrm{Q}_2 + \frac{1}{2} \mathrm{R}_2 \rightleftharpoons \mathrm{PQR} \) using \( K_1 \) and \( K_2 \). Since \( 2 \mathrm{PQ} \rightleftharpoons \mathrm{P}_2 + \mathrm{Q}_2 \) is the reverse of the formation of \( \mathrm{PQ} \) from \( \frac{1}{2} \mathrm{P}_2 + \frac{1}{2} \mathrm{Q}_2 \), the equilibrium constant for \( \frac{1}{2} \mathrm{P}_2 + \frac{1}{2} \mathrm{Q}_2 \rightleftharpoons \mathrm{PQ} \) is \( K_1^{-1/2} \).
4Step 4: Combine Equilibrium Constants
The formation of \( \mathrm{PQR} \) from \( \frac{1}{2} \mathrm{P}_2 + \frac{1}{2} \mathrm{Q}_2 + \frac{1}{2} \mathrm{R}_2 \) can be seen as the combination of two reactions:1. \( \frac{1}{2} \mathrm{P}_2 + \frac{1}{2} \mathrm{Q}_2 \rightleftharpoons\mathrm{PQ} \) with \( K = K_1^{-1/2} \)2. \( \mathrm{PQ} + \frac{1}{2} \mathrm{R}_2 \rightleftharpoons \mathrm{PQR} \) with \( K = K_2 \)So, \( K_3 = K_1^{-1/2} \times K_2 \).
5Step 5: Calculate \( K_3 \)
Now substitute the values:\[K_3 = (2.5 \times 10^5)^{-1/2} \times 5 \times 10^{-3}\]Calculate \( (2.5 \times 10^5)^{-1/2} \):\[(2.5 \times 10^5)^{1/2} = 500\]\[(2.5 \times 10^5)^{-1/2} = \frac{1}{500}\]Thus,\[K_3 = \frac{1}{500} \times 5 \times 10^{-3} = 1 \times 10^{-3}\]
6Step 6: Verify Calculation
Re-check each step in the calculation to ensure the application of laws of chemical equilibrium is correct, and all mathematical operations were performed correctly.
Key Concepts
Equilibrium ConstantReaction KineticsThermodynamics in Chemistry
Equilibrium Constant
The concept of the equilibrium constant is central to understanding chemical reactions at equilibrium. An equilibrium constant, denoted as \( K \), is a numerical value that describes the ratio of the concentrations of products to reactants at equilibrium for a given reaction.
This ratio is derived from the balanced chemical equation for the reaction. Each concentration is raised to the power of its corresponding stoichiometric coefficient.
Furthermore, equilibrium constants do not give information about the speed or duration required for the equilibrium to be reached.
This ratio is derived from the balanced chemical equation for the reaction. Each concentration is raised to the power of its corresponding stoichiometric coefficient.
- For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is expressed as: \[ K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
Furthermore, equilibrium constants do not give information about the speed or duration required for the equilibrium to be reached.
Reaction Kinetics
Reaction kinetics involves the study of the rate at which chemical reactions proceed and the factors that affect these rates. It’s different from equilibrium, which does not provide insights into the time taken to reach equilibrium but rather the proportions of products and reactants once equilibrium is achieved.
Several factors influence reaction rates:
- Concentration of Reactants: Higher concentrations generally lead to faster reaction rates due to more frequent collisions.
- Temperature: Increasing temperature usually increases reaction rates, as particles gain kinetic energy and collide more frequently and with greater energy.
- Catalysts: Catalysts increase reaction rates by lowering the activation energy required for the reaction to occur, without being consumed in the process.
- Surface Area: Larger surface areas can enhance reaction rates in reactions involving solids, as there are more sites available for reaction.
Thermodynamics in Chemistry
Thermodynamics is the branch of chemistry that focuses on the study of energy transformations. In the context of chemical reactions, it helps us understand why certain reactions occur and in which direction they are spontaneous.There are several laws and concepts within chemical thermodynamics that guide our understanding:
It is crucial in scenarios involving energy management and efficiency optimization in chemical processes.
- First Law of Thermodynamics: Energy cannot be created or destroyed; it can only change forms. This is often called the principle of conservation of energy.
- Second Law of Thermodynamics: In any spontaneous process, the total entropy of a system and its surroundings always increases.
- Gibbs Free Energy (\( G \)): Determines whether a process will occur spontaneously at constant temperature and pressure. A negative change in Gibbs Free Energy (\( \Delta G < 0 \)) indicates a spontaneous process.
It is crucial in scenarios involving energy management and efficiency optimization in chemical processes.
Other exercises in this chapter
Problem 24
In a reversible reaction, the catalyst (a) decreases activation energy of forward reaction (b) increases activation energy of forward reaction (c) decreases act
View solution Problem 25
A chemical reaction is catalysed by a catalyst \(\mathrm{X}\). Hence, \(\mathrm{X}\) (a) increases activation energy of the reaction (b) does not affect equilib
View solution Problem 27
For equilibrium reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+14.6 \mathrm{~J}\), increase in tem- per
View solution Problem 28
Which of the following favours the backward reaction in a chemical equilibrium? (a) decreasing the concentration of one of the reactants (b) increasing the conc
View solution