Problem 218
Question
The time for half life period of a certain reaction \(\mathrm{A} \longrightarrow\) products is 1 hour. When the initial concentration of the reactant ' \(A\) ', is \(2.0 \mathrm{~mol} \mathrm{~L}^{-1}\), how much time does it take for its concentration to come from \(0.50\) to \(0.25 \mathrm{~mol} \mathrm{~L}^{-1}\) if it is a zero order reaction? [2010] (a) \(4 \mathrm{~h}\) (b) \(0.5 \mathrm{~h}\) (c) \(0.25 \mathrm{~h}\) (d) \(1 \mathrm{~h}\)
Step-by-Step Solution
Verified Answer
The time taken is 0.25 hours, thus the correct answer is (c) \(0.25 \mathrm{~h}\).
1Step 1: Understanding Zero Order Reactions
In a zero-order reaction, the rate of reaction is independent of the concentration of reactants. The rate law is expressed as \( r = k \), where \( k \) is the rate constant.
2Step 2: Zero Order Reaction Half-life
For a zero-order reaction, the half-life formula is \( t_{1/2} = \frac{[A]_0}{2k} \). We know \( t_{1/2} = 1 \) hour, and the initial concentration \( [A]_0 = 2.0 \mathrm{\, mol \, L^{-1}} \). We can use this information to find \( k \).
3Step 3: Calculate Rate Constant \( k \)
Rearrange the half-life formula to find \( k \):\[ k = \frac{[A]_0}{2t_{1/2}} = \frac{2.0}{2 \times 1} = 1 \mathrm{\, mol \, L^{-1} \, h^{-1}} \].
4Step 4: Determine Time for Concentration Change
The formula to find the time \( t \) for a zero order reaction when concentration changes is \( t = \frac{[A]_0 - [A]}{k} \). Here, \( [A]_0 = 0.50 \mathrm{\, mol \, L^{-1}} \) and \( [A] = 0.25 \mathrm{\, mol \, L^{-1}} \).
5Step 5: Substitute Values and Solve for \( t \)
Substitute the known values in the formula:\[ t = \frac{0.50 - 0.25}{1} = 0.25 \mathrm{\, h} \].
6Step 6: Conclusion
After calculating, the time taken for the concentration to decrease from \(0.50\) to \(0.25 \mathrm{\, mol \, L^{-1}}\) is \(0.25\) hours.
Key Concepts
Half-lifeReaction RateRate ConstantChemical Kinetics
Half-life
The term "half-life" refers to the time taken for the concentration of a reactant to reduce to half of its initial amount.
It provides valuable insights into the speed at which a reaction proceeds.
For a zero order reaction, the half-life is calculated using the formula \( t_{1/2} = \frac{[A]_0}{2k} \), where:
This signifies that it will take 1 hour for the concentration to decrease from 2 mol/L to 1 mol/L.
Understanding half-life helps in predicting how long a reaction will take to reach a desired concentration.
It provides valuable insights into the speed at which a reaction proceeds.
For a zero order reaction, the half-life is calculated using the formula \( t_{1/2} = \frac{[A]_0}{2k} \), where:
- \([A]_0\) is the initial concentration of the reactant.
- \(k\) is the rate constant.
This signifies that it will take 1 hour for the concentration to decrease from 2 mol/L to 1 mol/L.
Understanding half-life helps in predicting how long a reaction will take to reach a desired concentration.
Reaction Rate
The reaction rate measures how quickly the reactants are converted into products.
In zero-order reactions, the reaction rate remains constant and does not depend on the concentration of the reactant.
This makes zero-order reactions unique since most reaction rates are influenced by concentration changes.The equation for the rate of a zero-order reaction is expressed as:
In our specific problem, since the rate is independent of the concentration, even as the concentration decreases, the rate stays at 1 mol/L/hr until all reactants are consumed or the reaction is terminated.
In zero-order reactions, the reaction rate remains constant and does not depend on the concentration of the reactant.
This makes zero-order reactions unique since most reaction rates are influenced by concentration changes.The equation for the rate of a zero-order reaction is expressed as:
- \( r = k \)
In our specific problem, since the rate is independent of the concentration, even as the concentration decreases, the rate stays at 1 mol/L/hr until all reactants are consumed or the reaction is terminated.
Rate Constant
The rate constant \( k \) is a crucial factor in the interpretation of reaction kinetics and represents the speed of the reaction.
It is instrumental in devising the rate law and understanding the dynamics of the reaction process.
For our zero-order reaction, the rate constant was calculated using the formula derived from the half-life:
This value specifically tells us that the reaction reduces by 1 mol of reactant per liter every hour, regardless of the amount of reactant present initially.
Having a clear value of \( k \) allows chemists to predict and analyze the behavior of the reaction over time.
It is instrumental in devising the rate law and understanding the dynamics of the reaction process.
For our zero-order reaction, the rate constant was calculated using the formula derived from the half-life:
- \( k = \frac{[A]_0}{2t_{1/2}} \)
This value specifically tells us that the reaction reduces by 1 mol of reactant per liter every hour, regardless of the amount of reactant present initially.
Having a clear value of \( k \) allows chemists to predict and analyze the behavior of the reaction over time.
Chemical Kinetics
Chemical kinetics focuses on understanding the speed and mechanisms by which chemical reactions proceed.
It encompasses various factors like temperature, concentration, and the presence of catalysts that influence reaction rates.
In the realm of kinetics, zero-order reactions bring about interesting phenomena because of their unique characteristics:
Understanding these concepts prepares us to address and predict the behavior of reactions in diverse chemical systems.
It encompasses various factors like temperature, concentration, and the presence of catalysts that influence reaction rates.
In the realm of kinetics, zero-order reactions bring about interesting phenomena because of their unique characteristics:
- The reaction rate is consistent over time.
- The reaction doesn't slow down as the reactants are used up.
Understanding these concepts prepares us to address and predict the behavior of reactions in diverse chemical systems.
Other exercises in this chapter
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