A substance reacts according to I order kinetics and rate constant for the reaction is \(1 \times 10^{-2} \mathrm{sec}^{-1}\). If its initial concentration is \(1 M\) (a) What is initial rate? (b) What is rate after 1 minute?

Short Answer

Expert verified
The initial rate is 1 × 10^{-2} M/sec, and the rate after 1 minute is approximately 5.488 × 10^{-3} M/sec.

Step by step solution

01

Determine the Initial Rate

For a reaction of first-order kinetics, the rate of the reaction can be determined using the rate law, which is the rate constant (k) times the concentration of the reactant ([A]). The initial rate is given by the formula rate = k * [A_initial]. Plugging in the given values, rate = (1 × 10^{-2} sec^{-1}) * (1 M).
02

Calculate Initial Rate

Multiplying the rate constant by the initial concentration yields the initial rate. So, initial rate = (1 × 10^{-2} sec^{-1}) * (1 M) = 1 × 10^{-2} M/sec.
03

Determine the Rate After 1 Minute

The rate of a first-order reaction at any time 't' can be found using the integrated rate law, which for a first-order reaction is [A] = [A_initial] * e^{-kt}. To find the concentration after 1 minute (60 seconds), we plug in the values: [A] = (1 M) * e^{-(1 × 10^{-2} sec^{-1}) * (60 sec)}.
04

Calculate the Concentration After 1 Minute

Use a calculator to compute the exponential term: [A] = (1 M) * e^{-0.6} ≈ (1 M) * 0.5488 = 0.5488 M.
05

Calculate Rate After 1 Minute

The rate after 1 minute is found by multiplying the rate constant by the concentration at that time: rate = k * [A]. So, rate after 1 minute = (1 × 10^{-2} sec^{-1}) * 0.5488 M = 5.488 × 10^{-3} M/sec.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Rate Law
Understanding the reaction rate law is crucial to studying the dynamics of how chemicals transform over time. Simply put, the rate law expresses the speed of a chemical reaction in terms of the concentration of its reactants. For a first-order reaction, the rate law has a direct relationship with the concentration of a single reactant. Mathematically, this relationship is represented as the rate equals the rate constant multiplied by the reactant's concentration, or in symbols, \( \text{rate} = k \times [A] \).

In practice, if you know the rate law for a reaction, you can predict how quickly a reactant will be consumed at any given concentration, as shown in the original exercise where the initial rate was determined using the given concentration and the rate constant.
Rate Constant
The rate constant is a proportionality factor in the rate law that influences how quickly a reaction proceeds. It is represented by the symbol \( k \) and can change with temperature, but is independent of reactant concentrations for a given reaction under constant conditions. For the exercise provided, the rate constant \( k \) is given as \(1 \times 10^{-2} \mathrm{sec}^{-1}\).

The rate constant gives us insight into the intrinsic speed of the reaction, and when coupled with reactant concentration, it allows us to calculate the initial reaction rate. The higher the value of \( k \) for a reaction at a given temperature, the faster the reaction takes place.
Integrated Rate Equation
Moving beyond the initial rate, we need to understand the integrated rate equation. This equation gives the concentration of a reactant at any time \( t \) during a first-order reaction, allowing us to see how the concentration decreases exponentially over time. The equation for a first-order reaction is \( [A] = [A_{\text{initial}}] \times e^{-kt} \), where \( [A] \) is the concentration at time \( t \) and \( [A_{\text{initial}}] \) is the initial concentration.

As shown in the exercise, this equation is used to determine the concentration of a reactant after a certain period has elapsed, which then helps to find the rate at that later time. Using this approach provides a powerful tool for predicting the future concentration of a reactant.
Half-life of Reactions
The concept of half-life in the context of reaction kinetics is a measure of how long it takes for half of the reactant in a chemical reaction to be consumed. For a first-order reaction, the half-life is a constant that does not depend on the initial concentration and is represented by the formula \( t_{1/2} = \frac{\ln(2)}{k} \).

The half-life provides a convenient way to understand the timescale of a reaction. For example, if the half-life is short, the reactant concentration decreases rapidly, indicating a fast reaction. Conversely, a long half-life means the reaction is slower. Being able to calculate half-life for a first-order reaction is particularly useful because it remains consistent throughout the reaction, thus serving as a straightforward and reliable measure of reaction speed.

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Most popular questions from this chapter

\(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposes according to equation; \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) (a) What does \(-\frac{d\left[\mathrm{~N}: \mathrm{C}_{5}\right]}{d t}\) denote? (b) What does \(\frac{d\left[\mathrm{O}_{2}\right]}{d t}\) denote? (c) What is the unit of rate of this reaction?

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) takes place according to I order as $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} $$ Calculate : (a) The rate constant, if instantaneous rate is \(1.4 \times 10^{-6}\) mol litre \(^{-1}\) \(\mathrm{sec}^{-1}\) when concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.04 \mathrm{M}\) (b) The rate of reaction when concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(1.20 \mathrm{M}\) (c) The concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) when the rate of reaction will be \(2.45 \times 10^{-5} \mathrm{~mol}\) litre \(^{-1} \mathrm{sec}^{-\mathrm{i}}\).

Show that time required to complete \(99.9 \%\) completion of a first order reaction is \(1.5\) times to \(90 \%\) completion.

Find the order of reaction for the rate expression rate \(=K[A][B]^{2 / 3}\). Also suggest the units of rate and rate constant for this expression.

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