Question

For the reaction $\mathrm{A}\to$ products, successive half-lives are observed to be $10.0,20.0,$ and 40.0 $\min$ for an experiment in which $[\mathrm{A}{]}_0=0.10M.$ Calculate the concentration of $\mathrm{A}$ at the following times. a. 80.0 $\min$ b. 30.0 $\min$

Short answer

a. At 80.0 minutes: $[\mathrm{A}]\approx 3.93\times {10}^{-3}M$ b. At 30.0 minutes: $[\mathrm{A}]\approx 0.0126M$

Step-by-step solution

Write the first-order integrated rate law

In a first-order reaction, the integrated rate law is given by: $$ln\frac{[\mathrm{A}{]}_0}{[\mathrm{A}]}=kt$$ Where: - $[\mathrm{A}]$ is the concentration of A at time t, - $[\mathrm{A}{]}_0$ is the initial concentration of A, - k is the rate constant, and - t is the time.

Determine the rate constant (k) using first half-life

We know that the half-life of a first-order reaction is given by: $$t_{1/2}=\frac{0.693}{k}$$ So, we can solve for k: $$k=\frac{0.693}{t_{1/2}}$$ Given the first half-life, $t_{1/2}=10.0minutes$, we can plug this value to calculate the rate constant k: $$k=\frac{0.693}{10.0minutes}=0.0693mi{n}^{-1}$$

Calculate the concentration of A at 80.0 minutes

Now that we have the rate constant (k), we can use the integrated rate law to find the concentration of A at 80.0 minutes. We know $[\mathrm{A}{]}_0=0.10M$, $k=0.0693mi{n}^{-1}$, and $t=80.0min$. Plugging these values into the integrated rate law equation: $$ln\frac{0.10M}{[\mathrm{A}]}=(0.0693mi{n}^{-1})(80.0min)$$ Solving for [\mathrm{A}]: $$[\mathrm{A}]=0.10M\times e^{-(0.0693mi{n}^{-1})(80.0min)}$$ $$[\mathrm{A}]=0.10M\times e^{-5.544}$$ $$[\mathrm{A}]\approx 3.93\times {10}^{-3}M$$

Calculate the concentration of A at 30.0 minutes

Similarly, we can use the integrated rate law to find the concentration of A at 30.0 minutes. We know $[\mathrm{A}{]}_0=0.10M$, $k=0.0693mi{n}^{-1}$, and $t=30.0min$. Plugging these values into the integrated rate law equation: $$ln\frac{0.10M}{[\mathrm{A}]}=(0.0693mi{n}^{-1})(30.0min)$$ Solving for [\mathrm{A}]: $$[\mathrm{A}]=0.10M\times e^{-(0.0693mi{n}^{-1})(30.0min)}$$ $$[\mathrm{A}]=0.10M\times e^{-2.079}$$ $$[\mathrm{A}]\approx 0.0126M$$ Now we have the concentration of A at the given times: a. At 80.0 minutes: $[\mathrm{A}]\approx 3.93\times {10}^{-3}M$ b. At 30.0 minutes: $[\mathrm{A}]\approx 0.0126M$

Key concepts

Integrated rate law

In chemistry, understanding the change in concentration of reactants over time is crucial. This is where the integrated rate law comes into play. For a first-order reaction, the integrated rate law is expressed as $$ln\frac{[A_0]}{[A]}=kt$$Here:

• $[A]$ denotes the concentration of the reactant at any time $t$.
• $[A_0]$ is the initial concentration.
• $k$ represents the rate constant, which is a measure of how fast the reaction occurs.
• $t$ is the time elapsed.

The integrated rate law allows chemists to predict how long a reactant will last at a certain concentration or how long it takes to reach a desired concentration. Thus, by applying this equation, understanding reaction kinetics becomes more practical, making it a fundamental concept in the realm of chemical reactions.

Half-life

The half-life of a reaction is the time taken for the concentration of a reactant to decrease to half of its initial value. In first-order reactions, half-life is a crucial aspect because it remains constant over time, distinguishing it from zero-order and second-order reactions. The half-life $t_{1/2}$ for a first-order reaction can be calculated using:$$t_{1/2}=\frac{0.693}{k}$$Here, $k$ is the rate constant.

This unique property of constant half-life implies that every time period of duration equal to the half-life will reduce the initial concentration of the reactant by 50%. Hence, for any subsequent half-lives, the concentration of the reactant reduces exponentially. In our exercise, the half-lives observed were 10, 20, and 40 minutes, demonstrating this constant degradation pattern typical of first-order kinetics.

Rate constant

The rate constant $k$ plays a pivotal role in understanding the speed of a chemical reaction. For first-order reactions, $k$ is extracted using the relationship with the half-life, which is formulated as:$$k=\frac{0.693}{t_{1/2}}$$The rate constant is expressed in time inverse units, such as $mi{n}^{-1}$, implying how quickly or slowly a reaction proceeds.

In first-order kinetics, knowing $k$ allows us to calculate the change in concentration over time using the integrated rate law. It also provides insight into reaction mechanisms and conditions that might affect the reaction speed. In our specific example exercise, the calculated rate constant $k$ for the reaction was $0.0693mi{n}^{-1}$, which was determined from the first observed half-life of 10 minutes. Recognizing and calculating the rate constant is fundamental for making predictions about a chemical system's behavior over time.