Question

Two substances, 'A' and 'B' are initially present as \(\left[A_{0}\right]=8\left[B_{0}\right]\) and \(t_{1 / 2}\) for the firstorder decomposition of 'A' and 'B' are 10 and \(20 \mathrm{~min}\), respectively. If they start decomposing at the same time, after how much time, the concentration of both of them would be same? (a) \(20 \mathrm{~min}\) (b) \(40 \mathrm{~min}\) (c) \(60 \mathrm{~min}\) (d) \(200 \mathrm{~min}\)

Short answer

After calculating, one finds that the time when the concentrations of A and B are the same is 40 minutes, so the answer is (b) 40 minutes.

Step-by-step solution

Use First-Order Kinetics Formulas

For a first-order reaction, the concentration of a substance at any time 't' can be determined using the formula \([A] = [A_0]e^{-kt}\) where \([A_0]\) is the initial concentration, 'k' is the rate constant of the reaction and 't' is the time. The rate constant 'k' can be calculated using the half-life formula for first-order reactions: \( k = \frac{0.693}{t_{1/2}}\). Calculate the rate constants for both substances A and B.

Calculate Rate Constants for A and B

For substance A with a half-life (\(t_{1 / 2}\)) of 10 min, the rate constant (\(k_A\)) is: \( k_A = \frac{0.693}{10}\).For substance B with a half-life (\(t_{1 / 2}\)) of 20 min, the rate constant (\(k_B\)) is: \( k_B = \frac{0.693}{20}\).Calculate both rate constants.

Set the Equations Equal to Solve for Time

At the time when the concentrations are equal (\([A] = [B]\)), for substance A: \( [A] = [A_0]e^{-k_At}\).And for substance B: \( [B] = [B_0]e^{-k_Bt}\).Use the initial condition \(\left[A_{0}\right]=8\left[B_{0}\right]\) and set both equations equal to solve for 't'.

Solve the Equation for t

With \([A_0] = 8[B_0]\), the equation becomes \( 8[B_0]e^{-k_At} = [B_0]e^{-k_Bt}\).Divide both sides by \([B_0]\) and solve the resulting equation: \( 8e^{-k_At} = e^{-k_Bt}\).Take the natural logarithm of both sides to solve for 't'.

Calculate the Time t

Applying the natural logarithm, we get: \(\ln(8) - k_At = -k_Bt \).Use the calculated rate constants to find the value for 't'.

Find the Answer

Isolate 't' and substitute the values of \(k_A\) and \(k_B\) into the equation to find the time when the concentrations are the same.

Key concepts

Chemical Kinetics

Chemical kinetics is the study of the rates of chemical processes and the factors affecting them. It's crucial for understanding how reactions occur and for the design of chemical reactions in industry and research. A fundamental aspect of chemical kinetics is the reaction rate, a measure of how quickly reactants are transformed into products over time. Rates can be influenced by various factors including the concentration of reactants, temperature, and presence of a catalyst.

When dealing with reactions, it's essential to understand that reactions can follow different orders. A first-order reaction depends linearly on the concentration of one reactant. For example, if we double the concentration, the reaction rate also doubles. This is vital when solving problems involving the decomposition of substances since the math behind the behavior of these reactions is quite specific and directly related to their order.

Half-Life of Reaction

The half-life of a reaction, often represented by the symbol \( t_{1/2} \), is the time it takes for the concentration of a reactant to decrease by half its initial value. This concept is particularly insightful for first-order reactions because the half-life remains constant regardless of the concentration.

Understanding the half-life offers a convenient way to perceive how quickly a reactant is depleted over time. In educational contexts, such as the exercise provided, students can use half-life to determine rate constants and predict the behavior of reactants over time. Knowing the half-lives of two substances decomposing simultaneously enables us to compare their rates of decomposition and eventually find the point in time where their concentrations equalize, which is exactly the target of the given exercise.

Rate Constants

Rate constants are a critical component of the kinetics equation for a chemical reaction. Represented by the letter \(k\), the rate constant relates the concentration of reactants to the rate of the reaction. It is determined experimentally and can provide insight into the speed of a reaction at a given temperature.

For first-order reactions, the relationship between the rate constant and the half-life is given by the formula \( k = \frac{0.693}{t_{1/2}} \). This formula allows us to convert the half-life information into a rate constant, which is necessary for calculating concentration changes over time. In the exercise described, calculating the rate constants for substances A and B is a pivotal step since it sets the foundation for determining when their concentrations will match.

Natural Logarithm in Kinetics

The natural logarithm plays a significant role in the mathematics of first-order reaction kinetics. Exponential decay in a first-order reaction is described by the equation \( [A] = [A_0]e^{-kt} \), where the natural exponential function \( e \) describes how concentration changes over time.

To isolate variables for solving kinetics problems, we often take the natural logarithm of both sides of the equation; for instance, converting \( 8e^{-k_At} = e^{-k_Bt} \) into a linear form using natural logarithms makes it easier to solve for time \(t\). This is because the natural logarithm of the exponential function \( e \) simplifies the equation, allowing us to apply algebra to find unknowns. In the context of our exercise, the natural logarithm is key to finding the time at which substances A and B have the same concentration.