Question

The second order reaction, \(2 \mathrm{Mn}(\mathrm{CO})_{5} \rightarrow \mathrm{Mn}_{2}(\mathrm{CO})_{10}\) has a rate constant equal to \(3.0 \times 10^{9} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). If the initial concentration of \(\mathrm{Mn}(\mathrm{CO})_{5}\) is \(1.0 \times 10^{-5} \mathrm{M}\), how long will it take for \(90 \%\) of the reactant to disappear? a. \(3.0 \times 10^{-4} \mathrm{~s}\) b. \(3.9 \times 10^{-6} \mathrm{~s}\) c. \(3.0 \times 10^{4} \mathrm{~s}\) d. \(4.3 \times 10^{-3} \mathrm{~s}\)

Short answer

Option (a): \(3.0 \times 10^{-4}\) seconds.

Step-by-step solution

Understand the Reaction Order

The reaction is given as a second-order reaction. The general form of a second-order reaction, for two reactants A and B with the same reactant involved, is:\[ 2A \rightarrow \text{products} \] where the rate of reaction \( r \) is expressed as \( r = k[A]^2 \). Here, \( [A] \) is the concentration of \( \mathrm{Mn(CO)}_5 \) and \( k \) is the rate constant.

Set Up the Integrated Rate Equation

For a second-order reaction, the integrated rate equation is:\[ \frac{1}{[A]} = kt + \frac{1}{[A]_0} \] where \( [A]_0 \) is the initial concentration and \( t \) is the time required.

Determine Final Concentration

We know that 90% of the reactant disappears, meaning 10% is left. Thus, the final concentration \([A]\) is:\[ [A] = 0.10 \times [A]_0 = 0.10 \times 1.0 \times 10^{-5} \text{ M} = 1.0 \times 10^{-6} \text{ M} \]

Substitute the Values into the Rate Equation

Substitute \( k = 3.0 \times 10^9 \text{ M}^{-1} \text{ s}^{-1} \), \([A]_0 = 1.0 \times 10^{-5} \text{ M}\), and \([A] = 1.0 \times 10^{-6} \text{ M}\) into the integrated rate equation:\[ \frac{1}{1.0 \times 10^{-6}} = 3.0 \times 10^9 \cdot t + \frac{1}{1.0 \times 10^{-5}} \] Simplify:\[ 10^6 = 3.0 \times 10^9 \cdot t + 10^5 \]

Solve for Time (t)

Rearrange the equation from Step 4 to solve for \( t \):\[ 10^6 - 10^5 = 3.0 \times 10^9 \cdot t \]\[ 9.0 \times 10^5 = 3.0 \times 10^9 \cdot t \]\[ t = \frac{9.0 \times 10^5}{3.0 \times 10^9} \]\[ t = 3.0 \times 10^{-4} \text{ s} \]

Verify and Interpret the Answer

Compare the calculated time \( t = 3.0 \times 10^{-4} \text{ s} \) with the given options. The closest match is option (a). The reaction takes approximately \(3.0 \times 10^{-4}\) seconds for 90% of \(\mathrm{Mn(CO)}_5\) to disappear.

Key concepts

Integrated Rate Equation

In the study of chemical kinetics, the integrated rate equation is a crucial tool to understand how the concentration of reactants changes over time for a reaction. For second-order reactions, where the rate depends on the square of the concentration of the reactant, the integrated rate equation is expressed as:\[\frac{1}{[A]} = kt + \frac{1}{[A]_0}\]Here, \([A]\) is the concentration of the reactant at time \(t\), \([A]_0\) is the initial concentration of the reactant, and \(k\) is the rate constant of the reaction. Integrated rate equations allow chemists to calculate the time it takes for a certain proportion of a reactant to be consumed, which is particularly useful in predicting the lifespan of reactants in various chemical processes.
By knowing this equation, we can accurately determine the amount of time necessary for a desired conversion, which in turn helps in scaling up chemical reactions for industrial applications.

Reaction Kinetics

Reaction kinetics is the branch of chemistry that studies the rates of chemical processes. It provides insights into how different conditions, such as concentration and temperature, affect the speed of a reaction.
In the context of the exercise, we are particularly interested in second-order reactions, where the rate of reaction is proportional to the square of the concentration of one reactant. The kinetic expression for a second-order reaction involving the same reactant is depicted as:

• Rate Law: \(r = k[A]^2\)

Where \(r\) represents the rate of the reaction, and \([A]\) is the concentration of the reactant.
Understanding reaction kinetics not only helps in determining how quickly reactions occur but also guides chemists in optimizing conditions to maximize yield and efficiency in chemical manufacturing processes. This is particularly critical in sectors such as pharmaceuticals and materials science, where reaction rates directly affect production and product quality.

Rate Constant

The rate constant \(k\) is a fundamental parameter that determines the rate of a reaction in relation to the concentration of its reactants. For a second-order reaction, the unit of the rate constant is \(\text{M}^{-1} \text{s}^{-1}\). The rate constant is independent of the concentration of reactants but is affected by other factors such as temperature and the presence of catalysts.
For second-order reactions, a larger rate constant indicates a faster reaction at a given concentration.
Understanding the role of the rate constant helps chemists forecast how a reaction will proceed under various conditions and enables them to tailor processes to achieve desired reaction rates. Calculating and interpreting the rate constant is essential for designing efficient and effective chemical processes.
The exercise specifies a rate constant of \(3.0 \times 10^9\ \text{M}^{-1} \text{s}^{-1}\) for the given reaction, which contributes to understanding how quickly the concentration of \(\text{Mn(CO)}_5\) will reduce as the reaction proceeds.

Chemical Concentration Calculations

Calculating the concentration of reactants over time is a key part of understanding reaction dynamics and kinetics. It allows chemists to predict how much reactant remains at any point in time during a reaction.
For the second-order reaction in the exercise, we start with an initial concentration \([A]_0\) and use the integrated rate equation to find out how much remains at a given time. The exercise requires determining the final concentration after a significant portion of the reactant has been used up. Specifically, it considers how long it will take for 90% of \(\text{Mn(CO)}_5\) to transform into products:

• Initial Concentration: \(1.0 \times 10^{-5}\ \text{M}\)
• Final Concentration: 10% of the initial, which calculates to \(1.0 \times 10^{-6}\ \text{M}\)

Using these values alongside the rate equation, we can solve for time \(t\) required for this transformation.
Calculating the changes in chemical concentration during reactions helps in not only understanding reaction rates but also in determining the efficiency and feasibility of chemical processes on an industrial scale.