Question

The rate law for the reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ at some temperature is $$ \text {Rate} =-\frac{\Delta[\text { NOBr }]}{\Delta t}=k[\mathrm{NOBr}]^{2} $$ a. If the half-life for this reaction is 2.00 s when \([\mathrm{NOBr}]_{0}=\) \(0.900 M,\) calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to 0.100\(M ?\)

Short answer

a. The rate constant 'k' for the reaction is 0.5556 M\(^{-1}\)s\(^{-1}\). b. The time required for the concentration of NOBr to decrease to 0.100 M is 14.19 seconds.

Step-by-step solution

Identify the reaction order and equations

The given rate law equation relates the rate of the reaction to the concentration of NOBr as follows: Rate = \(−\frac{\Delta[\text{NOBr}]}{\Delta t}\) = \(k[\text{NOBr}]^2\) This is a second order reaction, as the concentration of NOBr is raised to the power of 2. For a second order reaction, the equation relating half-life to the rate constant and initial concentration is: \( t_{1/2} = \frac{1}{k[\text{NOBr}_0]} \)

Calculate the value of k

We are given that the half-life of the reaction is 2.00 s and the initial concentration of NOBr is 0.900 M. Let's plug these values into the half-life equation to calculate the rate constant 'k'. \( 2.00 = \frac{1}{k(0.900)} \) Now, solve for k: \( k = \frac{1}{2.00 × 0.900} = \frac{1}{1.80} \) \( k = 0.5556\) M\(^{-1}\)s\(^{-1}\) The rate constant 'k' for this reaction is 0.5556 M\(^{-1}\)s\(^{-1}\).

Determine the time for NOBr concentration to decrease to 0.100 M

Now that we have the rate constant, we can use the second-order integrated rate law to find the time required for the concentration of NOBr to decrease to 0.100 M. The second-order integrated rate law equation is: \( \frac{1}{[\text{NOBr}]} - \frac{1}{[\text{NOBr}_0]} = kt \) Let's plug in the known values: \( \frac{1}{0.100} - \frac{1}{0.900} = 0.5556 \times t \) Solve for 't': \( 9.00 - 1.11 = 0.5556t \) \( t = \frac{7.89}{0.5556} = 14.19 \, \text{s} \) The time required for the concentration of NOBr to decrease to 0.100 M is 14.19 seconds.

Key concepts

Second order reactions

In the study of chemical kinetics, reactions are often classified by their reaction order. This refers to the power to which the concentration of a reactant is raised in the rate law expression. A second order reaction means that the rate of reaction is proportional to the square of the concentration of one reactant. For example, the reaction given in the exercise: \[ 2 \text{NOBr}(g) \rightarrow 2 \text{NO}(g) + \text{Br}_2(g) \] shows a rate law of: \[ \text{Rate} = -\frac{\Delta[\text{NOBr}]}{\Delta t} = k[\text{NOBr}]^2 \] This indicates it is a second order reaction with respect to NOBr, meaning the rate changes with the square of its concentration.Second order reactions are characterized by specific equations to calculate various parameters, like the half-life. In this case, the half-life (\( t_{1/2} \)) for second order reactions depends on the initial concentration \( [\text{NOBr}_0] \) and the rate constant \( k \):\[ t_{1/2} = \frac{1}{k[\text{NOBr}_0]} \]This equation helps determine how quickly a concentration decreases by half in second order reactions, which differs from first order reactions where half-life is constant irrespective of concentration.

Rate constant (k)

The rate constant, symbolized by \( k \), is a crucial component in rate laws and reflects how fast a reaction occurs. Unlike the reaction order, which is about how reactant concentrations affect the reaction rate, \( k \) itself encapsulates conditions like temperature and catalysts that influence this rate.For second order reactions, the rate constant has units of \( \text{M}^{-1}\text{s}^{-1} \) and can be determined from experimental data, as shown in the exercise:

• Half-life \( t_{1/2} \) was 2.00 s,
• Initial concentration \([\text{NOBr}_0] = 0.900 \text{ M} \).

By plugging these into the half-life equation for second order reactions \( \left( t_{1/2} = \frac{1}{k[\text{NOBr}_0]} \right) \), we solve for \( k \):\[ 2.00 = \frac{1}{k \times 0.900} \]\[ k = \frac{1}{1.80} = 0.5556 \text{ M}^{-1}\text{s}^{-1} \]This kind of calculation helps not just in understanding the speed of reactions but also in predicting how reactions might progress over time under specific conditions.

Integrated rate laws

Integrated rate laws are mathematical expressions that allow chemists to determine the concentration of reactants over time during a reaction. They are derived from the differential rate laws and vary with reaction order.For second order reactions, the integrated rate law can be written as:\[ \frac{1}{[\text{A}]} - \frac{1}{[\text{A}_0]} = kt \] where:

• \([\text{A}]\) is the concentration of reactant at time \( t \),
• \([\text{A}_0] \) is the initial concentration,
• \( k \) is the rate constant,
• \( t \) is the time elapsed.

In the solution provided in the exercise, this integrated rate law was used to determine how long it takes for \([\text{NOBr}]\) to decrease from 0.900 M to 0.100 M:\[ \frac{1}{0.100} - \frac{1}{0.900} = 0.5556 \times t \]The focus here is on integrating the concentrations over time, allowing for the calculation of the reaction’s progress and aiding in experimental design and predicted outcomes.