Question

The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent \(\left(\mathrm{CCl}_{4}\right)\) at a certain temperature: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) $$ \begin{array}{cc} {\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{M})} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 0.92 & 0.95 \times 10^{-5} \\ 1.23 & 1.20 \times 10^{-5} \\ 1.79 & 1.93 \times 10^{-5} \\ 2.00 & 2.10 \times 10^{-5} \\ 2.21 & 2.26 \times 10^{-5} \end{array} $$ Determine graphically the rate law for the reaction and calculate the rate constant.

Short answer

The rate of reaction is first order with respect to concentration of N2O5 as a linear relationship is observed between the initial rate and the concentration of N2O5. The rate law equation for the reaction is \(rate = k[\mathrm{N}_{2} \mathrm{O}_{5}]\). The rate constant (k) can be calculated to be approximately \(k = \frac{0.95 \times 10^{-5}}{0.92} M^{-1}s^{-1}\).

Step-by-step solution

Find the relationship between concentration and reaction rate.

First, plot the given data using the concentration of N2O5 on the x-axis and the initial rate on the y-axis. If the points roughly form a straight line passing through the origin, then the reaction is first order in N2O5. By looking at the provided data, we may infer that the reaction is first order as the initial rate increases linearly with an increase in concentration.

Determine the rate law.

As the relation is linear, the form of the rate law equation we have is \[rate = k[\mathrm{N}_{2} \mathrm{O}_{5}]^n\]. From the graph, we found that reaction is first order with respect to N2O5 (n = 1). So, the rate law for the reaction becomes \[rate = k[\mathrm{N}_{2} \mathrm{O}_{5}]\].

Calculate the rate constant (k).

Now, using one of the given data points from the table, for example the first one where \[[\mathrm{N}_{2} \mathrm{O}_{5}] = 0.92 M\] and the initial rate is \(0.95 \times 10^{-5} M/s\), insert these values into the rate law equation. Solving for k gives \(k = \frac{rate}{[\mathrm{N}_{2} \mathrm{O}_{5}]} = \frac{0.95 \times 10^{-5}}{0.92} M^{-1}s^{-1} \). The value of k (rate constant) obtained should align closely for the other data points, as long as they line is relatively straight.

Key concepts

Rate Law

In reaction kinetics, the rate law helps us understand how the concentration of reactants affects the speed or rate of a reaction. It is an expression that correlates the rate of a chemical reaction to the concentration of its reactants. The equation is typically written as \[ rate = k[A]^m[B]^n \] where:

• *rate* is the reaction rate,
• *k* is the rate constant,
• *[A]* and *[B]* are concentrations of reactants,
• *m* and *n* are the orders of the reaction with respect to reactants A and B respectively.

The rate law derived in the exercise is \[ rate = k[\mathrm{N}_{2} \mathrm{O}_{5}] \]and suggests that the reaction rate is proportional to the concentration of \( \mathrm{N}_{2} \mathrm{O}_{5} \). By determining the rate law, we can predict how the concentration influences the reaction's speed at a given temperature.

Reaction Order

Reaction order tells us how the rate of a chemical reaction depends on the concentration of its reactants. It is determined by the powers to which the concentration terms are raised in the rate law. In our exercise, we observed that the reaction is first order with respect to \( \mathrm{N}_{2} \mathrm{O}_{5} \). This conclusion is drawn from the linear relationship between concentration and rate shown graphically. A first-order reaction means:

• The rate of reaction is directly proportional to the concentration of that reactant.
• As the concentration of \( \mathrm{N}_{2} \mathrm{O}_{5} \) doubles, the reaction rate also doubles, so, for small time intervals, the decrease in \( \mathrm{N}_{2} \mathrm{O}_{5} \) is consistent with the concentration remaining proportional to the decay.

This concept can further help in predicting future concentrations or times needed for the reaction to reach completion.

Rate Constant

The rate constant, denoted as \( k \), is a crucial factor in the rate law equation as it provides the speed of a reaction at a certain temperature. To calculate it, we rearrange the rate law equation from the exercise: \[ k = \frac{rate}{[\mathrm{N}_2 \mathrm{O}_5]} \]Using the provided data, we see how the rate constant remains consistent through different trials when accurately measuring concentrations and rate. A constant \( k \) implies the relationship between concentration and rate is stable under the studied conditions. Key points about the rate constant:

• Its units vary depending on the overall order of the reaction. For a first-order reaction like ours, it's \( s^{-1} \).
• It changes with temperature, meaning the speed of reaction can increase or decrease.

Understanding \( k \) helps in designing and controlling processes involving chemical reactions, ensuring efficiency or predictability in real-world applications.