Question

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M}\) ?

Short answer

(a) The rate law for the reaction is: Rate = k [\(\mathrm{N}_{2}\mathrm{O}_{5}\)]. (b) For a concentration of 0.0240 M, the rate of reaction is \(1.16 \times 10^{-4} \mathrm{M} \cdot \mathrm{s}^{-1}\). (c) When the concentration of N2O5 is doubled to 0.0480 M, the rate of reaction doubles to \(2.31 \times 10^{-4} \mathrm{M} \cdot \mathrm{s}^{-1}\).

Step-by-step solution

Identify given information

We are given that the decomposition of N2O5 follows this balanced chemical equation: \(2 \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2} + \mathrm{O}_{2}\) The rate law for the reaction is first order in N2O5. At a temperature of \(64^{\circ} \mathrm{C}\), the rate constant (k) is \(4.82 \times 10^{-3} \mathrm{s}^{-1}\).

Write the rate law for the reaction

The rate law for a reaction can be written as: Rate = k [A]^m [B]^n ... As the rate law is first order in N2O5, the rate of reaction can be written as: Rate = k [\(\mathrm{N}_{2}\mathrm{O}_{5}\)]^1

Calculate the rate of reaction for a given concentration

For part (b), the concentration of N2O5 is given as 0.0240 M. We can now calculate the rate of reaction by plugging in the rate constant and concentration into the rate law formula: Rate = \((4.82 \times 10^{-3} \mathrm{s}^{-1})(0.0240 \mathrm{M})\) Rate = \(1.16 \times 10^{-4} \mathrm{M} \cdot \mathrm{s}^{-1}\)

Calculate the rate of reaction when the concentration is doubled

For part (c), the concentration of N2O5 is given as 0.0480 M (double the initial concentration). We can calculate the new rate of reaction by plugging in the new concentration into the rate law formula: Rate = \((4.82 \times 10^{-3} \mathrm{s}^{-1})(0.0480 \mathrm{M})\) Rate = \(2.31 \times 10^{-4} \mathrm{M} \cdot \mathrm{s}^{-1}\) Notice that the rate of reaction has doubled when the concentration of N2O5 has doubled. This is expected in a first-order reaction.

Key concepts

Rate Law

The rate law is an essential concept in chemical kinetics that describes the relationship between the concentration of reactants and the rate of a chemical reaction. In simpler terms, it tells us how quickly a reaction will occur depending on how much of each reactant is present.

Understanding the rate law begins with identifying which reactants influence the rate and how each reactant's concentration affects it. It is often expressed in the form:

• Rate = k[A]^m[B]^n...

Here, "k" is the rate constant, "[A]" and "[B]" are the concentrations of reactants, and "m" and "n" are the exponents indicating the order of the reaction with respect to each reactant. The sum of these exponents gives the overall order of the reaction.

In a first-order reaction, like the decomposition of \(\mathrm{N}_{2}\mathrm{O}_{5}\), the rate law would be simplified to only depend on one reactant:
Rate = k[\(\mathrm{N}_{2}\mathrm{O}_{5}\)]^1. This indicates that the reaction rate is directly proportional to the concentration of \(\mathrm{N}_{2}\mathrm{O}_{5}\).

First Order Reaction

A first-order reaction is a type of chemical reaction where the rate depends linearly on the concentration of a single reactant. This means that the rate increases directly as the concentration of that reactant increases.

For our example, the decomposition of \(\mathrm{N}_{2}\mathrm{O}_{5}\), is given as a first-order reaction, which can be represented as:

• Rate = k[\(\mathrm{N}_{2}\mathrm{O}_{5}\)]

This implies a few key characteristics:

• If the concentration of \(\mathrm{N}_{2}\mathrm{O}_{5}\) doubles, the reaction rate also doubles.
• It's straightforward to calculate the rate for different concentrations.
• The half-life of a first order reaction remains constant regardless of the concentration.

These traits make first-order reactions simpler to understand and predict compared to higher-order reactions.

In examining the original exercise, when the concentration of \(\mathrm{N}_{2}\mathrm{O}_{5}\) increased from 0.0240 M to 0.0480 M, the calculated rate also doubled, confirming the first-order nature of the reaction.

Rate Constant

The rate constant, represented as "k," is a crucial element in the study of chemical kinetics. It is a proportionality constant that links the concentration of reactants to the rate of a reaction.

For each reaction, the value of "k" is determined by the specific conditions, such as temperature. In our given example, the rate constant for the decomposition of \(\mathrm{N}_{2}\mathrm{O}_{5}\) at 64°C is \(4.82 \times 10^{-3}\ \mathrm{s}^{-1}\). This value helps quantify how fast the reaction proceeds under the stated conditions.

The units of "k" vary depending on the reaction order:

• For a first-order reaction, as in this case, the units are in s^{-1}.
• For second-order reactions, the units are in M^{-1}s^{-1}.

The value of the rate constant provides insight not only into the speed of reaction but also allows scientists to predict the reaction behavior over time.

Knowing "k" enables calculations of reaction rates at various reactant concentrations, giving us the ability to predict how long a reaction will take or how it will respond to changes in conditions.