Problem 29 - Chemistry The Central Science

Question

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} .\) 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 \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?\) (c) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M} ?\) (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to \(0.0120 \mathrm{M} ?\)

Step-by-Step Solution

Answer

(a) The rate law for the reaction is: Rate = k[N2O5] (b) The rate of reaction when [N2O5] = 0.0240 M is \(1.156 \times 10^{-4} \mathrm{M/s}\). (c) When the concentration of N2O5 is doubled to 0.0480 M, the rate also doubles to \(2.3136 \times 10^{-4} \mathrm{M/s}\). (d) When the concentration of N2O5 is halved to 0.0120 M, the rate halves to \(5.784 \times 10^{-5} \mathrm{M/s}\).

Step 1: Write the rate law for the reaction

Since we know that the reaction is first-order in N2O5, we can write the rate law as follows: Rate = k[N2O5]

Step 2: Calculate the rate of the reaction when [N2O5] = 0.0240 M

We are given the rate constant, k, as \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). Using the rate law from Step 1 and the given concentration of [N2O5] = 0.0240 M, we can calculate the rate: Rate = (4.82 x 10^-3 s^-1)(0.0240 M) = \(1.156 \times 10^{-4} \mathrm{M/s}\)

Step 3: Determine the effect of doubling the concentration of N2O5 on the rate

When the concentration of N2O5 is doubled to 0.0480 M, we can use the same rate law to calculate the new rate: New Rate = (4.82 x 10^-3 s^-1)(0.0480 M) = \(2.3136 \times 10^{-4} \mathrm{M/s}\) Notice that the new rate is exactly double the original rate (Step 2). This shows that when the concentration of a first-order reactant is doubled, the rate of the reaction also doubles.

Step 4: Determine the effect of halving the concentration of N2O5 on the rate

When the concentration of N2O5 is halved to 0.0120 M, we can use the same rate law to calculate the new rate: New Rate = (4.82 x 10^-3 s^-1)(0.0120 M) = \(5.784 \times 10^{-5} \mathrm{M/s}\) Notice that the new rate is exactly half the original rate (Step 2). This shows that when the concentration of a first-order reactant is halved, the rate of the reaction also halves.

Key Concepts

Reaction Rate

The rate of a chemical reaction tells us how quickly a reaction occurs. It is like a speedometer for reactions, giving us an idea of how fast reactants are being used up or products are being made. For example, in the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \), the reaction rate helps determine how fast \( \mathrm{N}_2\mathrm{O}_5 \) turns into \( \mathrm{NO}_2 \) and \( \mathrm{O}_2 \).
The reaction rate is usually expressed in terms of concentration change over time, such as moles per liter per second \( (\mathrm{M/s}) \). This gives us a numeric value that represents the rate at which the reactants transform. Understanding this helps predict and control chemical reactions, especially those important in industries or labs.

Rate Constant

The rate constant \( (k) \) is a specific value that connects the reaction rate to the concentrations of reactants in a rate law equation. It is unique for each reaction at a given temperature. For the decomposition of \(\mathrm{N}_2\mathrm{O}_5\), the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\).
- This constant remains consistent so long as the temperature does not change.- A higher \(k\) value typically indicates a faster reaction at a given concentration of reactants.- In our reaction, \(k\) helps us predict the rate by simply multiplying it by the concentration of \(\mathrm{N}_2\mathrm{O}_5\). Knowing \(k\) makes it straightforward to calculate the reaction rate using the rate law.

First Order Reaction

First order reactions have a rate that is directly proportional to the concentration of one reactant. In simpler terms, if you double the amount of reactant, you double the rate of reaction.
For our example, the reaction rate law is given by \( \text{Rate} = k[\mathrm{N}_2\mathrm{O}_5] \). This indicates a first-order relationship with the concentration of \( \mathrm{N}_2\mathrm{O}_5 \). - If \([\mathrm{N}_2\mathrm{O}_5] = 0.0240\mathrm{M}\), what happens to the rate when you increase it to \(0.0480\mathrm{M}\)? It doubles!- If \([\mathrm{N}_2\mathrm{O}_5] = 0.0240\mathrm{M}\), and you decrease it to \(0.0120\mathrm{M}\), you cut the rate in half.This relationship helps us easily predict how changes in concentration affect how fast the reaction goes.

Concentration Effect

Concentration changes can significantly impact the speed of a reaction, especially in first-order reactions. Let's see why.
By altering the concentration of \( \mathrm{N}_2\mathrm{O}_5 \), you are essentially changing how often reactant molecules collide, thus affecting the reaction rate.

• Doubling \([\mathrm{N}_2\mathrm{O}_5]\) from \(0.0240\mathrm{M}\) to \(0.0480\mathrm{M}\), means doubling the opportunities for \(\mathrm{N}_2\mathrm{O}_5\) molecules to react.
• Halving \([\mathrm{N}_2\mathrm{O}_5]\) means fewer molecules are available to react, slowing down the process to half its original rate.

This illustrates the principle that in first-order reactions, changes in concentration have a direct effect on how fast the reaction occurs. This understanding is crucial in fields such as chemistry and engineering, where control over reaction rates is often needed.