Problem 44 - Chemistry: The Central Science in SI Units, Global Edition

Question

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g), \quad\) at \(\quad 70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L} .(\mathbf{a})\) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min} ?\) (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010 \mathrm{~mol}\) ? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

(a) There will be \(0.0190\) moles of N2O5 remaining after \(5\) minutes. (b) It will take approximately \(7.42\) minutes for the quantity of N2O5 to drop to \(0.010 \,\text{mol}\). (c) The half-life of N2O5 at \(70^{\circ} \mathrm{C}\) is approximately \(1.7\) minutes.

Step 1: (a) Moles of N2O5 remaining after 5 minutes

We are given the initial concentration and the rate constant for the decomposition of N2O5. We will use the first-order rate law equation to calculate the remaining amount of N2O5 after 5 minutes: \[a_t = a_0e^{-kt}\] where: - \(a_t\) is the amount of N2O5 at time t - \(a_0\) is the initial amount of N2O5 - k is the first-order rate constant (6.82 x 10^-3 s^-1) - t is the time (5 minutes = 300 seconds) Plugging the values into the equation, we have: \[a_t = (0.0250) e^{-(6.82 \times 10^{-3})(300)}\] Calculating the result, we get: \[a_t = 0.0190 \,\text{mol}\] Hence, there will be 0.0190 moles of N2O5 remaining after 5 minutes.

Step 2: (b) Time to drop the quantity of N2O5 to 0.010 mol

We will use the same first-order rate law equation to find the time it takes for the amount of N2O5 to reach 0.010 mol. Rearranging the equation to solve for time: \[t = -\frac{\ln(\frac{a_t}{a_0})}{k}\] Plugging the values into the equation, we have: \[t = -\frac{\ln(\frac{0.010}{0.0250})}{6.82 \times 10^{-3}}\] Calculating the result, we get: \[t = 445 \,\text{seconds}\] Converting the time to minutes, we have: \[\text{Time} = \frac{445}{60} \, \text{min}\] Hence, it will take approximately 7.42 minutes for the quantity of N2O5 to drop to 0.010 mol.

Step 3: (c) Half-life of N2O5 at 70°C

The half-life of a first-order reaction can be determined using the following equation: \[t_{1/2} = \frac{0.693}{k}\] We already have the rate constant, so plugging the value into the equation, we get: \[t_{1/2} = \frac{0.693}{6.82 \times 10^{-3}}\] Calculating the result, we get: \[t_{1/2} = 102 \, \text{seconds}\] Converting this to minutes, we have: \[\text{Half-Life} = \frac{102}{60} \, \text{min}\] Hence, the half-life of N2O5 at 70°C is approximately 1.7 minutes.

Key Concepts

Understanding the Rate Constant

The rate constant is a crucial component in understanding reaction kinetics, especially for first-order reactions. It is a constant that signifies the speed at which a reaction occurs. In first-order kinetics, the rate of reaction is directly proportional to the concentration of a single reactant. For the decomposition of \( \mathrm{N}_2\mathrm{O}_5\), the given rate constant is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). A high rate constant means the reaction proceeds faster, while a lower rate constant indicates a slower process.

The units of the rate constant for a first-order reaction are always \( \mathrm{s}^{-1} \). This implies that the reaction rate is measured in terms of how it changes per second. Understanding the rate constant helps you predict how quickly reactants are converted into products under specific conditions, such as temperature, which in this case is 70°C.

So when looking at reactions, always consider the rate constant to assess how fast a reaction might occur and how this influences the reactant's concentration over time.

What is a Decomposition Reaction?

A decomposition reaction is a type of chemical reaction in which one compound breaks down into two or more simpler substances. In the given problem, we are dealing with the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \), which decomposes into nitrogen dioxide \( \mathrm{NO}_2 \) and oxygen \( \mathrm{O}_2 \).

This specific decomposition is represented by the equation:

• \(2 \mathrm{N}_2\mathrm{O}_5(g) \longrightarrow 4 \mathrm{NO}_2(g) + \mathrm{O}_2(g)\)

The process occurs in the gaseous state and is crucial for understanding how complex molecules can break down into simpler molecules or atoms.

Decomposition reactions can be influenced by various factors, such as temperature or the presence of a catalyst, which can affect the rate at which the breakdown occurs. Understanding how and why substances decompose is fundamental in predicting the behavior of chemical compounds under different environmental conditions.

Calculating the Half-Life in First-Order Reactions

The half-life of a reaction is the time it takes for half of the reactant to be consumed in a chemical reaction. For first-order reactions, the half-life is particularly straightforward because it is constant and does not depend on the initial concentration of the reactant.

To calculate the half-life of \( \mathrm{N}_2\mathrm{O}_5 \) at 70°C, we use the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Where \( k \) is the rate constant. Substituting the given rate constant \( 6.82 \times 10^{-3} \mathrm{~s}^{-1} \) into the equation gives:
\[ t_{1/2} = \frac{0.693}{6.82 \times 10^{-3}} = 102 \text{ seconds} \]

This result can be converted to minutes, resulting in approximately 1.7 minutes. Understanding the half-life helps predict how long a reactant will last in the system, which is vital for both practical applications and theoretical studies in chemistry.