Problem 135 - Chemistry: An Atoms-Focused Approach

Question

The following is an important reaction in the formation of photochemical smog: $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ The reaction is first order in \(\mathrm{NO}\) and \(\mathrm{O}_{3} .\) The rate constant of the reaction is \(80 M^{-1} \mathrm{s}^{-1}\) at \(25^{\circ} \mathrm{C}\) and \(3000 M^{-1} \mathrm{s}^{-1}\) at \(75^{\circ} \mathrm{C}\) a. If this reaction were to occur in a single step, would the rate law be consistent with the observed order of the reaction for \(\mathrm{NO}\) and \(\mathrm{O}_{3} ?\) b. What is the value of the activation energy of the reaction? c. What is the rate of the reaction at \(25^{\circ} \mathrm{C}\) when \([\mathrm{NO}]=3 \times 10^{-6} \mathrm{Mand}\left[\mathrm{O}_{3}\right]=5 \times 10^{-9} \mathrm{MP}\) d. Predict the values of the rate constant at \(10^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

a. The rate law is consistent with the observed reaction orders, as it is first order with respect to both NO and O3. b. The activation energy of the reaction is approximately 76946 J/mol. c. The rate of the reaction at 25°C is approximately 1.2×10^(-12) M s^(-1). d. The rate constant at 10°C is approximately 24.45 M^(-1)s^(-1) and at 35°C, it is approximately 135.48 M^(-1)s^(-1).

Step 1: Rate Law and Reaction Order

The rate law is given by: $$\text{Rate} = k[\mathrm{NO}]^{m}[\mathrm{O}_{3}]^{n}$$ where \(k\) is the rate constant, \(m\) and \(n\) are the reaction order with respect to \(\mathrm{NO}\) and \(\mathrm{O}_{3}\), respectively. According to the question, the reaction is first order with respect to both \(\mathrm{NO}\) and \(\mathrm{O}_{3}\). That means \(m = 1\) and \(n = 1\). The rate law is indeed consistent with the observed reaction orders. #b. Calculating Activation Energy# We will use the Arrhenius equation to find the activation energy (\(E_a\)) of the reaction. The Arrhenius equation is: $$k = Ae^{-\frac{E_a}{RT}}$$ Where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in kelvin. We have two temperature points, and the rate constants for those temperatures are given. By taking the ratio of the rate constant at the two temperatures, we can obtain the activation energy. First, we isolate \(E_a\): $$\frac{k_2}{k_1} = e^{\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \Rightarrow E_{a} = R\left(\frac{T_1T_2}{T_2 - T_1}\right)\ln{\left(\frac{k_2}{k_1}\right)}$$

Step 2: Values and Calculation

Now, we substitute the given values: $$R = 8.314\,Js^{-1} \mathrm{mol}^{-1}\,K^{-1}, \quad T_1 = 25^{\circ}C = 298\,K, \quad T_2 = 75^{\circ}C = 348\,K,$$ $$k_1 = 80\,M^{-1}s^{-1}, \quad k_2 = 3000\,M^{-1}s^{-1}$$ Substituting the values, we get: $$E_a = 8.314\left(\frac{298\times348}{348-298}\right)\ln{\left(\frac{3000}{80}\right)} \approx 76946\, J\mathrm{mol}^{-1}$$ The activation energy of the reaction is approximately \(76946\, J\mathrm{mol}^{-1}\). #c. Calculating the reaction rate at given concentrations# We can calculate the reaction rate using the rate law and given concentrations of \(\mathrm{NO}\) and \(\mathrm{O}_{3}\): $$\text{Rate} = k[\mathrm{NO}][\mathrm{O}_{3}]$$

Step 3: Values and Calculation

Now, substituting the given values: $$k = 80\,M^{-1}s^{-1}, \quad [\mathrm{NO}] = 3\times10^{-6}\,M, \quad [\mathrm{O}_{3}] = 5\times10^{-9}\,M$$ $$\text{Rate} = 80\times(3\times10^{-6})(5\times10^{-9}) \approx 1.2\times10^{-12}\,M \,s^{-1}$$ The rate of the reaction at \(25^{\circ}C\) is approximately \(1.2\times10^{-12}\,M \,s^{-1}\). #d. Predicting rate constants at given temperatures# Again, we will use the Arrhenius equation to predict the rate constants at \(10^{\circ}C\) and \(35^{\circ}C\). $$k = Ae^{-\frac{E_a}{RT}}$$ We will use the \(k\) and \(E_a\) values from the \(25^{\circ}C\) case as the reference values. We will plug in the values of \(T\) for both cases and find the corresponding \(k\) values.

Step 4: 10°C Case

For \(10^{\circ}C\) or \(283\,K\): $$k_{10} = 80\,e^{\frac{76946(1/298 - 1/283)}{8.314}} \approx 24.45\,M^{-1}s^{-1}$$ The rate constant at \(10^{\circ}C\) is approximately \(24.45\,M^{-1}s^{-1}\).

Step 5: 35°C Case

For \(35^{\circ}C\) or \(308\,K\): $$k_{35} = 80\,e^{\frac{76946(1/298 - 1/308)}{8.314}} \approx 135.48\,M^{-1}s^{-1}$$ The rate constant at \(35^{\circ}C\) is approximately \(135.48\,M^{-1}s^{-1}\).

Key Concepts

Rate Law

Rate law is a mathematical expression that shows the relationship between the concentration of reactants and the reaction rate. For the reaction: \(\text{NO}(g) + \text{O}_{3}(g) \rightarrow \text{NO}_{2}(g) + \text{O}_{2}(g)\), the rate law is given as: ```\text{Rate} = k[\text{NO}]^{m}[\text{O}_{3}]^{n}```- **\(k\)** is the rate constant.- **\(m\) and \(n\)** are the orders of the reaction with respect to \(\text{NO}\) and \(\text{O}_{3}\).According to the provided information, this reaction is first order with respect to both \(\text{NO}\) and \(\text{O}_{3}\). This implies that both \(m\) and \(n\) are 1, leading to a rate law:```\text{Rate} = k[\text{NO}][\text{O}_{3}]```Understanding rate law is crucial as it tells us how changes in concentration affect the rate of the reaction.

Activation Energy

Activation energy, denoted as \(E_a\), is the minimum amount of energy required to start a chemical reaction. It is a barrier that reactants must overcome to transform into products. The Arrhenius equation describes the dependence of the rate constant \(k\) on the temperature \(T\) and \(E_a\):```k = Ae^{-\frac{E_a}{RT}}```Coordinates:- **\(A\)** is the pre-exponential factor.- **\(R\)** is the universal gas constant.When calculating \(E_a\), we use the rate constants at different temperatures. With the given rate constants and temperatures in this example, we used:```mathE_a = R\left(\frac{T_1T_2}{T_2-T_1}\right)\ln{\left(\frac{k_2}{k_1}\right)}```Substituting values, we find \(E_a\) to be approximately \(76946 \, J\,\text{mol}^{-1}\). Activation energy helps explain why reactions are slow at low temperatures and faster at high temperatures.

Arrhenius Equation

The Arrhenius equation provides a formula to calculate the rate constant \(k\) of a reaction as a function of temperature \(T\) and activation energy \(E_a\). The equation is written as:```k = Ae^{-\frac{E_a}{RT}}```- **\(k\)** increases with an increase in temperature or a decrease in \(E_a\).- Taking the natural logarithm of both sides transforms it into a linear equation:```math\ln{k} = \ln{A} - \frac{E_a}{RT}```With known values of \(k\) at two temperatures, \(E_a\) can be estimated. This relationship explains how temperature can significantly affect the reaction rate, as it exponentially increases \(k\) making the reaction proceed faster.

Reaction Rate Calculation

Calculating the reaction rate involves using the rate law expression to determine how fast reactants are converted into products. In this problem, with the reaction rate law:```\text{Rate} = k[\text{NO}][\text{O}_{3}]```The given concentrations of reactants allow the calculation of the reaction rate at a specified temperature:- **\([\text{NO}]=3 \times 10^{-6} \text{M}\)** - **\([\text{O}_{3}]=5 \times 10^{-9} \text{M}\)**- **\(k = 80 \, M^{-1}s^{-1}\)**Substituting these into the rate law yields a rate of approximately ```1.2 \times 10^{-12} \, M \,s^{-1}```This shows that, despite being slow under these conditions, photochemical reactions still contribute significantly to atmospheric changes over time.