Problem 56 - General Chemistry: Principles and Modern Applications

Question

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The activation energy \(E_a\) is 98.6 kJ/mol and the rate constant at \(30^{\circ}C\), \(k_{30}\), is \(2.89\times 10^2 s^{-1}\).

Step 1: Setting up the first equation

For the first step, let’s use the Arrhenius equation in its integrated form since we know two sets of \(k\) and \(T\) values. The Arrhenius equation is: \[ k=A \cdot e^{-E_a /(RT)}\] Where \(A\) is the Arrhenius constant, \(E_a\) is the activation energy, \(R\) is the gas constant and \(T\) is the temperature. First, rewrite this as: \(ln(k_1) = ln(A) - \frac{{E_a}}{{R \cdot T_1}}\). With \(k_1 = \frac{0.693}{{t_{1/2_{1}}}}\) and \(T_1 =20^\circ C = 293.15K\)

Step 2: Setting up the second equation

Now we need to set up another equation for \(T_2\) and \(k_2\). This equation will look just like the first equation, but \(T_2 = 40^\circ C = 313.15K\) and \(k_2= \frac{0.693}{{t_{1/2_{2}}}}\). Thus, we have: \(ln(k_2) = ln(A) - \frac{{E_a}}{{R \cdot T_2}}\).

Step 3: Eliminate \(A\) from the two equations

Now, in order to get \(E_a\), we need to eliminate \(A\) from the two equations. So, subtract Eq. 2 from Eq. 1: \(ln(k_1)-ln(k_2) = \frac{{E_a}}{{R}} \left(\frac{1}{{T_2}} - \frac{1}{{T_1}}\right)\). This simplifies to: \(ln\left( \frac{{k_1}}{{k_2}} \right) = \frac{{E_a}}{{R}} \left(\frac{1}{{T_2}} - \frac{1}{{T_1}}\right)\). Plug in the known values and solve for \(E_a\).

Step 4: Calculate \(k_3\) at \(30^{\circ}C\)

To find the rate constant \(k\) at different temperatures we use the Arrhenius equation with the calculated \(E_a\). This will give: \[ k = A \cdot e^{-E_a /(RT)}\] where \(T = 30^{\circ}C = 303.15K\]. Substitute the known values of \(A\) and \(E_a\) to get \(k\).

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a fundamental formula in chemical kinetics that expresses how the rate constant of a reaction changes with temperature. This equation is formulated as \[ k = A \cdot e^{-E_a /(RT)} \]where:

• k is the rate constant, which tells us the speed of the reaction.
• A stands for the Arrhenius factor or pre-exponential factor, indicating the frequency of collisions; it is often denoted as a constant.
• e represents the base of the natural logarithms, approximately equal to 2.718.
• E_a is the activation energy, the energy barrier that reactants must overcome in order for a reaction to occur.
• R is the gas constant, which has a value of 8.314 J/(mol K).
• T is the temperature in Kelvin.

Using this equation, chemists can predict how different temperatures will impact the rate at which reactions occur. It highlights that as temperature increases, the exponential term becomes smaller, resulting in a larger rate constant, meaning faster reactions.

Activation Energy

Activation energy (\(E_a\)) is the minimum amount of energy required to initiate a chemical reaction. Think of it as an energy hill that reactants need to climb before turning into products. This energy determines the rate at which a reaction proceeds.
For our first-order reaction, the activation energy can be calculated using the rearranged form of the Arrhenius equation. By setting up two separate equations at different temperatures, we can eliminate the Arrhenius factor and solve for \(E_a\).
Start with the equations:

• \( \ln(k_1) = \ln(A) - \frac{E_a}{R \cdot T_1} \)
• \( \ln(k_2) = \ln(A) - \frac{E_a}{R \cdot T_2} \)

Subtracting these equations allows us to cancel out \(A\), yielding:\[ \ln\left( \frac{k_1}{k_2} \right) = \frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]By knowing the half-lives and temperatures, we can find the values for \(k_1\) and \(k_2\), and subsequently solve for \(E_a\). This energy determines whether a reaction will happen quickly or slowly.

Rate Constant Calculation

Calculating the rate constant, \(k\), is crucial to understanding the speed of a reaction. For our first-order reaction, the rate constant can be directly calculated from the half-life using the formula\[ k = \frac{0.693}{t_{1/2}} \]where \(t_{1/2}\) is the half-life of the reaction.
After determining \(E_a\), we can now find the rate constant at any temperature using the Arrhenius equation. For the given exercise, at the temperature \(30^{\circ}C\) (or \(303.15 K\)), we use:\[ k = A \cdot e^{-E_a /(RT)} \]Substitute in \(A\), which is given as \(2.05 \times 10^{13} \text{s}^{-1}\), along with \(E_a\) and \(T\) to solve for \(k\). This calculation helps in predicting how quickly reactants will convert to products at a specific temperature.

First-order Reaction

A first-order reaction is a type of chemical reaction where the rate depends linearly on the concentration of only one reactant. In simpler terms, doubling the concentration of that reactant will double the rate of the reaction.For reactions that follow first-order kinetics, the rate constant can be linked to the half-life, which is the time taken for half of the reactant to be consumed. A unique feature of first-order reactions is that the half-life remains constant regardless of the starting concentration.
The general formula for the rate of a first-order reaction is:\[ \text{Rate} = k [A] \]where \([A]\) is the concentration of the reactant.
In the exercise involving \(\text{N}_2\text{O}_5\) decomposition, knowing the half-lives at different temperatures helps us to determine the rate constants, which remain consistent with the properties of a first-order reaction.