Question

Consider the hypothetical reaction \(\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(g) \longrightarrow\) \(2 \mathrm{AB}(g)\), where the rate law is: $$ -\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{~A}_{2}\right]\left[\mathrm{B}_{2}\right] $$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is \(2.45 \times 10^{-4} \mathrm{~L} / \mathrm{mol}\). \(\mathrm{s}\), and at \(508^{\circ} \mathrm{C}\) the rate constant is \(0.891 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\). What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C}\) ?

Short answer

The activation energy for the given reaction is approximately \(74.2 \mathrm{kJ/mol}\). The rate constant at \(375^{\circ} \mathrm{C}\) is approximately \(2.32 \times 10^{-3} \mathrm{~L} / \mathrm{mol}. \mathrm{s}\).

Step-by-step solution

Convert temperatures to Kelvin.

To work with the Arrhenius equation, we need to convert all temperatures to Kelvin: $$ T_1 = 302 + 273.15 = 575.15\mathrm{~K} $$ $$ T_2 = 508 + 273.15= 781.15\mathrm{~K} $$ $$ T_3 = 375 + 273.15= 648.15\mathrm{~K} $$

Calculate the activation energy (Ea).

We have two rate constants, \(k_1\) and \(k_2\), at two different temperatures, \(T_1\) and \(T_2\). We can use this information to find the activation energy as follows: First, apply the Arrhenius equation for both rate constants: $$ k_1 = Ae^{\frac{-E_a}{R T_1}} $$ $$ k_2 = Ae^{\frac{-E_a}{R T_2}} $$ Divide the first equation by the second equation to remove the pre-exponential factor: $$ \frac{k_1}{k_2} = e^{\frac{-E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})} $$ Now, solve for the activation energy, \(E_a\): $$ E_a = -R \frac{\log(\frac{k_1}{k_2})}{\frac{1}{T_1} - \frac{1}{T_2}} $$ Using \(k_1 = 2.45 \times 10^{-4} \mathrm{~L} / \mathrm{mol}.\mathrm{s}\), \(k_2 = 0.891 \mathrm{~L} / \mathrm{mol}.\mathrm{s}\), and the converted temperatures, we get: $$ E_a \approx 74172 \mathrm{J/mol} $$ The activation energy for the reaction is approximately \(74.2 \mathrm{kJ/mol}\).

Calculate the rate constant at \(375^{\circ} \mathrm{C}\).

Now we need to find the rate constant, \(k_3\), at temperature \(T_3\). To do this, we can use the Arrhenius equation and the calculated activation energy. From Step 1, $$ k_3 = Ae^{\frac{-E_a}{R T_3}} $$ We need to find the pre-exponential factor, \(A\). To do this, we can use one of the given rate constants, say, \(k_1\), and the activation energy we just found: $$ k_1 = Ae^{\frac{-E_a}{R T_1}} \Rightarrow A = k_1 e^{\frac{E_a}{R T_1}} $$ Now, plug this value of \(A\) back into the formula for \(k_3\): $$ k_3 = k_1 e^{\frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_3})} $$ Using \(k_1 = 2.45 \times 10^{-4} \mathrm{~L} / \mathrm{mol}.\mathrm{s}\), and the converted temperatures, we get: $$ k_3 \approx 2.32 \times 10^{-3} \mathrm{~L} / \mathrm{mol}. \mathrm{s} $$ The rate constant for this reaction at \(375^{\circ} \mathrm{C}\) is approximately \(2.32 \times 10^{-3} \mathrm{~L} / \mathrm{mol}. \mathrm{s}\).

Key concepts

Activation Energy

Activation energy, represented as \(E_a\), is the minimum amount of energy that reacting particles must have to undergo a chemical transformation. During a reaction, reactants must overcome an energy barrier before forming products. This energy barrier is due to the breaking of old bonds and the forming of new ones.

Let’s connect this to the hypothetical reaction \(A_2(g) + B_2(g) \longrightarrow 2AB(g)\) from the exercise. The rate at which this reaction occurs is heavily dependent on the activation energy. A higher \(E_a\) means that fewer molecules will have sufficient energy to react at a given temperature, leading to a slower reaction rate. Conversely, a lower \(E_a\) results in more molecules having the energy to react, thus increasing the reaction rate.

In solving the exercise, a crucial step was to compute the activation energy using the known rate constants at two different temperatures. The calculated activation energy helps us understand not only the kinetics of the reaction but also how the reaction rate would change with temperature variations.

Rate Constant

The rate constant \(k\) is a parameter in the rate law of a chemical reaction that indicates the speed of the reaction at a given temperature and is influenced by the activation energy. The rate law for the given reaction is \( -\frac{\Delta[A_2]}{\Delta t} = k[A_2][B_2] \). In this equation, the rate constant \(k\) is unique to each reaction and can significantly change as the temperature changes, which is why knowing how to find it at various temperatures is critical.

In the context of the exercise, we determined the value of \(k\) at a new temperature using the Arrhenius equation. The correctness of this calculation is essential, as it’s a fundamental aspect determining the speed of the reaction at that specific temperature. This impacts not only academic exercises but also real-world applications such as the design of chemical reactors and the prediction of reaction behavior under different temperature conditions.

Temperature Conversion

In chemistry, temperature plays a pivotal role in the kinetics of reactions. Temperature conversion is necessary when using the Arrhenius equation since it requires temperatures to be expressed in Kelvin rather than Celsius. The Kelvin scale is an absolute temperature scale with a null point at the total absence of thermal energy, making it ideal for scientific calculations.

To convert from Celsius to Kelvin—a key step in the exercise—the relation \(T(K) = T(\degree C) + 273.15\) is used. Without proper conversion, any subsequent calculations involving temperature, such as those to find the activation energy or the rate constant, would yield incorrect results. Temperature conversion is one of the fundamental skills in chemistry that allows for precise and accurate interpretation of thermal phenomena in reactions.