Question

How much faster would a reaction proceed at \(46^{\circ} \mathrm{C}\) than at \(28^{\circ} \mathrm{C}\) if the activation energy of the reaction is \(121 \mathrm{~kJ} / \mathrm{mol} ?\)

Short answer

Answer: The reaction would proceed approximately 2.47 times faster at 46°C than at 28°C.

Step-by-step solution

Understand the Arrhenius equation

The Arrhenius equation describes the temperature dependence of reaction rates, and it is given as: \(k(T) = Ae^{-\frac{E_a}{RT}}\) Where \(k(T)\) is the rate constant at temperature \(T\), \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant (8.314 J/mol·K) and \(T\) is the temperature in Kelvin.

Convert temperatures to Kelvin

To use the Arrhenius equation, we need to convert the given temperatures from Celsius to Kelvin. The formula for doing this is: \(T_{Kelvin} = T_{Celsius} + 273.15\) \(T_1 = 28^{\circ}\mathrm{C} = 28 + 273.15 = 301.15 \mathrm{K}\) \(T_2 = 46^{\circ}\mathrm{C} = 46 + 273.15 = 319.15 \mathrm{K}\)

Calculate the rate constants at both temperatures

Using the Arrhenius equation, we can calculate the rate constants at \(T_1\) and \(T_2\). However, since we are looking for the ratio of how much faster the reaction proceeds at \(T_2\) compared to \(T_1\), we can simplify the calculation by cancelling out the pre-exponential factor \(A\). Thus, for the ratio of the rate constants, we have: \(\frac{k(T_2)}{k(T_1)} = \frac{e^{-\frac{E_a}{R\cdot T_2}}}{e^{-\frac{E_a}{R\cdot T_1}}}\)

Substitute values and simplify

Substitute the given values for activation energy \(E_a\), gas constant \(R\), and temperatures \(T_1\) and \(T_2\) in the equation, and simplify it. \(\frac{k(T_2)}{k(T_1)} = \frac{e^{-\frac{121\times 10^3 \mathrm{J}/\mathrm{mol}}{8.314\ \mathrm{J}/\mathrm{mol} \cdot \mathrm{K}\cdot 319.15\ \mathrm{K}}}}{e^{-\frac{121\times 10^3 \mathrm{J}/\mathrm{mol}}{8.314\ \mathrm{J}/\mathrm{mol} \cdot \mathrm{K}\cdot 301.15\ \mathrm{K}}}}\) \(\frac{k(T_2)}{k(T_1)} = e^{\left(\frac{-121\times 10^3}{8.314 \cdot 301.15} + \frac{121\times 10^3}{8.314 \cdot 319.15}\right)}\)

Solve the expression

Solve the above expression to find the ratio of rate constants at \(T_2\) and \(T_1\). \(\frac{k(T_2)}{k(T_1)} \approx 2.47\) Thus, the reaction would proceed approximately 2.47 times faster at \(46^{\circ}\mathrm{C}\) than at \(28^{\circ}\mathrm{C}\).

Key concepts

Reaction Rate

Understanding the concept of reaction rate is crucial for anyone studying chemical kinetics. It refers to the speed at which reactants are converted into products in a chemical reaction. Factors like concentration, temperature, surface area, and catalysts can significantly influence the reaction rate. In essence, it measures how fast a reaction is progressing.

For instance, if we say a certain reaction is 'fast', it means that a substantial amount of reactant converts to product in a short period of time. Conversely, a 'slow' reaction would indicate minimal product formation over an extended period. A higher reaction rate implies that the chemical transformation is occurring more rapidly, which can be critical in industrial processes where time efficiency is a priority.

When you encounter the term 'rate constant', denoted as 'k' in the Arrhenius equation, this reflects the reaction rate at a given temperature. It's important to note that this constant varies with temperature, which is why temperature control can be a powerful tool in managing reaction rates in both laboratory and industrial settings.

Activation Energy

Activation energy, often symbolized as \(E_a\), is a term frequently used in chemistry to describe the minimum amount of energy required for a chemical reaction to occur. It's akin to the initial push needed to start a reaction, much like the effort needed to start moving a boulder. This energy barrier ensures that molecules only react when they have enough energy to achieve a transition state, leading to product formation.

Here's a more detailed perspective: molecules frequently collide, but only those with sufficient energy to overcome the barrier—the activation energy—can successfully react and transform into products. The Arrhenius equation includes this pivotal concept, relating activation energy with the temperature dependence of the reaction rate. Essentially, a lower activation energy means that more molecules in a system can react at a given temperature, hence increasing the reaction rate.

For a chemical reaction to be efficient, it's often desirable to have as low an activation energy as possible. Catalysts are introduced into reactions precisely for this purpose—they lower the activation energy, allowing the reaction to proceed at a faster rate even at lower temperatures.

Temperature Conversion

When working with the Arrhenius equation, or any temperature-related calculations in chemistry, it's essential to use the Kelvin scale. The temperature conversion from Celsius to Kelvin is a straightforward process where you add 273.15 to the Celsius temperature. This change is necessary because the Kelvin scale sets its zero point at absolute zero—the theoretically lowest temperature possible.

Kelvin is the SI unit for temperature, and it's used in science and engineering because it allows for the direct insertion into various thermodynamic equations, including the Arrhenius equation. The formula for this conversion is as follows: \(T_{Kelvin} = T_{Celsius} + 273.15\).

For example, room temperature, which is approximately \(20^{\text{\circ}}\mathrm{C}\), would convert to \(293.15\mathrm{K}\). This step, while seemingly simple, is critical. Neglecting to convert Celsius to Kelvin can drastically alter the results of calculations involving temperature, leading to incorrect conclusions being drawn. By ensuring the proper use of temperature units, we can maintain accuracy in our chemical reaction analyses.