Question

The activation energy for the decomposition of \(\mathrm{HI}(g)\) to \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) is \(186 \mathrm{kJ} / \mathrm{mol} .\) The rate constant at \(555 \space\mathrm{K}\) is \(3.52 \times\) \(10^{-7} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} .\) What is the rate constant at \(645 \space\mathrm{K} ?\)

Short answer

The rate constant at 645 K for the decomposition of HI(g) to H2(g) and I2(g) is approximately \(1.99 \times 10^{-5} L/mol·s\).

Step-by-step solution

Write down the Arrhenius equation and the given values

The Arrhenius equation relates the activation energy, temperature, and rate constant: \[ k = Ae^{-\frac{E_a}{RT}} \] Where: - \(k\) is the rate constant - \(A\) is the pre-exponential factor or frequency factor - \(E_a\) is the activation energy - \(R\) is the gas constant (8.314 J/mol·K) - \(T\) is the temperature in Kelvin We are given: - Activation energy, \(E_a = 186 \space kJ/mol = 186,000 J/mol\) - Rate constant at 555 K, \(k_1 = 3.52 \times 10^{-7} L/mol·s\) - Temperature at 555 K, \(T_1 = 555 K\) - Temperature at 645 K, \(T_2 = 645 K\) We want to find the rate constant at 645 K, \(k_2\).

Divide the Arrhenius equation for the two different temperatures

By taking the ratio of the Arrhenius equation at 645 K and 555 K, we can eliminate the pre-exponential factor \(A\): \[\frac{k_2}{k_1} = \frac{Ae^{-\frac{E_a}{R T_2}}}{Ae^{-\frac{E_a}{R T_1}}}\]

Simplify the equation and solve for \(k_2\)

Simplify the equation in Step 2: \[\frac{k_2}{k_1} = \frac{e^{-\frac{E_a}{R T_2}}}{e^{-\frac{E_a}{R T_1}}}\] Now take the natural logarithm of both sides: \[ln(\frac{k_2}{k_1}) = ln(\frac{e^{-\frac{E_a}{R T_2}}}{e^{-\frac{E_a}{R T_1}}}) = ln(e^{-\frac{E_a}{R T_2}})-ln(e^{-\frac{E_a}{R T_1}})\] Using the property of logarithms to rewrite the equation again: \[-\frac{E_a}{R T_2} + \frac{E_a}{R T_1} = ln(\frac{k_2}{k_1})\] Solve for \(k_2\): \[k_2 = k_1 \cdot e^{\frac{-E_a}{R T_2} + \frac{E_a}{R T_1}}\]

Insert the given values and calculate \(k_2\)

Insert the given values into the equation from Step 3 and calculate \(k_2\): \[k_2 = (3.52 \times 10^{-7} L/mol·s) \cdot e^{\frac{-186,000 J/mol}{(8.314 J/mol·K)(645 K)} + \frac{186,000 J/mol}{(8.314 J/mol·K)(555 K)}}\] \[k_2 \approx 1.99 \times 10^{-5} L/mol·s\] So the rate constant at 645 K is approximately \(1.99 \times 10^{-5} L/mol·s\).

Key concepts

Understanding Activation Energy

Activation energy, symbolized by \(E_a\), is the minimum energy barrier that reacting molecules must overcome for a chemical reaction to occur. Imagine it as the initial push needed to start rolling a boulder up a hill. Without sufficient energy, the reaction will not proceed. In essence, activation energy helps determine how fast a reaction will proceed by affecting the number of molecules that can successfully transform into products.

High activation energy means a slower reaction because fewer molecules will have the energy to surmount the energy barrier. Conversely, a low activation energy results in a quicker reaction, as more molecules can participate successfully. This concept is crucial when predicting or controlling reaction rates, as it shows why some reactions need additional conditions, like heat or catalysts, to proceed effectively.

Rate Constant Calculation using the Arrhenius Equation

The Arrhenius equation is vital in understanding how the rate constant \(k\) changes with temperature. The equation is represented as \(k = Ae^{-\frac{E_a}{RT}}\), where \(A\) is the frequency factor, \(E_a\) is activation energy, \(R\) is the universal gas constant, and \(T\) is temperature in Kelvin.

This equation shows that even small changes in temperature can lead to significant changes in the reaction rate. Calculating the rate constant involves using given data, such as activation energy and temperature, to find how quickly a reaction proceeds at different temperatures. In practice, if you have a known rate constant at one temperature, you can predict its value at another by considering the Arrhenius equation. Utilizing logarithms simplifies solving for \(k_2\) when given the initial rate constant (\(k_1\)), the two temperatures, and \(E_a\).

The exercise exemplifies this by determining \(k_2\) at 645 K using the value from 555 K. The worked solution shows that even without knowing the frequency factor \(A\), you can calculate changes in \(k\) by comparing ratios, making this an essential tool in chemistry.

Examining Temperature Dependence of Reaction Rates

Temperature has a profound effect on chemical reaction rates. Typically, as temperature increases, so does the reaction rate. This relation is embedded in the Arrhenius equation: a higher \(T\) leads to a larger value for the exponential term, which increases \(k\) and, subsequently, the reaction rate.

Temperature influences molecular movements. At higher temperatures, molecules move faster, increasing collision frequency and energy, which boosts the likelihood of surpassing the activation energy barrier. The reaction rate, therefore, is not only dependent on the presence of reactants but is significantly accelerated by temperature increases.

This concept’s significance is demonstrated in various applications. In industries, controlling temperature is a key strategy to optimize reaction rates for efficient production. The exercise illustrates these ideas by showing how the rate constant increases significantly when temperature rises from 555 K to 645 K, highlighting that understanding temperature dependence is crucial for predicting and designing chemical processes effectively.