Problem 17 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The rate of a reaction decreased by \(3.555\) times when the temperature was changed from \(40^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\). The activation energy (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) of the reaction is \(.\) Take; \(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) In \(\overline{3.555}=1.268\)

Step-by-Step Solution

Answer

The activation energy is approximately 99.38 kJ/mol.

Step 1: Understand the Arrhenius Equation

The Arrhenius equation describes how the rate constant (\(k\)) of a reaction depends on temperature and activation energy. It is given by\[k = Ae^{-\frac{E_a}{RT}}\]where \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

Step 2: Use the Modified Arrhenius Equation

When comparing reaction rates at two different temperatures, we use the ratio of the rate constants:\[\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]where \(k_1\) is the rate constant at temperature \(T_1\) and \(k_2\) is at \(T_2\). Given that the rate decreases by 3.555 times, \(\frac{k_2}{k_1} = \frac{1}{3.555}\).

Step 3: Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin:\[T_1 = 40^{\circ}C + 273.15 = 313.15\,\text{K}\]\[T_2 = 30^{\circ}C + 273.15 = 303.15\,\text{K}\]

Step 4: Substitute Values into Equation

Substitute the known values into the equation from Step 2:\[\ln \left(\frac{1}{3.555}\right) = \frac{E_a}{8.314}\left(\frac{1}{303.15} - \frac{1}{313.15}\right)\]Given ln(3.555) = 1.268, it follows that:\[\ln\left(\frac{1}{3.555}\right) = -1.268\]

Step 5: Simplify and Solve for Activation Energy

Now simplify and solve the equation:\[-1.268 = \frac{E_a}{8.314} \left(0.003300 - 0.003194\right)\]Calculate the difference:\[0.003300 - 0.003194 = 0.000106\]Thus the revised equation is:\[-1.268 = \frac{E_a}{8.314} \times 0.000106\]Solve for \(E_a\):\[E_a = \frac{-1.268 \times 8.314}{0.000106} \approx 99377.358\,\text{J/mol}\]

Step 6: Convert Activation Energy to kJ/mol

Since we are asked for the activation energy in kJ/mol, convert from J/mol:\[\frac{99377.358}{1000} \approx 99.38\,\text{kJ/mol}\]

Key Concepts

Activation Energy

Every chemical reaction requires a certain amount of energy to proceed; this is what we call activation energy, often denoted as \(E_a\). It represents the energy barrier that must be overcome for reactants to transform into products. Think of it as a hill that molecules have to "climb" before they can "roll" down to form products.

• Activation energy is a crucial factor because it determines how "difficult" it is for a reaction to occur.
• Reactions with high activation energies tend to be slower at a given temperature because fewer molecules possess the requisite energy.
• On the other hand, if the activation energy is low, less energy is required, allowing the reaction to proceed more rapidly.

The Arrhenius equation mathematically relates this concept using the formula:\[E_a = -R imes rac{d( ext{ln }k)}{d(1/T)}\]Here, \(R\) is the gas constant and \(k\) is the rate constant. The negative sign reflects that as temperature increases, the rate constant, and hence reaction rate, generally increases.
Understanding \(E_a\) helps in predicting how much heat is needed for a reaction to commence and how altering conditions can affect reaction speeds.

Temperature Dependence of Reaction Rate

Temperature dramatically influences how fast a chemical reaction occurs. This relationship is captured well by the Arrhenius equation, indicating that reaction rates increase with temperature.

• When you increase the temperature, molecules move faster and collide more frequently, with more energy.
• This increase in activity boosts the number of successful collisions, leading to a faster reaction rate.
• A 10°C increase can nearly double or triple the reaction rate, depending on the reaction specifics.

The exercise example highlights how dropping the temperature from 40°C to 30°C caused the reaction rate to decrease by 3.555 times. This outcome is precisely what we expect given the temperature's exponential impact on reaction rates as modeled by the Arrhenius equation.
Let's express this quantitatively through the modified Arrhenius equation:\[\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]This equation shows that even a simple temperature change can significantly alter the ratio of reaction rates \(k_2/k_1\), underscoring the sensitivity of reaction rates to thermal changes.

Rate Constant

The rate constant, denoted as \(k\), is a crucial component of the Arrhenius equation and a foundational parameter in chemical kinetics. It helps us describe how quickly a reaction progresses under specified conditions.

• The rate constant is not truly constant because it varies with temperature.
• It reflects both the frequency of collisions and the fraction of collisions that are successful, leading to reaction.
• The rate constant increases with temperature, showing the same dependency as reaction rates.

In the Arrhenius equation, \(k\) is expressed as:\[k = A e^{-\frac{E_a}{RT}}\]Here, \(A\) represents the frequency factor, which accounts for the frequency of collisions and their orientation. The exponential component ties the rate constant to temperature and activation energy.
The exercise employs this idea by comparing the rate constants at different temperatures and calculating the activation energy. The variation of \(k\) embodies the reaction's temperature sensitivity, providing valuable insight into reaction kinetics across different conditions.