Problem 222 - A Complete Resource Book in Chemistry for JEE Main

Question

The rate of a reaction doubles when its temperature changes from \(300 \mathrm{~K}\) to \(310 \mathrm{~K}\). Activation energy of such a reaction will be: \(\left(\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right.\) and \(\left.\log 2=0.301\right)\) (a) \(58.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(60.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(48.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The activation energy is approximately 53.6 kJ/mol (option c).

Step 1: Understand the Arrhenius Equation

The problem involves determining activation energy using the Arrhenius equation. The equation is given by\[k = A e^{-E_a / (RT)}\]where:- \(k\) is the rate constant,- \(A\) is the frequency factor,- \(E_a\) is the activation energy,- \(R\) is the gas constant (8.314 \(J \, K^{-1} \, mol^{-1}\)),- \(T\) is the temperature in Kelvin.

Step 2: Express Rate Change in Terms of Activation Energy

The rate of reaction doubles as the temperature changes from 300 K to 310 K. Using the rate constant ratio \[ \frac{k_2}{k_1} = 2 \]and the modified Arrhenius equation for two temperatures:\[\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]The task is now to substitute the given values into this equation to determine \(E_a\).

Step 3: Substitute Given Values and Solve for Activation Energy

Given:- \(\log 2 = 0.301\)- \(T_1 = 300 \, K\)- \(T_2 = 310 \, K\)- \(R = 8.314 \, J \, K^{-1} \, mol^{-1}\)Substitute these values into the equation:\[0.301 = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{300} - \frac{1}{310}\right)\]First, calculate \(\frac{1}{300} - \frac{1}{310}\):\[\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} \approx 0.0001075\]

Step 4: Calculate Activation Energy

Substitute the value from the previous step back into our expression:\[0.301 = \frac{E_a}{19.14742} \times 0.0001075\]Solve for \(E_a\):\[E_a = \frac{0.301 \times 19.14742}{0.0001075} \approx 53852.5 \, J/mol\]Convert units from J/mol to kJ/mol:\[E_a = 53.9 \, kJ/mol\]Rounding to match the options, we find that the closest value is 53.6 \(kJ/mol\) (option c).

Key Concepts

Activation Energy

Activation energy is a pivotal concept in chemical kinetics. It represents the minimum energy required for a chemical reaction to occur. This energy barrier must be overcome for reactants to transform into products. Think of it as the initial push needed to get a ball rolling downhill—without sufficient force, it will simply stay put.
For any reaction to proceed, molecules must collide with enough energy to surpass this barrier. If they do not, the reaction will not occur. Thus, activation energy dictates the speed of the reaction. A high activation energy means that fewer molecules have the necessary energy, leading to a slower reaction.

• Lower activation energy = Faster reaction
• Higher activation energy = Slower reaction

Arrhenius Equation

The Arrhenius equation is a formula that establishes a quantitative relationship between the rate constant of a reaction and temperature. The equation is expressed as:\[k = A e^{-E_a / (RT)}\]where:

• \( k \) is the rate constant
• \( A \) is the frequency factor, which is related to the frequency of collisions between reactant molecules
• \( E_a \) is the activation energy
• \( R \) is the universal gas constant (8.314 J\(K^{-1}mol^{-1}\))
• \( T \) is the temperature in Kelvin

This equation highlights that the rate constant increases exponentially with an increase in temperature or a decrease in activation energy. It emphasizes how crucial temperature is in influencing the speed of chemical reactions.

Temperature Dependence of Reaction Rate

The rate of a chemical reaction is profoundly impacted by temperature changes. Generally, as temperature increases, the rate of the reaction increases as well. This happens because an increase in temperature results in molecules moving more energetically, resulting in more frequent and forceful collisions.\[\text{Rate change} = 2 \text{ when } T_1 = 300K \text{ and } T_2 = 310K\]This is because higher temperatures provide molecules with more energy, increasing the number of collisions that can exceed the activation energy barrier. In simpler terms, heating the reaction can speed it up. However, it's essential to remember that not all reactions double their rate with a 10 K temperature increase; it depends on the specific energy dynamics of the reaction.

• Higher temperatures generally = faster reactions
• Lower temperatures generally = slower reactions