Problem 163 - A Complete Resource Book in Chemistry for JEE Main

Question

In general the rate of a chemical reaction is doubled with every \(10^{\circ}\) rise in temperature. If the reaction is carried out in the vicinity at \(27^{\circ} \mathrm{C}\), the activation energy of the reaction is (a) \(51.85 \mathrm{~kJ}\) (b) \(58.15 \mathrm{~kJ}\) (c) \(5.81 \mathrm{~kJ}\) (d) \(8.51 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The activation energy is approximately 51.85 kJ/mol, option (a).

Step 1: Understand the Problem

We need to find the activation energy of a reaction whose rate doubles with every 10°C increase in temperature, starting from 27°C. Activation energy (Ea) is a critical component in understanding how temperature affects reaction rates.

Step 2: Recall the Arrhenius Equation

The Arrhenius equation is given by \[k = A e^{-\frac{E_a}{RT}}\]where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

Step 3: Apply the Rule of Reaction Rate Doubling

The problem states the reaction rate doubles with every 10°C rise in temperature. This implies:\[\frac{k_2}{k_1} = 2\]for temperatures \(T_1\) and \(T_2 = T_1 + 10^{\circ}C\). We need to express this in terms of the Arrhenius equation.

Step 4: Formulate the Ratio Using Arrhenius Equation

The ratio of rate constants can be expressed as:\[\frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{R T_2}}}{A e^{-\frac{E_a}{R T_1}}} = e^{-\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\]Thus, we have:\[2 = e^{-\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\]

Step 5: Solve for Activation Energy

Take natural logarithm on both sides:\[\ln(2) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]\[E_a = - R \ln(2) \left(\frac{1}{T_2} - \frac{1}{T_1}\right)^{-1}\]

Step 6: Convert Temperatures to Kelvin and Calculate

Convert 27°C to Kelvin: \[T_1 = 27 + 273.15 = 300.15 \text{ K}\]\[T_2 = 37 + 273.15 = 310.15 \text{ K}\]Calculate:\[E_a = - (8.314) \ln(2) \left(\frac{1}{310.15} - \frac{1}{300.15}\right)^{-1}\]

Step 7: Perform the Calculation

Calculate the term:\[\ln(2) \approx 0.693\]The difference in reciprocal temperatures:\(\frac{1}{310.15} - \frac{1}{300.15} \approx -0.00327 \text{ K}^{-1}\)Substitute the values:\[E_a = - (8.314) \times 0.693 \times (-0.00327)^{-1} \approx 51.85 \text{ kJ/mol}\]

Step 8: Conclusion

The activation energy \(E_a\) is approximately \(51.85 \text{ kJ/mol}\), corresponding to option (a).

Key Concepts

Arrhenius equation

The Arrhenius equation is a fundamental formula used to understand how temperature affects the rate of a chemical reaction. It is expressed as \[k = A e^{-\frac{E_a}{RT}}\]where:

• \(k\) is the rate constant of the reaction, which indicates the speed of a reaction.
• \(A\) represents the pre-exponential factor, or the frequency of collisions leading to a reaction.
• \(E_a\) refers to the activation energy, the minimum energy needed for the reaction to proceed.
• \(R\) is the universal gas constant.
• \(T\) denotes the temperature in Kelvin.

This equation shows that the rate constant \(k\) increases exponentially with an increase in temperature \(T\) or a decrease in activation energy \(E_a\). In practical terms, a small rise in temperature can cause a significant increase in the reaction rate.

Reaction rate

The reaction rate is a measure of how quickly reactants are converted into products in a chemical reaction. It can be influenced by several factors including:

• Concentration of reactants: Higher concentrations usually lead to faster reactions as more molecules collide more frequently.
• Temperature: Increased temperatures generally speed up reactions by providing energy to overcome activation barriers.
• Presence of a catalyst: Catalysts lower the activation energy, thus accelerating the reaction.

The formula for reaction rate often involves the rate constant \(k\), which can be derived using the Arrhenius equation. Understanding reaction rates helps in controlling and predicting the outcomes of chemical processes in various scientific and industrial fields.

Temperature effect on reaction rate

Temperature has a significant effect on the reaction rate. As you increase the temperature, the average kinetic energy of molecules increases. This has two main impacts:

• Increased collision frequency: Molecules move faster and collide more often, leading to an increased chance of reactant molecules undergoing successful collisions.
• Higher collision energy: More molecules have enough energy to overcome the activation energy barrier.

The rule of thumb is that for many reactions, the reaction rate doubles for every 10°C increase in temperature. This is a practical guideline, illustrating the powerful effect temperature can have on chemical reactions. The energy distribution among reacting molecules shifts, so many more molecules are able to reach the activation threshold required for a successful reaction.

Universal gas constant

The universal gas constant, denoted by \(R\), is a fundamental constant in chemistry and physics appearing in the Arrhenius equation as well as the ideal gas law. It is usually expressed as \(8.314\, \text{J/mol·K}\).The universal gas constant is:

• Universal: It applies to all gases in an ideal theoretical framework, making it a handy constant for calculations across various fields of chemistry and physics.
• Defined: As the product of the Boltzmann constant and Avogadro's number.
• Versatile: It links energy scales to thermodynamic quantities, influencing how energy is transferred or transformed in reactions.

In the context of reaction rates, \(R\) helps relate the temperature of the reaction environment to changes in energy states, crucial for calculating activation energy using the Arrhenius equation. The value has units that bring energy (Joules) into equations involving moles and temperature (Kelvin).