Problem 10 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The rate coefficient \((k)\) for a particular reactions is \(1.3 \times 10^{-4} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\), and \(1.3 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(150{ }^{\circ} \mathrm{C}\). What is the energy of activation \(\left(E_{\mathrm{a}}\right)\) (in \(\mathrm{kJ}\) ) for this reaction? \((\mathrm{R}=\) molar gas constant \(=\) \(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) ) (a) 16 (b) 60 (c) 99 (d) 132

Step-by-Step Solution

Answer

The activation energy is approximately 60 kJ/mol.

Step 1: Identify the formula for Activation Energy

We will use the Arrhenius equation in the natural logarithmic form for two different temperatures. The equation is given by: \[ \ln \left( \frac{k_1}{k_2} \right) = \frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]\ where \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively, \( R \) is the gas constant, and \( E_a \) is the activation energy.

Step 2: Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula: \( T(K) = T(^{\circ}C) + 273.15 \). Thus, for \( 100^{\circ}C \), \( T_1 = 373.15 \) K, and for \( 150^{\circ}C \), \( T_2 = 423.15 \) K.

Step 3: Substitute Known Values into the Formula

We have \( k_1 = 1.3 \times 10^{-4} \; \text{M}^{-1}\text{s}^{-1} \) at \( T_1 = 373.15 \; \text{K} \), \( k_2 = 1.3 \times 10^{-3} \; \text{M}^{-1}\text{s}^{-1} \) at \( T_2 = 423.15 \; \text{K} \), and \( R = 8.314 \; \text{J K}^{-1} \text{mol}^{-1} \). Substitute these values into the Arrhenius formula: \[ \ln \left( \frac{1.3 \times 10^{-4}}{1.3 \times 10^{-3}} \right) = \frac{E_a}{8.314} \left( \frac{1}{423.15} - \frac{1}{373.15} \right) \]

Step 4: Calculate the Natural Logarithm

Calculate \( \ln \left( \frac{1.3 \times 10^{-4}}{1.3 \times 10^{-3}} \right) = \ln(0.1) = -2.302 \).

Step 5: Simplify the Temperature Difference

Calculate \( \frac{1}{423.15} - \frac{1}{373.15} \), which equals approximately \(-0.000289 \; \text{K}^{-1} \).

Step 6: Solve for Activation Energy

Rearrange the formula to solve for \( E_a \): \[ E_a = \frac{-2.302 \times 8.314}{-0.000289} \]. Calculate \( E_a \) and convert it to kJ by dividing by 1000: \[ E_a = \frac{2.302 \times 8.314}{0.000289} \approx 66.91 \; \text{kJ/mol} \].

Step 7: Round Activation Energy to Nearest Value

Round \( 66.91 \) kJ/mol to the nearest value given in the options: the closest is \( 60 \) kJ/mol, therefore the answer is approximately \( 60 \) kJ/mol.

Key Concepts

Arrhenius Equation

The Arrhenius Equation is crucial in chemical kinetics. It relates the rate of chemical reactions with temperature and helps us calculate Activation Energy. At its core, this equation connects how fast a reaction proceeds with the energy required to initiate it and the temperature at which it occurs. It is expressed as:\[ k = A e^{-\frac{E_a}{RT}} \]Here, \( k \) is the rate coefficient (or rate constant), \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvin.In the context of determining activation energy between two temperatures, the equation can be manipulated to its logarithmic form to compare two rate constants like this:\[ \ln \left( \frac{k_1}{k_2} \right) = \frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]This rearrangement allows us to calculate the activation energy \( E_a \) by knowing the rate coefficients at two different temperatures.

Rate Coefficient

The rate coefficient, also known as the rate constant, is a fundamental part of reaction kinetics. This value captures the speed at which a chemical reaction proceeds under specific conditions. For any given reaction, if the rate coefficient is known for one temperature (e.g., 100°C), understanding how it changes with another temperature (e.g., 150°C) is possible through the Arrhenius Equation.In our example, the rate coefficients at two temperatures were:- At 100°C, \( k_1 = 1.3 \times 10^{-4} \; \text{M}^{-1}\text{s}^{-1} \)- At 150°C, \( k_2 = 1.3 \times 10^{-3} \; \text{M}^{-1}\text{s}^{-1} \)The significant increase in the rate coefficient indicates that the reaction speeds up as the temperature rises. This direct relationship between the rate constant and temperature underpins the necessity of calculating activation energy, as it helps to quantify how temperature affects reaction speeds.

Temperature Conversion

Temperature plays a key role in chemical reactions, and for calculations involving the Arrhenius Equation, it's important to convert temperatures from Celsius to Kelvin. The Kelvin scale is the standard in scientific equations because it starts at absolute zero, providing a consistent basis for calculations.The conversion process is simple:\[ T(\text{K}) = T(^{\circ}\text{C}) + 273.15 \]For instance, to convert 100°C to Kelvin:

• 100°C + 273.15 = 373.15 K

Likewise, 150°C converts as follows:

• 150°C + 273.15 = 423.15 K

Temperature conversion ensures the accuracy and consistency of reaction calculations and is especially vital when using the Arrhenius Equation, which requires absolute temperatures.

Natural Logarithm Calculation

Understanding natural logarithms is essential when working with the Arrhenius Equation in its logarithmic form. The natural logarithm (\(\ln\)) is a mathematical operation that finds the power to which the base 'e' (approximately 2.718) must be raised to obtain a given number.In the given solution, the natural logarithm of the rate coefficient ratio was calculated as follows:\[ \ln \left( \frac{1.3 \times 10^{-4}}{1.3 \times 10^{-3}} \right) = \ln(0.1) = -2.302 \]This calculation is a key step in rearranging the Arrhenius Equation to solve for activation energy. By accurately calculating the natural log, we determine the precise impact of temperature changes on reaction rates, thereby facilitating the computation of activation energies.