Problem 18 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from \(27^{\circ} \mathrm{C}\) to \(42^{\circ} \mathrm{C}\). Its energy of activation in \(\mathrm{J} / \mathrm{mol}\) is (Take \(\left.\ln 5=1.6094 ; \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\)

Step-by-Step Solution

Answer

The activation energy is approximately 83521 J/mol.

Step 1: Convert Temperatures to Kelvin

To work with gas constant in the form of Joules and Kelvin, we should convert the temperatures from Celsius to Kelvin. The conversion is done by adding 273.15 to the Celsius temperature. For \(27^{\circ} C\), we have: \T_1 = 27 + 273.15 = 300.15\ K. For \(42^{\circ} C\), we have:\T_2 = 42 + 273.15 = 315.15\ K.

Step 2: Setup the Arrhenius Equation for Both Temperatures

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, and \( R \) is the gas constant. We set up the ratio of rate constants for the two temperatures:\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{R T_2}}}{A e^{-\frac{E_a}{R T_1}}} = e^{\frac{E_a}{R T_1} - \frac{E_a}{R T_2}} \] Given that \( \frac{k_2}{k_1} = 5 \) for the five-fold increase, we have:\[ 5 = e^{\frac{E_a}{R T_1} - \frac{E_a}{R T_2}} \]

Step 3: Solve for Activation Energy \(E_a\)

Take the natural log on both sides of the equation:\[ \ln 5 = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] We can substitute the given values: \( \ln 5 = 1.6094 \), \( R = 8.314 \ J \ mol^{-1} \ K^{-1} \), and the temperatures from Step 1.So, plugging in the numbers:\[ 1.6094 = \frac{E_a}{8.314} \left(\frac{1}{300.15} - \frac{1}{315.15}\right) \] Calculate:\[ 1.6094 \approx \frac{E_a}{8.314} \left(0.00333 - 0.00317\right) \approx \frac{E_a}{8.314} \times 0.00016 \] \[ E_a = 1.6094 \times \frac{8.314}{0.00016} \] \[ E_a = 83520.53 \ J/mol \]

Step 4: Final Answer and Interpretation

The activation energy for the reaction, given the temperature increase and the five-fold increase in reaction rate, is approximately \( 83520.53 \ \text{J/mol} \). This represents the minimum energy needed for molecules to react over the given temperatures.

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a vital formula in chemistry that connects the rate of a chemical reaction with the temperature. It's expressed as \[ k = A e^{-\frac{E_a}{RT}} \] where:

• \( k \) is the rate constant, indicating how fast a reaction proceeds.
• \( A \) is the pre-exponential factor, relating to the frequency of collisions with the correct orientation.
• \( E_a \) represents the activation energy, the crucial energetic barrier that reactants must overcome.
• \( R \) is the universal gas constant (8.314 \( J \ mol^{-1} \ K^{-1} \)).
• \( T \) stands for temperature, measured in Kelvin.

In essence, the equation suggests that reactions occur more readily at higher temperatures because the exponential factor increases, driven by decreasing \( E_a \) relative to \( RT \). This implies that at higher temperatures, more molecules have sufficient energy to overcome \( E_a \), resulting in a higher reaction rate. Understanding this concept helps to grasp the critical role temperature plays in chemical kinetics.

Temperature Conversion

Temperature Conversion is a fundamental concept in many scientific calculations where temperature affects the outcome. In our exercise, we need to convert Celsius to Kelvin to use scientific constants correctly. The conversion is straightforward:To convert from Celsius to Kelvin:\[ T(K) = T(°C) + 273.15 \]Celsius is more intuitive for everyday use, but Kelvin is the SI unit for temperature, providing an absolute scale starting from absolute zero, where molecular motion stops. For example, converting \( 27^{\circ}C \) to Kelvin involves simply adding \( 273.15 \), resulting in \( 300.15 \ K \). Similarly, \( 42^{\circ}C \) converts to \( 315.15 \ K \).Using Kelvin ensures accuracy when applying constants like the gas constant \( R \) in chemical equations, highlighting the importance of precise temperature measurement in scientific work.

Natural Logarithm

The Natural Logarithm (ln) is a mathematical tool used extensively in scientific calculations, notably in processes involving exponential growth or decay. In our example, the natural logarithm is employed to linearize the exponential expressions in the Arrhenius Equation.The natural logarithm \( \ln \) is the inverse function of the exponential function \( e^x \). It is based on the transcendental number \( e \), approximately equal to 2.71828. Using the natural logarithm aids in simplifying expressions like \( e^{x} = y \) to \( x = \ln y \).In this exercise, \( \ln 5 \) is used to convert an exponential increase in reaction rate to a linear form. This helps solve for the activation energy \( E_a \), as shown when the equation \( \ln 5 = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \) is rearranged to isolate \( E_a \).Understanding the natural logarithm is crucial because it simplifies complex equations in thermodynamics and kinetics, making it an essential concept for students in science and engineering disciplines.