QNEET-2024-Ch-42: NEET 2024 Chemistry - Class 12 NEET-2024-Ch-42 Solution: The rate of a reaction quadruples when temperature changes from …

To solve this problem, we will use the Arrhenius equation, which relates the rate constant ( k ) of a chemical reaction to the temperature ( T ) and the activation energy ( E a ). The equation in its logarithmic form is: log _ k = log _ A _ E a 2.303 R T where: k is the rate constant, A is the pre-exponential factor (frequency factor), E a is the activation energy, R is the universal gas constant ( 8.314 J mol _ 1 K _ 1 ), T is the temperature in Kelvin, 2.303 is the factor to convert from natural log to common log. According to the problem, the rate of the reaction quadruples ( k 2 = 4 k 1 ) when the temperature increases from 27 _ C (which equals 27 + 273.15 = 300.15 K ) to 57 _ C (which equals 57 + 273.15 = 330.15 K ). Using the logarithmic form of the Arrhenius equation, for two different temperatures we have: log _ k 1 = log _ A _ E a 2.303 R _ 300.15 log _ k 2 = log _ A _ E a 2.303 R _ 330.15 With k 2 = 4 k 1 , substituting in the values and taking their difference: log _ 4 k 1 _ log _ k 1 = ( log _ A _ E a 2.303 _ 8.314 _ 330.15 ) _ ( log _ A _ E a 2.303 _ 8.314 _ 300.15 ) log _ 4 = E a 2.303 _ 8.314 ( 1 300.15 _ 1 330.15 ) Given that log _ 4 = 0.6021 , we now solve for E a : 0.6021 = E a 2.303 _ 8.314 ( 1 300.15 _ 1 330.15 ) First, calculate 1 300.15 _ 1 330.15 : 1 300.15 � 0.003332 and 1 330.15 � 0.003029 1 300.15 _ 1 330.15 = 0.003332 _ 0.003029 = 0.000303 So, 0.6021 = E a 19.141872 _ 0.000303 E a 19.141872 = 0.6021 0.000303 = 1987.05 Solve for E a : E a = 1987.05 _ 19.141872 = 38067 J/mol � 38.067 kJ/mol Therefore, the activation energy E a is 38.07 kJ/mol , which best matches Option A: 38.04 kJ/mol