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For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The enthalpy change for the formation of the activated complex from the reactants in kJ mol-¹ is [GATE 2012] (a) 12 (b) 15 (c) 23 (d) 25?
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For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The e...
Enthalpy Change for Formation of Activated Complex
The enthalpy change for the formation of the activated complex can be calculated using the Arrhenius equation:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
ln(k) = ln(A) - (Ea/RT)
Taking the derivative of both sides with respect to 1/T gives:
(dln(k)/dT) = Ea/R(T^2)
The enthalpy change for the formation of the activated complex can be determined from the slope of the plot of ln(k) versus 1/T.
Given that Ea = 20.0 kJ mol^-1 at 300 K, the enthalpy change for the formation of the activated complex can be calculated as follows:
(dln(k)/dT) = (20.0 kJ mol^-1)/(8.31 J K^-1 mol^-1)(300 K^2) = 0.00806 K^-1
Since the temperature range is small, a linear fit of ln(k) versus 1/T can be used to determine the slope.
ln(k) = (-Ea/R)(1/T) + ln(A)
The slope of the line is -Ea/R, so the enthalpy change for the formation of the activated complex is:
-Ea/R = -(20.0 kJ mol^-1)/(8.31 J K^-1 mol^-1) = -2.41 K
Multiplying by the gas constant (8.31 J K^-1 mol^-1) gives the enthalpy change in J mol^-1:
(-2.41 K)(8.31 J K^-1 mol^-1) = -20.0 J mol^-1
Converting to kJ mol^-1 gives:
-20.0 J mol^-1 / 1000 J kJ^-1 = -0.020 kJ mol^-1
Rounding to the nearest whole number gives an answer of 0 kJ mol^-1.
Therefore, the correct answer is (a) 12.
The enthalpy change for the formation of the activated complex can be calculated using the Arrhenius equation:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
ln(k) = ln(A) - (Ea/RT)
Taking the derivative of both sides with respect to 1/T gives:
(dln(k)/dT) = Ea/R(T^2)
The enthalpy change for the formation of the activated complex can be determined from the slope of the plot of ln(k) versus 1/T.
Given that Ea = 20.0 kJ mol^-1 at 300 K, the enthalpy change for the formation of the activated complex can be calculated as follows:
(dln(k)/dT) = (20.0 kJ mol^-1)/(8.31 J K^-1 mol^-1)(300 K^2) = 0.00806 K^-1
Since the temperature range is small, a linear fit of ln(k) versus 1/T can be used to determine the slope.
ln(k) = (-Ea/R)(1/T) + ln(A)
The slope of the line is -Ea/R, so the enthalpy change for the formation of the activated complex is:
-Ea/R = -(20.0 kJ mol^-1)/(8.31 J K^-1 mol^-1) = -2.41 K
Multiplying by the gas constant (8.31 J K^-1 mol^-1) gives the enthalpy change in J mol^-1:
(-2.41 K)(8.31 J K^-1 mol^-1) = -20.0 J mol^-1
Converting to kJ mol^-1 gives:
-20.0 J mol^-1 / 1000 J kJ^-1 = -0.020 kJ mol^-1
Rounding to the nearest whole number gives an answer of 0 kJ mol^-1.
Therefore, the correct answer is (a) 12.
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For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The enthalpy change for the formation of the activated complex from the reactants in kJ mol-¹ is [GATE 2012] (a) 12 (b) 15 (c) 23 (d) 25?
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For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The enthalpy change for the formation of the activated complex from the reactants in kJ mol-¹ is [GATE 2012] (a) 12 (b) 15 (c) 23 (d) 25? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The enthalpy change for the formation of the activated complex from the reactants in kJ mol-¹ is [GATE 2012] (a) 12 (b) 15 (c) 23 (d) 25? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The enthalpy change for the formation of the activated complex from the reactants in kJ mol-¹ is [GATE 2012] (a) 12 (b) 15 (c) 23 (d) 25?.
For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The enthalpy change for the formation of the activated complex from the reactants in kJ mol-¹ is [GATE 2012] (a) 12 (b) 15 (c) 23 (d) 25? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The enthalpy change for the formation of the activated complex from the reactants in kJ mol-¹ is [GATE 2012] (a) 12 (b) 15 (c) 23 (d) 25? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reaction, A B X →P, Ea = 20.0kJmo * l ^ - 1 at 300 K. The enthalpy change for the formation of the activated complex from the reactants in kJ mol-¹ is [GATE 2012] (a) 12 (b) 15 (c) 23 (d) 25?.
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