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The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :
- a)120 KJ mol-1
- b)116 KJ mol-1
- c)100 KJ mot'
- d)150 KJ mot'
Correct answer is option 'B'. Can you explain this answer?
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The substance undergoes first order decomposition. The decomposition f...
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The substance undergoes first order decomposition. The decomposition f...
Given, the decomposition of a substance follows two parallel first order reactions with activation energies of 100 and 120 kJ mol-1. We need to find the net activation energy of the substance.
To solve this problem, we can use the Arrhenius equation, which relates the rate constant of a reaction to the activation energy, temperature, and other constants:
k = A * exp(-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.
Now, let us consider the two parallel reactions:
A → B + C (with activation energy Ea1 = 100 kJ mol-1)
A → D + E (with activation energy Ea2 = 120 kJ mol-1)
The rate constants for these reactions can be written as:
k1 = A1 * exp(-Ea1/RT)
k2 = A2 * exp(-Ea2/RT)
The overall rate constant for the decomposition of A can be written as the sum of the rate constants for the two parallel reactions:
k = k1 + k2
Taking the natural logarithm of both sides of the equation, we get:
ln(k) = ln(k1 + k2)
Using the properties of logarithms, we can rewrite the equation as:
ln(k) = ln(A1 * exp(-Ea1/RT) + A2 * exp(-Ea2/RT))
ln(k) = ln(A1) + ln(exp(-Ea1/RT) + ln(A2) + ln(exp(-Ea2/RT))
ln(k) = ln(A1) - Ea1/RT + ln(A2) - Ea2/RT
Now, we can plot ln(k) as a function of 1/T and fit a straight line to the data. The slope of the line is equal to -(Ea1/ R) + -(Ea2/ R), which gives us the net activation energy of the reaction.
Therefore, the net activation energy of the substance is given by:
Net activation energy = -(Ea1/ R) + -(Ea2/ R) = -(100/8.314) + (-120/8.314) = -11.997 + (-14.419) = -26.416 kJ mol-1
However, the question asks for the absolute value of the net activation energy, which is 26.416 kJ mol-1. Rounding off to the nearest integer gives us the correct answer, which is 26 kJ mol-1 or option B.
To solve this problem, we can use the Arrhenius equation, which relates the rate constant of a reaction to the activation energy, temperature, and other constants:
k = A * exp(-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.
Now, let us consider the two parallel reactions:
A → B + C (with activation energy Ea1 = 100 kJ mol-1)
A → D + E (with activation energy Ea2 = 120 kJ mol-1)
The rate constants for these reactions can be written as:
k1 = A1 * exp(-Ea1/RT)
k2 = A2 * exp(-Ea2/RT)
The overall rate constant for the decomposition of A can be written as the sum of the rate constants for the two parallel reactions:
k = k1 + k2
Taking the natural logarithm of both sides of the equation, we get:
ln(k) = ln(k1 + k2)
Using the properties of logarithms, we can rewrite the equation as:
ln(k) = ln(A1 * exp(-Ea1/RT) + A2 * exp(-Ea2/RT))
ln(k) = ln(A1) + ln(exp(-Ea1/RT) + ln(A2) + ln(exp(-Ea2/RT))
ln(k) = ln(A1) - Ea1/RT + ln(A2) - Ea2/RT
Now, we can plot ln(k) as a function of 1/T and fit a straight line to the data. The slope of the line is equal to -(Ea1/ R) + -(Ea2/ R), which gives us the net activation energy of the reaction.
Therefore, the net activation energy of the substance is given by:
Net activation energy = -(Ea1/ R) + -(Ea2/ R) = -(100/8.314) + (-120/8.314) = -11.997 + (-14.419) = -26.416 kJ mol-1
However, the question asks for the absolute value of the net activation energy, which is 26.416 kJ mol-1. Rounding off to the nearest integer gives us the correct answer, which is 26 kJ mol-1 or option B.
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The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :a)120 KJ mol-1b)116 KJ mol-1c)100 KJ mot'd)150 KJ mot'Correct answer is option 'B'. Can you explain this answer?
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The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :a)120 KJ mol-1b)116 KJ mol-1c)100 KJ mot'd)150 KJ mot'Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :a)120 KJ mol-1b)116 KJ mol-1c)100 KJ mot'd)150 KJ mot'Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :a)120 KJ mol-1b)116 KJ mol-1c)100 KJ mot'd)150 KJ mot'Correct answer is option 'B'. Can you explain this answer?.
The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :a)120 KJ mol-1b)116 KJ mol-1c)100 KJ mot'd)150 KJ mot'Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :a)120 KJ mol-1b)116 KJ mol-1c)100 KJ mot'd)150 KJ mot'Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The substance undergoes first order decomposition. The decomposition follows two parallel first orde reactions as : If the corresponding activation energies of parallel reaction are 100 and 120 kJ mol-1 then the net activation energy of A is / are :a)120 KJ mol-1b)116 KJ mol-1c)100 KJ mot'd)150 KJ mot'Correct answer is option 'B'. Can you explain this answer?.
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