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.
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