Figure shows a graph in lo g 10 k v s T 1 where k is rate constant and T is temperature. The straight line BC has slope, tan θ = − 2.303 1 and an intercept of 5 on Y -axis. Thus E a , the energy of activation is: Use R = 2 cal mol^-1 K^-1.

lo g 10 k m E a = lo g 10 A − 2.303 R E a T 1 = − 2.303 R E a = − 2.303 1 K = R ( 1 K ) = 2 cal mo l − 1 .

Start Free Mock LoginAlready registered? Login here

Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsArrhenius EquationMedium2 minQB

ChemistryMediumsingle choice

Figure shows a graph in $lo g_{10} k v s \frac{1}{T}$ where k is rate constant and T is temperature. The straight line BC has slope, $tan θ = - \frac{1}{2.303}$ and an intercept of 5 on Y -axis. Thus $E_{a}$ , the energy of activation is: Use R = 2 cal mol^-1 K^-1.

Options:

Answer:

Solution:

$lo g_{10} k m E_{a} = lo g_{10} A - \frac{E _{a}}{2.303 R} \frac{1}{T} = - \frac{E _{a}}{2.303 R} = - \frac{1}{2.303} K = R (1 K) = 2 cal mo l^{- 1} .$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Arrhenius Equation

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...