Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsZero Order ReactionsMedium2 minPYQ_2017

ChemistryMediumsingle choice

Two reactions $A_{1}$ and $A_{2}$ have identical pre-exponential factors. The activation energy of $A_{1}$ is more than $A_{2}$ by $10 kJ {mol}^{- 1}$ . If $k_{1}$ and $k_{2}$ are the rate constants for reactions $A_{1}$ and $A_{2}$ , respectively at $300 K$ , then $\ln (\frac{k_{2}}{k_{1}})$ is equal to

$(R = 8 .314 J {mol}^{- 1} K^{- 1})$

Options:

Answer:

Solution:

From Arrhenius equation

$k = {Ae}^{- Ea / RT}$

$k_{1} = {Ae}^{- {Ea}_{1} / RT}$

$k_{2} = {Ae}^{- {Ea}_{2} / RT}$

$\ln \frac{k_{2}}{k_{1}} = e^{({Ea}_{1} - {Ea}_{2}) / RT}$

$\ln \frac{k_{2}}{k_{1}} = \frac{{Ea}_{1} - {Ea}_{2}}{RT} = \frac{10 \times 1000}{8 .314 \times 300} = 4$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Zero Order Reactions

⏱ 2mℹ️ Source: PYQ_2017

Doubts & Discussion

Loading discussions...