Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsArrhenius EquationMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

A substance ' X ' dissociates by two parallel reactions with different ' Arrhenius constants $(A_{1}$ and $A_{2})$ " and different "Activation energies" ( $E_{a_{1}}$ and $E_{a_{2}}$ ) as shown.

$X (g) \frac{A _{1} = 2 \times 1 0 ^{14} s ^{- 1}}{E _{a_{1}} = 12 K cal / mol} Y (g); X (g) \frac{A _{2} = 2 \times 1 0 ^{14} \times e ^{- 2} s ^{- 1}}{E _{a_{2}} = 9 K cal / mol} Z (g)$

[Given : $R = 2 Cal / mol - K$ ]

If the temperature is increased, then-

Options:

Answer:

Solution:

Using the Arrhenius equation:

$k = A e^{- E_{a} / RT}$

On increasing the temperature, both $k_{1}$ and $k_{2}$ increase, so the rates of formation of Y and Z both increase.

Since $E_{a 1} = 12 k c a l / m o l > E_{a 2} = 9 k c a l / m o l$ , the reaction with higher activation energy (Y) shows a greater percentage increase in rate with temperature.

Correct Option: (B)

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Arrhenius Equation

⏱ 2mℹ️ Source: QB

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