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Concentration Terms

For a reactio...

Question

For a reaction $A ⟶ B$ , the rate constant $k_{1} = A_{1} e^{- {E a}_{1} / (R T)}$ and for the reaction $X ⟶ Y$ , the rate constant $k_{2} = A_{2} e^{- {E a}_{2} / (R T)}$ .

If $A_{1} = 10^{8}, A_{2} = 10^{10}$ and ${E a}_{1} = 600 c a l / m o l, {E a}_{2} = 1800 c a l / m o l$ , then the temperature at which $k_{1} = k_{2}$ is:

(Given: $R = 2 c a l / k m o l$ )

1200 K

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1200 \times 4.606 K

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\frac{1200}{4.606} K

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\frac{600}{4.606} K

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Solution

The correct option is D $\frac{600}{4.606} K$

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Similar questions

A \to B

k_{1} = A_{1} e^{- E a_{1} / R T}

P \to Q

k_{2} = A_{2} e^{- E a_{2} / R T}

A_{1} = 10^{8}, A_{2} = 10^{10}

E a_{1} = 600, E a_{2} = 1200

k_{1} = k_{2}

E_{a_{1}}

E_{a_{2}}

k_{1}

k_{2}

k_{1} > k_{2}

k_{1}

E_{a_{1}} = 180 k J / m o l

k_{2}

E_{a_{2}} = 80 k J / m o l

k_{3}

E_{a_{3}} = 50 k J / m o l

k = {(\frac{k_{1} k_{2}}{k_{3}})}^{2 / 3}

k_{1}, k_{2} & k_{3}

E a_{1}, E a_{2} a n d E a_{3}

k = \frac{k_{1} k_{3}}{k_{2}} . A s s u m i n g A_{a v} = \frac{A_{1} A_{3}}{A_{2}}

A K - \to P r o d u c t s, E_{a_{1}} = 180 K J / m o l e, E_{a_{2}} = 80 K J / m o l, E_{a_{3}} = 50 K J / m o l .

k = {(\frac{K_{1} - K_{2}}{K_{3}})}^{2 / 3} .

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