CameraIcon

SearchIcon

MyQuestionIcon

4

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Arrhenius Equation

Two reactions...

Question

Two reactions $R_{1}$ and $R_{2}$ have identical pre-exponential factors. Activation energy of $R_{1}$ exceeds that of $R_{2}$ by $10 k J m o l^{- 1}$ . If $k_{1}$ and $k_{2}$ are rate constant for reactions $R_{1}$ and $R_{2}$ respectively at $300 K$ , then $l n (k_{2} / k_{1})$ is equal to: $(R = 8.314 J m o l^{- 1} K^{- 1})$

A

12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 4
We know $K = A e^{- \frac{E_{a}}{R T}}$

$K_{1} = A e^{- \frac{E_{a 1}}{R T}}$

$K_{2} = A e^{- \frac{E_{a 2}}{R T}}$

$\frac{K_{2}}{K_{1}} = e^{\frac{E_{a 1} - E_{a 2}}{R T}}$

$l n (\frac{K_{2}}{K_{1}}) = \frac{E_{a 1} - E_{a 2}}{R T}$

$=$ $\frac{10 \times 1000}{8.314 \times 300}$

$= 4.00$

The correct option is C.

Suggest Corrections

thumbs-up

0

Similar questions

Q. Two reactions

R_{1}

R_{2}

have identical pre-exponential factors. Activation energy of

R_{1}

exceeds that of

R_{2}

10 K j m o l^{- 1}

K_{1}

K_{2}

are rate constants for reactions

R_{1}

R_{2}

respectively at

300 K

(K_{2} / K_{1})

(R = 8.314 j m o l^{- 1} K^{- 1})

Q. Two reactions

R_{1}

R_{2}

have identical pre-exponential factors. Activation energy of

R_{1}

exceeds that of

R_{2}

10 k J m o l^{- 1}

k_{1}

k_{2}

are rate constants for reactions

R_{1}

R_{2}

respectively at 300 K, then

l n (\frac{k_{2}}{k_{1}})

is equal to :
(R = 8.314

J m o l^{- 1} K^{- 1}

)

Q. Two reactions

R_{1}

R_{2}

have identical pre-exponential factors. Activation energy of

R_{1}

exceeds that of

R_{2}

10 k J m o l^{- 1}

k_{1}

k_{2}

are rate constants for reactions

R_{1}

R_{2}

respectively at 300 K, then

l n (\frac{k_{2}}{k_{1}})

is equal to :
(R = 8.314

J m o l^{- 1} K^{- 1}

)

Q. Two reactions

R_{1}

R_{2}

have identical pre-exponential factors. Actiation energy of

R_{1}

exceeds that of

R_{2}

10 k J^{- 1}

K_{1}

K_{2}

are rate constants for reactions

R_{1}

R_{1}

respectively at 300 K, then In(

K_{1}

K_{2}

) is equal to :-

R = 8.314 J^{- 1} K^{- 1}

Q. For the following reactions, the correct reaction rates

r_{1}, r_{2}, r_{1}^{'}

r_{2}^{'}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Arrhenius Equation

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Arrhenius Equation

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon