You visited us 3 times! Enjoying our articles? Unlock Full Access!

Arrhenius Equation

Slope of stra...

Question

Slope of straight line in graph-B is -6670, the activation energy for the reaction is :

127.71 J

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

127.71 k J

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

12.771 J

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

12.771 k J

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

Open in App

Solution

The correct option is C $127.71 k J$
The expression for the slope of the graph -
$S l o p e = - \frac{E_{a}}{2.303 R} = - 6670.$

Hence, the expression for the activation energy becomes $E_{a} = 6670 \times 2.303 R = 6670 \times 2.303 \times 8.314 = 127710 J = 127.71 k J .$

Suggest Corrections

Similar questions

l o g k v e r s u s \frac{1}{T}

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}

- 5000.

(k J k^{- 1} m o l^{- 1})

l o g k

\frac{1}{T}

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}

- 5000.

(k J k^{- 1} m o l^{- 1})

l o g k v e r s u s \frac{1}{T}

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}

- 5000.

(k J K^{- 1} m o l^{- 1})

k

log k = c o n s t a n t - \frac{E_{a}}{2.303 R} \times \frac{1}{T}

E_{a}

log k

\frac{1}{T}

- 6670 k

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

log k

\frac{1}{T}

- 8000 K

Join BYJU'S Learning Program