First slide

Kinetics of chemical reactions

Question

For a reaction A $\to$ B at 27ºC , 1% reactant molecules possess activation energy, the energy of activation in K.cal/ mole is nearly(R = 2 cal K^-1 mol^-1)

Moderate

A

1.38
B

0.6
C

1.2
D

2.76

Solution

$\begin{array}{l} K = A \times e^{- E a / K T} \\ e^{- E a / R T} \to fraction of molecular possessing Ea \\ e^{- E a / R T} = \frac{1}{100} \\ \frac{- E a}{R T} = - 2 \times 2.303 \\ E a = \frac{2 \times 300 \times 2.303 \times 2}{1000} \\ = 4 \times 3 \times 0.2303 = 2.76 K . cal / mole \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App