A certain reaction rate increases 1000 folds in the presence of a cataylst at 27-C- The activation energy of the original pathway is 98 kJ-mol- What is the activation energy of the new pathway-

The correct option is B 80.77 kJ Given that rate increase by 1000 tines #160; ⇒K_2K_1=1000 logK_2K_1 = log 1000=3 3= E_1 E_22.303RT ...

116

You visited us 116 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Activation Energy

A certain rea...

Question

A certain reaction rate increases $1000$ folds in the presence of a cataylst at $27^{\circ} C$ . The activation energy of the original pathway is $98 k J / m o l$ . What is the activation energy of the new pathway?

80.77 k J

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

56.38 k J

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

24.67 k J

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

90.34 k J

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

Open in App

Solution

The correct option is B $80.77 k J$
Given that rate increase by 1000 tines

$\Rightarrow \frac{K_{2}}{K_{1}} = 1000$

$log \frac{K_{2}}{K_{1}} = log 1000 = 3$
$3 = \frac{E_{1} - E_{2}}{2.303 R T}$
$3 = \frac{98000 - E_{2}}{2.303 \times 8.314 \times 300}$
$E_{2} (c a t a l y s t) = 80.77 k J$ .

Suggest Corrections

Similar questions

Q. In presence of a catalyst, the activation energy is lowered by 3 kcal at

27^{o} C

. Hence, the rate of reaction will increase by:

Q. The catalyst provides a new pathway involving lower amount of activation energy, to make the reaction fast. Which of the following is/are the result of catalytic action?

Mark the incorrect statements.
(a)Catalyst provides an alternative pathway to reaction mechanism
(b)Catalyst raises the activation energy
(c)Catalyst lowers the activation energy
(d)Catalyst alters enthalpy change of the reaction

Q. In the presence of a catalyst, activation energy of a reaction is lowered by 2 kcal at

27^{\circ} C

. Hence, rate will be:

Q. The energy of activation for a reaction is

100

k J

{m o l}^{- 1}

. Presence of a catalyst lowers the energy of activation by

75

%. What will be the effect on rate of reaction at

20^{o} C

, other things being equal?

Join BYJU'S Learning Program

A certain reaction rate increases 1000 folds in the presence of a cataylst at 27∘C. The activation energy of the original pathway is 98 kJ/mol. What is the activation energy of the new pathway?

The correct option is B 80.77 kJGiven that rate increase by 1000 tines ⇒K2K1=1000logK2K1=log1000=33=E1−E22.303RT3=98000−E22.303×8.314×300E2(catalyst)=80.77 kJ.

A certain reaction rate increases $1000$ folds in the presence of a cataylst at $27^{\circ} C$ . The activation energy of the original pathway is $98 k J / m o l$ . What is the activation energy of the new pathway?

The correct option is B $80.77 k J$
Given that rate increase by 1000 tines

$\Rightarrow \frac{K_{2}}{K_{1}} = 1000$

$log \frac{K_{2}}{K_{1}} = log 1000 = 3$
$3 = \frac{E_{1} - E_{2}}{2.303 R T}$
$3 = \frac{98000 - E_{2}}{2.303 \times 8.314 \times 300}$
$E_{2} (c a t a l y s t) = 80.77 k J$ .