The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be Increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.
A
9.84 K
B
4.92 K
C
2.45 K
D
19.67 K
Explanation
For reaction A, T_{1} = 300 K, T_{2} = 310 K, k_{2} = 2 k_{1}
$$ \therefore $$ Increased temperature = (304.92 – 300) = 4.92 K
2
MCQ (Single Correct Answer)
JEE Main 2017 (Online) 9th April Morning Slot
The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :
(Assume activation energy and preexponential factor are independent of
temperature; ln 2 = 0.693; R = 8.314 J mol^{−1} K^{−1})
A
107.2 kJ mol^{$$-$$1}
B
53.6 kJ mol^{$$-$$1}
C
26.8 kJ mol^{$$-$$1}
D
214.4 kJ mol^{$$-$$1}
3
MCQ (Single Correct Answer)
JEE Main 2018 (Offline)
At 518^{o}C the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr,
was 1.00 Torr s^{–1} when 5% had reacted and 0.5 Torr s^{–1} when 33% had reacted. The order of the reaction is
A
0
B
2
C
3
D
1
Explanation
For a n^{th} order reaction, the rate of reaction at time t ,
Rate = K [P_{t}] ^{n}
Here P_{t} = pressure at time t, k = constant.
Note :
Here instead of concentration of product, pressure of product is given.
When 5% is reacted at a rate 1 Toss S^{$$-$$1}
Then un-reacted is 95%..
As initial pressure is 363 Torr
then after 5% reaction completed the pressure will be
N_{2}O_{5} decomposes to NO_{2} and O_{2} and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be :