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.
Answers
Answered by
24
Hey dear,
● Answer- 304.92 K
● Explanation-
# Given-
Ta1 = 300 K
Ta2 = 310 K
Tb1 = 300 K
Tb2 = x
Eb = 2 Ea
# Solution-
We have the relation,
Ea × (1/Ta1-1/Ta2) = Eb × (1/Tb1-1/Tb2)
Ea × (1/300-1/310) = 2Ea × (1/300-1/x)
(1/300-1/x) = 1/18600
1/x = 1/300-1/18600
1/x = 61/18600
x = 304.92 K
Temperature of reaction B need to be 304.92 K to double its rate.
Hope this is helpful...
● Answer- 304.92 K
● Explanation-
# Given-
Ta1 = 300 K
Ta2 = 310 K
Tb1 = 300 K
Tb2 = x
Eb = 2 Ea
# Solution-
We have the relation,
Ea × (1/Ta1-1/Ta2) = Eb × (1/Tb1-1/Tb2)
Ea × (1/300-1/310) = 2Ea × (1/300-1/x)
(1/300-1/x) = 1/18600
1/x = 1/300-1/18600
1/x = 61/18600
x = 304.92 K
Temperature of reaction B need to be 304.92 K to double its rate.
Hope this is helpful...
Answered by
4
Answer:
Answer is 4.92
Explanation:
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