Two reactions, A→ Products and B→ Products, have rate constants kA and kB at temperature T and activation energies EA and EB respectively. If kA>kB and EA<EB and assuming that collision frequency for both the reactions is same, then:
A
At higher temperature kA will be greater than kB
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B
At lower temperature kA and kB will be close to each other in magnitude
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C
As temperature rises, kA and kB will be close to each other in magnitude
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D
At lower temperature kA>kB
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Solution
The correct options are A At higher temperature kA will be greater than kB D As temperature rises, kA and kB will be close to each other in magnitude For both reactions:
kA=AAe−Ea/RTkB=ABe−E′a/RTAA=AB=A
Since Ea<E′a thus even at higher temperature kA>kB.
But kA and kB will converge at limT→∞ where kA=kB=A.