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Figure shows a graph in log K vs 1/T whe...

Figure shows a graph in log K vs `1/T` where K is rate constant and T is temperature. The straight line BC has slope `tan theta=1/2.303` and an intercept of 5 on y-axis. Thus `E_(a)` the energy of activation is ………………………Cal.

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The correct Answer is:
2
`logK=logA-((E_(a))/(2.303R))(1/T)impliesE_(a)=R=2` Cal
`30^(@)Cto40^(2)Cto50^(@)Cto60^(@)Cto70^(@)Cto80^(@)C`
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