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The rate constant of a reaction is given...

The rate constant of a reaction is given by `k = 2.1 xx 10^(10) exp(-2700//RT)`. It means that

A

`log k vs 1//T` will be a straight line with slope `= - (2700)/(2.303R)`

B

log vs 1/T will be a straight line with intercept on log k axis `= log 2.1 xx 10^(10)`

C

The number of effective collisions are `2.1 xx 10^(10) "cm"^(-2) sec^(-1)`

D

Half-life of the reaction increases with increase of temperature

Text Solution

Verified by Experts

The correct Answer is:
A, B
`kl = 2.1 xx 10^(10) e^(-2700//RT)`
`l nk=l n2.1xx10^(-1)+l n e^(-2700//RT)`
`l nk=l n2.1xx10^(10)-2700//RT`
`2.303 log k=2.303 log 2.1 xx 10^(10) -(2700)/(2.303 RT)`
`log k=(2700)/(2.303R)(1)/(T)+log2.1xx10^(10)c.f`.,
`y=mx+C`
`:. "m (slope)" = -(2700)/(2.303 R)`
and `C = log 2.1 xx 10^(10)`.
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