1.

The correct nature of plot for first order reaction is (are):A. B. C. D.

Answer» Correct Answer - A::D
`k = (2.303)/(t)log.(c_(0))/(c )`
`(k.t)/(2.303) = log c_(0) - log c`
`log c = (-(k)/(2.303))t + log c_(0)`
Thus, plot of `log c` vs `t` is a straight with negative slope.
`t_(1//2) = x` should remain constant.
`-(dc)/(dt) = k.c`,
`log(-(dc)/(dt)) = log k + log c`
So, the plot of `log(-(dc)/(dt))` vs `log c` should be a straight line of the slope equal to unity.


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