For a first order reaction, we obtain a straight line with positive slope, what we need to plot?

Options

A. `-log_(10)[A]vst`
B. `-log_(e)[A]vst`
C. `log_(10)[A]vs log t`
D. `[A] vst `

Answer

B

Explanation

Rate of first order reaction depends on the concentration raised to the power 1.
Rate `=(d[A])/(t)=-k[A]`
Negative sign shows reactant A is consumed with time.
`(d[A])/([A])=-kdt`
Now integrate
`ln[A]=-kt+c`
at `t=0`, `[A]=[A]_(0)` initial concentration.
`ln[A]=-kt+ln[A]_(0)`
If we compare the above equation with
`y=mx+c`
the graph has negative slope equal to `-k`
So, multiply both sides by `-1`.
`-ln[A]=kt-ln[A]_(0)`
Now the slope is positive. So we have to plot `-ln[A]` vst i.e., `-log_(e)[A]`vst.