# For a first order reaction , the plot of t against log C gives a straight with slope equal to :

$(A)\;\frac{K}{2.303} \\ (B)\;\frac{-K}{2.303} \\(C)\;\frac{ln K}{2.303} \\(D)\;\frac{-2.303}{K}$

For a first order reaction , $K_t=2.303 \log \bigg( \large\frac{a}{a-x}\bigg)$
or $t= \large\frac{2.303}{K}$$\log a - \large\frac{2.303 }{K}$$ \log (a-x)$
$y=c+mx$
Thus slope $m= \large\frac{-2.303}{K}$
If $\log (a-x)=\log a -\large\frac{Kt}{2.303}$
Then slope is $\large\frac{-K}{2.303}$
Hence b is the correct answer.