Question

Rate constant $k$ of the first order reaction when initial concentration $C_{0}$ and concentration $C_{t}$ at time $t$ is given by equation
$k t=\log C_{0}-\log C_{t}$
Graph is a straight line if we plot:

• A. $t$ vs $\log C_{0}$
• B. $t$ vs $\log C_{t}$
• C. $t^{-1}$ vs $\log C_{t}$
• D. $\log C_{0}$ vs $\log C_{t}$

Answer 1

Answer: (B)

$ K t=-\log _{e} C_{t}+\log _{e} C_{o} $

$ t =-\frac{1}{K} \log _{e} C_{t}+\log _{e} C_{o} $

$ y =m x+ c $ straight line, - ve slope