Rate constant k of the first order reaction when initial concentration C0 and concentration Ct at time t is given by equation k t= log C0- log Ct Graph is a straight line if we plot:

Question

Rate constant k of the first order reaction when initial concentration C0 and concentration Ct at time t is given by equation k t= log C0- log Ct Graph is a straight line if we plot: (A) t vs log C0 (B) t vs log Ct (C) t-1 vs log Ct (D)  log C0 vs log Ct

Tardigrade Admin · Accepted Answer

K t=- log e Ct+ log e Co t =-(1/K) log e Ct+ log e Co y =m x+ c straight line, - ve slope

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

$t$ vs $lo g C_{0}$

$t$ vs $lo g C_{t}$

$t^{- 1}$ vs $lo g C_{t}$

$lo g C_{0}$ vs $lo g C_{t}$

$K t = - lo g_{e} C_{t} + lo g_{e} C_{o}$

$t = - \frac{1}{K} lo g_{e} C_{t} + lo g_{e} C_{o}$

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

Q. Rate constant k of the first order reaction when initial concentration C0​ and concentration Ct​ at time t is given by equation kt=logC0​−logCt​ Graph is a straight line if we plot:

t vs logC0​

t vs logCt​

t−1 vs logCt​

logC0​ vs logCt​