Question

The given plots represent the variation of the concentration of a reactant R with time for two different reaction (i) and (ii) The respective orders of the reactions are

• A. 1, 0
• B. 1, 1
• C. 0, 1
• D. 0, 2

Answer 1

Answer: (A)

Integrated rate law for first order reaction is
$k=\frac{1}{t}1n\left(\frac{C_{o}}{C_{t}}\right)$
[Where $C_{t} \rightarrow $ concentration of reactant at time $t,C_{o} \rightarrow $ initial concentration of reactant]
$1n\left(\frac{C_{o}}{C_{t}}\right)=1nC_{o}-1nC_{t}=kt$
1n $\left(\right.C_{t}\left.\right)=-kt+1n\left(\right.C_{o}\left.\right)$
If $\left(\right.C_{t}\left.\right)=\left[\right.R\left]\right.=$ concentration of reactant
$1n\left[\right.R\left]\right.=-kt+1n\left(\right.C_{o}\left.\right)$
On comparing with $y=\text{mx}+C$
$\Rightarrow $ Slope $m=-k$ for graph between 1n (R) and t with intercept 1n $\left(\right.C_{o}\left.\right)$
$\Rightarrow $ First graph is of first order
Integrated rate law for zero order reactions is $\left[C_{t}\right]=\left[C_{o}\right]-kt$ or $\left[\right.C_{t}\left]\right.=-kt+\left[\right.C_{o}\left]\right.$ or $\left[\right.R\left]\right.=-kt+\left[\right.C_{o}\left]\right.\left(\right.C_{t}=R$ given)
On comparing with $y=\text{mx}+C$
$\Rightarrow $ slope $m=-k$ for graph between (R) and t with intercept $C_{o}$
$\Rightarrow $ Second graph is of zero order reactions.