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 $C_t\to$ concentration of reactant at time $t,C_o\to$ initial concentration of reactant]
$1n\left(\frac{C_o}{C_t}\right)=1n{C}_o-1n{C}_t=kt$
1n $(C_t)=-kt+1n(C_o)$
If $(C_t)=[R]=$ concentration of reactant
$1n[R]=-kt+1n(C_o)$
On comparing with $y=\text{mx}+C$
$\Rightarrow$ Slope $m=-k$ for graph between 1n (R) and t with intercept 1n $(C_o)$
$\Rightarrow$ First graph is of first order
Integrated rate law for zero order reactions is $\left[C_t\right]=\left[C_o\right]-kt$ or $[C_t]=-kt+[C_o]$ or $[R]=-kt+[C_o](C_t=R$ 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.