Question

Match the graphs given in column I with the order given in column II and mark the appropriate choice.

	Column I		Column II
(A)		(i)	Third order
(B)		(ii)	First order
(C)		(iii)	Zero order
(D)		(iv)	Second order

• A. $\left(A\right)$ $\to$ $\left(i\right)$, $\left(B\right)$ $\to$ $\left(ii\right)$, $\left(C\right)$ $\to$ $\left(iii\right)$, $\left(D\right)$ $\to$ $\left(iv\right)$
• B. $\left(A\right)$ $\to$ $\left(iii\right)$, $\left(B\right)$ $\to$ $\left(ii\right)$, $\left(C\right)$ $\to$ $\left(iv\right)$, $\left(D\right)$ $\to$ $\left(i\right)$
• C. $\left(A\right)$ $\to$ $\left(ii\right)$, $\left(B\right)$ $\to$ $\left(i\right)$, $\left(C\right)$ $\to$ $\left(iii\right)$, $\left(D\right)$ $\to$ $\left(iv\right)$
• D. $\left(A\right)$ $\to$ $\left(iv\right)$, $\left(B\right)$ $\to$ $\left(iii\right)$, $\left(C\right)$ $\to$ $\left(i\right)$, $\left(D\right)$ $\to$ $\left(ii\right)$

Answer 1

Answer: (B)

Formula that relate concentration and time is:-
$Kt=\frac{1}{n-1}\left[\frac{1}{A{o}^{n-1}}-\frac{1}{A{t}^{n-1}}\right],$
$t\propto {\frac{1}{Ao}}^{n-1}$
For zero order reaction, this equation will be
$At=Ao-Kt$
$Y=+c-mx$ so the first is zero order reaction graph,
[slope is negative]
For first order reaction, this equation will be
$\log [At]=\log [Ao]-Kt$
$Y=+c-mx$ so the second is first order reaction graph,
[slope is negative]
For $C$ and $D$ there is direct relation between time and concentration
$t\propto \frac{1}{Ao}$
for $C$ power of $\{A\}$ is $-1$ that means $1-n=-1$
So order $(n)=1+1=2$
for $D$ power of $\{A\}$ is $-2$ that means $1-n=-2$
So order $(n)=2+1=3$.