Question

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

	Column II
(i)	Third order
(ii)	First order
(iii)	Zero order
(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. $\left(A\right)\to \left(iii\right),\left(B\right)\to \left(ii\right),\left(C\right)\to \left(iv\right),\left(D\right)\to (iv$
• 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

The correct option is B $\left(A\right)\to \left(iii\right),\left(B\right)\to \left(ii\right),\left(C\right)\to \left(iv\right),\left(D\right)\to (iv$

Formula that relate concentration and time is:-

$Kt=\frac{1}{n-1}[\frac{1}{{Ao}^{n-1}}-\frac{1}{{At}^{n-1}}]$,

$t\propto {\frac{1}{Ao}}^{n-1}$

For zero order reaction, this equation will be

$At=Ao-Kt$

$Y=+c-mx$ so the first is zero order reaction graph,

$\left[slopeisnegative\right]$

For first order reaction, this equation will be

$log\left[At\right]=log\left[Ao\right]-Kt$

$Y=+c-mx$ so the second is first order reaction graph,

$\left[slopeisnegative\right]$

For $C$ and $D$ there is direct relation between time and concentration

$t\propto {\frac{1}{Ao}}^{n-1}$

for $C$ power of {A} is $-1$ that means $1-n=-1$

$Soorder\left(n\right)=1+1=2$

for $D$ power of {A} is $-2$that means $1-n=-2$

$Soorder\left(n\right)=2+1=3$.

So the correct option is [B].