Question

For a reaction $2A+B$ $\to$ $C+D$, half life of reaction $\left(t_{1/2}\right)$ and seven/eighth life of the reaction ($t_{7/8}$) was calculated. From the given data, the value of $[\frac{b}{a}-\frac{d}{c}]$ is:

($a,b,c$ and $d$ are as mentioned in the table)

Set	$A_0\times {10}^{-2}M$	$B_0\times {10}^{-2}M$	$t_{1/2}$ (min)	$t_{7/8}$ (min)
1	500	2	30	90
2	3	400	10	70
3	6	800	a	b
4	250	1	c	d

• A. 4
• B. 6
• C. 8
• D. 2

Answer 1

The correct option is A 4

Rate of reaction $=K\left[A{]}^{\alpha }\right[B{]}^{\beta }$
Case-1
[A] is in excess.
Rate of reaction $=K^{{}^{\prime }}[B{]}^{\beta }$
$t_{7/8}=3t_{1/2}$
Reaction is first order with respect to [B].
Case-2
[B] is in excess.
Rate of reaction $=K^{\prime \prime }[A{]}^{\alpha }$
$t_{7/8}=7t_{1/2}$
Reaction is second order with respect to [A].
Rate of reaction $=K\left[A{]}^2\right[B{]}^1$
Case-3
[B] is in excess.
Rate of reaction $=\left(K\right[B\left]\right)[A{]}^2$
Rate of reaction $=K^{{}^{\prime \prime \prime }}[A{]}^2$
$t_{1/2}=\frac{1}{K^{{}^{\prime \prime \prime }}{C}_A}=\frac{1}{K\left[B\right]\left[A\right]}$
$\left(t_{1/2}\right)\left[B\right]\left[A\right]=$ constant
${\left(t_{1/2}\right)\left[B\right]\left[A\right]}_{exp\left(ii\right)}={\left(t_{1/2}\right)\left[B\right]\left[A\right]}_{exp\left(iii\right)}$
$10\times 400\times 3=t_{1/2}\times 800\times 6$
$t_{1/2}=$ 2.5 min
$t_{7/8}=$ 17.5 min
Case-4
[A] is in excess.
Rate of reaction $=\left[K\right[A{]}^2\left]\right[B{]}^1$
$t_{1/2}=\frac{ln2}{K[A{]}^2}$
$t_{1/2}[A{]}^2=\frac{ln2}{K}=$ constant
$\left(t_{1/2}\right[A{]}^2{)}_{exp.\left(i\right)}=\left(\right[t_{1/2}\left]\right[A{]}^2{)}_{exp.\left(iv\right)}$
$30\times (500{)}^2=t_{1/2}\times (250{)}^2$
$t_{1/2}=$ 120 min
$t_{7/8}=$ 360 min
$\frac{17.5}{2.5}-\frac{360}{120}=4$