Question

For a gaseous phase reaction $2A+B_2\to 2AB$, the following rate data was obtained at $300K$

	Rate of disappearance of $B_2$ (mole/litre min)	Concentration
		[A]	[B]
i	$1.8\times 10^{-3}$	$0.015$	$0.15$
ii	$1.08\times 10^{-2}$	$0.09$	$0.15$
iii	$5.4\times 10^{-3}$	$0.015$	$0.45$

The rate constant for the reaction is

• A. $0.5mo{l}^{-1}{\min }^{-1}$ litre
• B. $0.8mo{l}^{-1}{\min }^{-1}$ litre
• C. $1.5mol{e}^{-1}{\min }^{-1}$ litre
• D. $2mo{l}^{-1}{\min }^{-1}$ litre

Answer 1

Answer: (B)

$r=\frac{-d(B{)}_2}{dt}=1.8\times 10^{-3}=K[A{]}^x{\left[B_2\right]}^y$
$1.8\times 10^{-3}=K[0.015{]}^x[0.15{]}^y...(i)$
$1.08\times 10^{-2}=K[0.09{]}^x[0.15{]}^y...(ii)$
$5.4\times 10^{-3}=K[0.015{]}^x[0.45{]}^y...(iii)$
(i)/(ii) divide
$\frac{1.8\times 10^{-3}=K[0.015{]}^x[0.15{]}^y}{1.08\times 10^{-2}=K[0.09{]}^x[0.15{]}^y}$
$(0.16)=[0.16{]}^x$
i.e., $[x=1]$
(i)/(iii) divide
$\frac{1.8\times 10^{-3}=K[0.015{]}^x[0.15{]}^y}{5.4\times 10^{-3}=K[0.015{]}^x[0.45{]}^y}$
$0.33=(0.33)y$
$\therefore$ i.e., $y=1$
$1.8\times 10^{-3}=K[0.015{]}^1[0.15{]}^1$
$K\Rightarrow \frac{1.8\times 10^{-3}}{0.00225}$
$\Rightarrow 0.8mo{l}^{-1}{\min }^{-1}$ litre