Question

For a gaseous phase reaction $2 A + B _{2} \rightarrow 2 AB$, the following rate data was obtained at $300\, K$

	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. $0.5 \,mol ^{-1} \min ^{-1}$ litre
• B. $0.8 \,mol ^{-1} \min ^{-1}$ litre
• C. $1.5 \,mole ^{-1} \min ^{-1}$ litre
• D. $2 \,mol ^{-1} \min ^{-1}$ litre

Answer 1

Answer: (B)

$r=\frac{-d(B)_{2}}{d t}=1.8 \times 10^{-3}=K[A]^{x}\left[B_{2}\right]^{y}$
$1.8 \times 10^{-3}=K[0.015]^{x}[0.15]^{y}\,\,\,...(i)$
$1.08 \times 10^{-2}=K[0.09]^{x}[0.15]^{y} \,\,\,...(ii)$
$5.4 \times 10^{-3}=K[0.015]^{x}[0.45]^{y}\,\,\,...(iii)$
(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 0.8 \,mol ^{-1} \min ^{-1} $ litre