Question

For the decomposition of a compound $AB$ at $600K$. the following data were obtained .The order for the decomposition of $AB$ is

$[AB]mold{m}^{-3}$	Rate of decomposition of AB in mol $d{m}^{-3}{s}^{-1}$
0.20	$2.75\times 10^{-8}$
0.40	$11.0\times 10^{-8}$
0.60	$24.75\times 10^{-8}$

• A. 1
• B. 2
• C. 1.5
• D. 0

Answer 1

Answer: (B)

$AB⟶$ Product

Rate $=k[AB{]}^n$

$(\text{ Rate }{)}_1=k[0.20{]}^n=2.75\times 10^{-8}...(i)$

${\text{ (Rate) }}_2=k[0.40{]}^n=11.0\times 10^{-8}...(ii)$

Dividing Eq. (ii) by Eq. (i),

$\frac{(\text{ Rate }{)}_2}{(\text{ Rate }{)}_1}=\frac{k[0.40{]}^n}{k[0.20{]}^n}=\frac{11.0\times 10^{-8}}{2.75\times 10^{-8}}$

$2^n=4$

Or $2^n=2^2$

Hence, $n=2$