Question

Fill in the blanks in the following table which treats reaction of a compound $A$ with a compound $B$, that is of the first-order with respect to $A$ and zero-order with respect to $B$.

Expt. No.	$\left[A\right]$ $\left(mol{L}^{-1}\right)$	$\left[B\right]$ $\left(mol{L}^{-1}\right)$	Initial rate $\left(mol{L}^{-1}{s}^{-1}\right)$
$1.$	$0.1$	$0.1$	$2.0\times {10}^{-2}$
$2.$	___	$0.2$	$4.0\times {10}^{-2}$
$3.$	$0.4$	$0.4$	___
$4.$	___	$0.2$	$2.0\times {10}^{-2}$

Answer 1

Rate $=K{\left[A\right]}^1{\left[B\right]}^0$

${\left(\frac{0.1}{x}\right)}^1{\left(\frac{0.1}{0.2}\right)}^0=\frac{2\times {10}^{-2}}{4\times {10}^{-2}}$

$0.1\times 2=x$

i) $x=0.2$

II) ${\left(\frac{0.1}{0.4}\right)}^1{\left(\frac{0.1}{0.4}\right)}^0=\left(\frac{2\times {10}^{-2}}{R}\right)$

$\frac{0.1}{0.4}=\frac{2\times {10}^{-2}}{R}$

$R=\frac{2\times {10}^{-2}\times 0.4}{0.1}=8\times {10}^{-2}mol{l}^{-1}{S}^{-1}$

iii) ${\left(\frac{0.1}{y}\right)}^1=\frac{2\times {10}^{-2}}{2\times {10}^{-2}}$

$y=0.1$