Question

For a particular $A+B\to C$ was studied at ${25}^{\circ }C$. The following results are obtained.

[A]	[B]	[C]
(mole/lit)	(moles/lit)	(mole lit ${}^{-1}se{c}^{-2}$)
$9\times {10}^{-5}$	$1.5\times {10}^{-2}$	$0.06$
$9\times {10}^{-5}$	$3\times {10}^{-3}$	$0.012$
$3\times {10}^{-5}$	$3\times {10}^{-3}$	$0.004$
$6\times {10}^{-5}$	x	$0.024$

Then the value of x is :

• A. $6\times {10}^{-3}moleslitr{e}^{-1}$
• B. $3\times {10}^{-3}moleslitr{e}^{-1}$
• C. $4.5\times {10}^{-3}moleslitr{e}^{-1}$
• D. $9\times {10}^{-3}moleslitr{e}^{-1}$

Answer 1

Answer: (D)

$9\times {10}^{-3}moleslitr{e}^{-1}$

$A+B\to C$

$rate=k\left[A\right]\left[B\right]$

$Experiment3\&2chosenforvalueofkas\left[B\right]issameinboth$

$\frac{r_3}{r_2}=\frac{0.004}{0.012}=k\frac{\left[{3\times 10}^{-5}\right]\left[{3\times 10}^{-3}\right]}{\left[{9\times 10}^{-5}\right]\left[{3\times 10}^{-3}\right]}$

$k=1usingthisrateconstantvalueinfindingx$

$\frac{r_4}{r_3}=\frac{0.024}{0.004}=k\frac{\left[{6\times 10}^{-5}\right]\left[x\right]}{\left[{3\times 10}^{-5}\right]\left[{3\times 10}^{-3}\right]}$

$\left[x\right]={9\times 10}^{-3}$