Question

The rate of gaseous reaction is given by $K\left[A\right]\left[B\right]$. If the volume of reaction vessel is reduced to 1/4 of initial volume the reaction rate relative to the original rate is?

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $8$
• D. $16$

Answer 1

The correct option is D $16$

As the $rate=K\left[A\right]\left[B\right]$ where $\left[A\right]=concofA$
$\left[B\right]=conc.ofB$

We know that $Concentration=\frac{numberofmoles}{Volume}$

therefore we decrease the volume of vessel to $1/4^{th}$ intial volume the conc of $AandB$ both will increase by $4times$ { conc is inversely proportional to volume } .
so final conc of $A$ will become $=4\left[A\right]$
final conc of $B$ will become $=4\left[B\right]$

$Finalrateofreaction=k.4\left[A\right]\times 4\left[B\right]$
$=16k\left[A\right]\left[B\right]$

Option D is correct.