Question

(a) A reaction is second order in $A$ and first order in $B$.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of $A$ three times?
(iii) How is the rate affected when the concentrations of both $A$ and $B$ are doubled?
(b) A first order reaction 40 minutes for 30% decomposition. Calculate $t_{1/2}$ for this reaction.
(Give log 1.428 = 0.1548)

Answer 1

Given that
Order of reactant $A$, $x$$=$$2$
Order of reactant $B$,$y$$=$$1$
The differential rate equation will be

$-\frac{d\left[R\right]}{dt}=k\left[A{]}^x\right[B{]}^y$
Plug the value in this formula
$-\frac{d\left[R\right]}{dt}$$=$$k\left[A{]}^2\right[B]$
Given that concentration of $A$ is increased three times
So that
New concentration of $A$ $=$3[A]
Put the values we get new rate of reaction
$-\frac{d\left[R\right]}{dt}$$=$$k\left[3A{]}^2\right[B]$
Hence, the rate of reaction will increase by 9 times.
(iii)
Given that
New concentrations of $A=2\left[A\right]$
New concentration of $B=2\left[B\right]$
Plug the values we get
$-\frac{d\left[R\right]}{dt}$$=$$k\left[2A{]}^2\right[2B]$
Hence the rate of reaction increased by 8 times

(b) $k=\frac{1}{t}ln\frac{100}{100-30}$

$=\frac{1}{40}\times 0.1548$

$=0.00387mi{n}^{-1}$