Question

Select the rate law that conforms to the following data for the reaction : $A+B\to Products$
$\begin{array}{cccc}ExpNo.& \left[A\right]\left(mol{L}^{-1}\right)& \left[B\right]\left(mol{L}^{-1}\right)& Initialrate\left(mol{L}^{-1}{S}^{-1}\right)\\ 1& 0.0012& 0.0035& 1\times {10}^{-4}\\ 2& 0.024& 0.070& 8\times {10}^{-3}\\ 3& 0.024& 0.035& 1\times {10}^{-3}\\ 4& 0.012& 0.070& 8\times {10}^{-3}\end{array}$

• A. $-\frac{d\left[A\right]}{dt}=k\left[A\right][B{]}^2$
• B. $-\frac{d\left[A\right]}{dt}=k\left[A{]}^2\right[B]$
  
• C. $-\frac{d\left[A\right]}{dt}=k\left[A\right][B{]}^3$
• D. $-\frac{d\left[A\right]}{dt}=k[B{]}^3$

Answer 1

The correct option is D

$-\frac{d\left[A\right]}{dt}=k[B{]}^3$

Rate = $k\left[A{]}^x\right[B{]}^y$ ; From exp 2 & 3.
$\frac{(rate{)}_2}{(rate{)}_3}=k\frac{\left[0.024{]}^x\right[0.070{]}^y}{\left[0.024{]}^x\right[0.035{]}^y}=\frac{8\times {10}^{-3}}{1\times {10}^{-3}}=8or{2}^y=8=2^3,y=3$
From exp 2 and 4,
$\frac{(rate{)}_2}{(rate{)}_1}=\frac{8\times {10}^{-3}}{8\times {10}^{-3}}=\frac{k\left[0.024{]}^x\right[0.070{]}^y}{k\left[0.012{]}^x\right[0.070{]}^y}=2^{x}or{2}^x=1\Rightarrow x=0or,-\frac{d\left[A\right]}{dt}=k[B{]}^3$