Question

Consider the reaction:$3A+B+C\to D+E$
Where the rate law is defined as: $-\frac{\Delta \left[A\right]}{\Delta t}=k\left[A{]}^2\right[B\left]\right[C]$
An experiment is carried out where $[B{]}_o=[C{]}_o=1.00M$ and $[A{]}_o=1.00\times {10}^{-4}M$, Calculate the value of $t_{\frac{1}{2}}$.

Answer 1

As the concentrations [B] and [C] are much greater than [A], this is a pseudo second order reaction, the new rate law is: $r=k^{\prime }[A{]}^2$
The integrated rate law is:
$\frac{1}{C_A}=k^{\prime }t+\frac{1}{C_{A_o}}$
Substituting, we get $k^{\prime }=\frac{2}{5}\times {10}^{4}L/molmin=4000L/molmin$
For calculating $t_{half}$
$\frac{1}{C_{A_o}/2}=4000t+\frac{1}{C_{A_o}}$
$C_{A_o}={10}^{-4}M$
On substituting, we get: $t_{half}=2.5min$