Question

For the reaction: $3A+B\to C+D$, mechanism is
Step 1: $A+B\stackrel{k_1}{\to }E$
Step 2: $E+2A\stackrel{k_2}{\to }C+D$
What is the total order of reaction considering steady state approximation?

Answer 1

From steady state approximation,$\frac{dE}{dt}=K_1\left[A\right]\left[B\right]-K_2\left[E\right][A{]}^2=0$
$\left[E\right]=\frac{K_1\left[A\right]\left[B\right]}{K_2[A{]}^2}=\frac{K_1}{K_2}\frac{\left[B\right]}{\left[A\right]}$
So, $-\frac{dB}{dt}=K_1\left[A\right]\left[B\right]$
And $-\frac{dA}{dt}=K_1\left[A\right]\left[B\right]+2K_2\left[E\right][A{]}^2$
$=K_1\left[A\right]\left[B\right]+2K_2\times \frac{K_1}{K_2}\frac{\left[B\right]}{\left[A\right]}[A{]}^2$
$=3K_1\left[A\right]\left[B\right]$
$-\frac{1}{3}\frac{dA}{dt}=K_1\left[A\right]\left[B\right]$
So, the order of reaction is $1+1=2$.