Question

For a reaction that has $\left[A\right]$ and $\left[B\right]$ both doubled, the rate goes up by a factor of $8$. The rate law is: $rate=k\left[A{]}^m\right[B{]}^2$.
What is the rate order with respect to $\left[A\right]$?

• A. Third order
• B. Second order
• C. First order
• D. Can not be determined with the given information

Answer 1

Answer: (C)

First order

$R=Rate=K{\left[A\right]}^n{\left[B\right]}^2⟶\left(i\right)$

When Rate$=8R,{[A}_t]=[2A],[B_t]=[2B]\therefore 8R=K{\left[2A\right]}^n{\left[2B\right]}^2\therefore 8R=4K{\left[2A\right]}^n{\left[B\right]}^2⟶(ii)$

Dividing (i) by (ii),

$\frac{1}{8}=\frac{1}{4}\times \frac{1}{2^n}\times \frac{{\left[A\right]}^n}{{\left[A\right]}^n}\therefore \frac{1}{2}=\frac{1}{2^n}\therefore m=1$

$\therefore$ Rte law is, Rate$=K{\left[A\right]}^1{\left[B\right]}^2$

$\therefore$ Rate order with respect to [A] is 1.