Question

For the reaction $H_2+{Br}_2⟶2HBr$, the rate expression is, rate $=K\left[H_2\right]{\left[{Br}_2\right]}^{\frac{1}{2}}$ which statement is true about this reaction?

• A. The reaction is of second order
• B. Order of the reaction is 3/2
• C. The unit of K is ${\sec }^{-1}$
• D. Molecularity of the reaction is 2

Answer 1

The correct option is B Order of the reaction is 3/2

$H_2+{Br}_2⟶2HBrRate=K\left[H_2\right]{\left[{Br}_2\right]}^{\frac{1}{2}}⟶\left(1\right)$

Order of reaction$=$sum of the power of concentration term in rate law.

So, order of reaction$=K{\left[H_2\right]}^1{\left[{Br}_2\right]}^{\frac{1}{2}}1+\frac{1}{2}=\frac{3}{2}$