Question

If decomposition of hydrogen peroxide is a first order reaction, it's rate law equation can be represented as

• A. $r = k [\mathrm{H_2O_2}]^2$
• B. $r = k [\mathrm{H_2O_2}]$
• C. $r = k [\mathrm{H_2O}][\mathrm{O_2}]^{1/2} / [\mathrm{H_2O_2}]$
• D. $r = k [\mathrm{H_2O_2}][\mathrm{H_2O_2}][\mathrm{O_2}]^{1/2}$

Hint:

Never rely blindly on balanced stoichiometric coefficients to write a rate law unless the reaction is explicitly stated to be elementary. Always use the experimental order given in the text of the problem!

Answer 1

The Correct Option is B

Step 1: Recall the meaning of order.
The rate law writes the rate as the rate constant times reactant concentrations raised to their orders. For a first-order reactant the exponent is $1$.
Step 2: Write the decomposition.
\[ 2\,\mathrm{H_2O_2(aq)} \rightarrow 2\,\mathrm{H_2O(l)} + \mathrm{O_2(g)}. \]
Step 3: Separate order from stoichiometry.
The coefficient $2$ in the balanced equation does not set the order; order is found from experiment, and here it is given as first order.
Step 4: Build the rate law.
First order in $\mathrm{H_2O_2}$ means the exponent on $[\mathrm{H_2O_2}]$ is $1$: \[ r = k\,[\mathrm{H_2O_2}]. \]
Step 5: Reject the distractors.
A square term would be second order, and the fraction-containing forms invent products and partial orders not stated.
Step 6: Conclude.
The rate law is $r = k\,[\mathrm{H_2O_2}]$, option (B).
\[ \boxed{r = k\,[\mathrm{H_2O_2}]} \]