Question

At \(40^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}_{2}(a q)\) will decompose according to the following reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{~g}) $$ The following data were collected for the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) at various times. $$ \begin{array}{|cc|} \hline \begin{array}{c} \text { Time } \\ (\mathbf{s}) \end{array} & \begin{array}{c} {\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]} \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 0 & 1.000 \\ \hline 2.16 \times 10^{4} & 0.500 \\ \hline 4.32 \times 10^{4} & 0.250 \\ \hline \end{array} $$ a. Calculate the average rate of decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) between 0 and \(2.16 \times 10^{4} \mathrm{~s}\). Use this rate to calculate the average rate of production of \(\mathrm{O}_{2}(g)\) over the same time period. b. What are these rates for the time period \(2.16 \times 10^{4} \mathrm{~s}\) to \(4.32 \times 10^{4} \mathrm{~s} ?\)

Short answer

a) The average rate of decomposition of \(H_2O_2\) between 0 and \(2.16 \times 10^4\) seconds is \(-2.31 \times 10^{-5}\, mol/L/s\). The average rate of production of \(O_2\) during this time period is \(1.16 \times 10^{-5}\, mol/L/s\). b) The average rate of decomposition of \(H_2O_2\) between \(2.16 \times 10^4\) and \(4.32 \times 10^4\) seconds is \(-1.16 \times 10^{-5}\, mol/L/s\). The average rate of production of \(O_2\) during this time period is \(5.79 \times 10^{-6}\, mol/L/s\).

Step-by-step solution

Calculate the rate of decomposition of H2O2 in the first time interval

According to the data given: Initial concentration of H2O2: \([H_2O_2]_i = 1.000 \, mol/L\) Initial time: \(t_i = 0\,s\) Final concentration of H2O2: \([H2O2]_f = 0.500 \, mol/L\) Final time: \(t_f = 2.16 \times 10^4 \, s\) Now we can apply the formula for the average rate of reaction: Average rate = \(\frac{[H_2O_2]_f - [H_2O_2]_i}{t_f - t_i}\)

Calculate the average rate of production of O2 in the first time interval

To find the average rate of production of O2, analyze the stoichiometry of the reaction: \(2H_2O_2(aq) \longrightarrow 2H_2O(l) + O_2(g)\) The ratio of H2O2 to O2 according to the reaction coefficients is 2:1. Thus, the rate of production of O2 gas will be half the rate of H2O2 decomposition.

Calculate the rates for the second time interval

Use the same approach to find the rates for the second time interval (2.16 x 10^4 s to 4.32 x 10^4 s). Initial concentration of H2O2: \([H_2O_2]_i = 0.500\, mol/L\) Initial time: \(t_i = 2.16 \times 10^4\,s\) Final concentration of H2O2: \([H2O2]_f = 0.250 \, mol/L\) Final time: \(t_f = 4.32 \times 10^4 \, s\) Apply the formula for the average rate of decomposition and use stoichiometry to calculate the rate of production of O2 in the second time interval, following the same steps as in the first time interval.

Key concepts

Rate of Reaction

The rate of reaction refers to how quickly or slowly the reactants in a chemical reaction form into products. It's a measure of the speed at which the concentration of a reactant or product changes over time. For our exercise on the decomposition of hydrogen peroxide (\(\text{H}_2\text{O}_2\)), the rate of reaction tells us how quickly \(\text{H}_2\text{O}_2\) is breaking down into water and oxygen gas.

• The average rate of reaction can be calculated using the formula:\[ \text{Average rate} = \frac{[\text{H}_2\text{O}_2]_f - [\text{H}_2\text{O}_2]_i}{t_f - t_i}\]

In a practical sense, this means taking the difference in concentration of \(\text{H}_2\text{O}_2\) at two different times, then dividing by the time difference to give a rate in mol/L/s. This tells us how fast the concentration of reactants is changing, and thus, how fast the reaction is occurring. The rates will vary depending on different time intervals as the reaction proceeds.

Stoichiometry

Stoichiometry is a concept that deals with the quantitative relationships of the reactants and products in a chemical reaction. It's all about understanding how much of something is needed or produced based on balanced chemical equations.

• In this decomposition reaction:\(2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2\), the stoichiometry is 2:2:1.
• This means that two moles of \(\text{H}_2\text{O}_2\) decompose to form two moles of water (\(\text{H}_2\text{O}\)) and one mole of oxygen (\(\text{O}_2\)).

Understanding stoichiometry allows us to relate the rates of disappearance of reactants to the rates of appearance of products. In this case, because the ratio of \(\text{H}_2\text{O}_2\) to \(\text{O}_2\) is 2:1, the rate of oxygen gas production is half the rate of hydrogen peroxide decomposition. Stoichiometry ensures we can precisely calculate these relationships.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. It is essentially the opposite of composition or synthesis reactions.

• The reaction of hydrogen peroxide (\(\text{H}_2\text{O}_2\)) decomposing into water and oxygen can be seen as:\(2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2\).
• This is a classic example of a decomposition reaction. The reactant \(\text{H}_2\text{O}_2\) breaks down to form two different products: water (\(\text{H}_2\text{O}\)) and oxygen gas (\(\text{O}_2\)).

Decomposition reactions generally require an input of energy in the form of heat, light, or electricity. In this exercise, the reaction occurs naturally at a moderate temperature of \(40^\circ\text{C}\). Understanding decomposition reactions is essential because they help explain processes such as the decay of organic matter and various industrial applications.