Chapter 16, Problem 59A

Interpretation Introduction

Interpretation:

The time required to produce 12 mL of oxygen gas at 288K needs to be determined.

Concept introduction:

Rate of a chemical reaction is the change in concentration of a reactant or product per unit of time, expressed as mol/(L.s). Different factors like nature of reactants, concentration, surface area and temperature affects the rate of the reaction. Rate of the reaction increases with increasing temperature.

Answer to Problem 59A

Time required to yield 12 mL of oxygen at 288K is $200\text{s}$.

Explanation of Solution

Consider the graph of the relative reaction rate and temperature of a reaction. The data marks are given as (Temperature, rate).

The given data points from the graph is as follows: (290K, 2), (310K, 8), (320K, 16) and (330K, 32). The temperature is raised by 10K. Therefore, it is clear that the reaction rate should double from 2 to 4.

At temperature 298K, $3\%$ of hydrogen peroxide decomposes to yield 12 mL of oxygen gas in 100 s. When the same experiment runs by adding temperature by 10K that means at 308 K, according to the graph the rate should be doubled.

Since At temperature 298K 12 mL oxygen gas is produced. Therefore,

At 308 K, $2\times 12=24\text{mL}$ oxygen gas is produced.

Since the reaction rate is doubled for every 10K increases in temperature.

Consider the reactant rate at 288 K. At a temperature of 298K, to produce 12 mL oxygen it takes 100 s. Since the reaction rate is halved for a decrease in temperature of 10K. Therefore, at 288K it takes twice as long to produce 12 mL of oxygen.

Therefore, the time required to yield 12 mL of oxygen at 288K is $2\times 100=200\text{s}$.

Conclusion

The time required to yield 12 mL of oxygen at 288K is $200\text{s}$.