Question

For the zero-order decomposition of HI on a gold surface $$\mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{I}_{2}(g)$$ it takes \(16.0 \mathrm{~s}\) for the pressure of HI to drop from \(1.00\) atm to \(0.200 \mathrm{~atm}\). (a) What is the rate constant for the reaction? (b) How long will it take for the pressure to drop from \(0.150 \mathrm{~atm}\) to \(0.0432\) atm? (c) What is the half-life of \(\mathrm{HI}\) at a pressure of \(0.500 \mathrm{~atm}\) ?

Short answer

Question: Calculate (a) the rate constant for the reaction, (b) the time required for the pressure to drop from 0.150 atm to 0.0432 atm, and (c) the half-life of HI at 0.500 atm. Answer: (a) The rate constant (k) is -0.0500 atm/s. (b) The time required for the pressure to drop from 0.150 atm to 0.0432 atm is 2.136 seconds. (c) The half-life of HI at 0.500 atm is 5.00 seconds.

Step-by-step solution

Calculate the rate constant (k)

Using the given data, we can calculate the rate constant (k) as follows: $$k=\frac{P_{final} - P_{initial}}{time}$$ $$k=\frac{0.200 \ \text{atm} - 1.00 \ \text{atm}}{16.0 \ \text{s}}$$ $$k = -\frac{0.8 \ \text{atm}}{16.0 \ \text{s}} = -0.0500 \ \frac{\text{atm}}{\text{s}}$$ Note that the rate constant is negative because the pressure is decreasing.

Calculate the time for the pressure to drop from 0.150 atm to 0.0432 atm

Using the rate constant we just found, we can find the time needed for the pressure to drop from 0.150 atm to 0.0432 atm: $$time = \frac{P_{final}-P_{initial}}{k}$$ $$time = \frac{0.0432\ \text{atm} - 0.150\ \text{atm}}{-0.0500\ \frac{\text{atm}}{\text{s}}}$$ $$time = \frac{-0.1068\ \text{atm}}{-0.0500\ \frac{\text{atm}}{\text{s}}} = 2.136\ \text{s}$$

Calculate the half-life of HI at a pressure of 0.500 atm

To find the half-life, we need to calculate the time that is required for the pressure to become half of the initial pressure: $$P_{initial}=0.500 \ \text{atm}$$ $$P_{final}=\frac{1}{2} P_{initial} = 0.250\ \text{atm}$$ Using the rate constant, we can calculate the half-life (t): $$t = \frac{P_{final} - P_{initial}}{k}$$ $$t = \frac{0.250 \ \text{atm} - 0.500 \ \text{atm}}{-0.0500 \ \frac{\text{atm}}{\text{s}}}$$ $$t = \frac{-0.250\ \text{atm}}{-0.0500\ \frac{\text{atm}}{\text{s}}} = 5.00\ \text{s}$$ So, the half-life of HI at a pressure of 0.500 atm is 5.00 seconds.

Key concepts

Rate Constant Calculation

Understanding the rate constant (k) is crucial for analyzing the kinetics of a chemical reaction. In zero-order kinetics, the reaction rate is constant and does not depend on the concentration of the reactants. This means that the rate of reaction is equal to the rate constant itself.

When tackling an exercise involving zero-order reaction kinetics, such as the decomposition of hydrogen iodide (HI) on a gold surface, we can calculate the rate constant using the formula:

\[k = \frac{P_{\text{final}} - P_{\text{initial}}}{\text{time}}\]

This equation simplifies the process of finding k by incorporating the change in pressure over time, as pressure can be used as a proxy for concentration in gaseous reactions. The negative sign in the calculation of the rate constant indicates a decrease in pressure as the reaction proceeds, which aligns with the consumption of HI.

For students, it is essential to note the specific conditions under which the reaction occurs, such as the substance being on a gold surface, as this might affect reaction kinetics. In practical scenarios, the value of k can help predict how quickly a reaction will reach completion under constant conditions, indicating its efficiency and potential for industrial applications.

Pressure Change in Reaction

Pressure plays a significant role when dealing with gases in chemical reactions. For a gaseous reactant like HI undergoing a decomposition reaction, the reaction's progress can be monitored through changes in pressure. In our example, as HI decomposes, the pressure it exerts decreases.

In zero-order reactions, since the rate is independent of the concentration of the reactants, you may expect a linear relationship between pressure and time. This allows us to predict the pressure at any given point in time, as demonstrated in the calculation of the time it takes for the pressure of HI to drop from an initial value to some final value:

\[\text{time} = \frac{P_{\text{final}} - P_{\text{initial}}}{k}\]

The absolute value of the rate constant controls the slope of the pressure versus time plot, which is a straight line for a zero-order reaction. This relationship is invaluable for industries where control of reaction conditions is vital, such as in the pharmaceutical industry where reactant concentrations must be precisely managed to maintain quality control.

Reaction Half-life

The half-life of a reaction, usually denoted as t\textsubscript{1/2}, is the time required for the quantity of a reactant to reach half of its initial value. In zero-order kinetics, the half-life is dependent on the initial concentration or pressure and can be calculated using the rate constant, as seen in our textbook exercise.

To find the half-life for the given reaction, we set the final pressure to be half of the initial pressure and use the formula where the negative sign takes into account the decreasing pressure:

\[t = \frac{P_{\text{final}} - P_{\text{initial}}}{k}\]
Understanding the half-life concept is advantageous because it provides insight into the timescale of the reaction under study. This is particularly important in fields like pharmacokinetics, where the half-life determines how frequently a drug should be administered to maintain its therapeutic effect. In the classroom, mastering half-life calculations equips students with a better grasp of reaction dynamics and helps them to apply these principles across various disciplines.