Question

The decomposition of hydrogen iodide on finely divided gold at \(150^{\circ} \mathrm{C}\) is zero order with respect to HI. The rate defined below is constant at \(1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) $$ \begin{array}{r} 2 \mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t}=k=1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{array} $$ a. If the initial HI concentration was \(0.250 \mathrm{~mol} / \mathrm{L}\), calculate the concentration of HI at 25 minutes after the start of the reaction. b. How long will it take for all of the \(0.250 M\) HI to decompose?

Short answer

a. The concentration of HI after 25 minutes is 0.070 mol/L. b. It takes approximately 34.7 minutes for all of the 0.250 mol/L HI to decompose.

Step-by-step solution

Identify the reaction order and integrated rate law

Since the decomposition of hydrogen iodide (HI) is zero-order with respect to HI, we can write the integrated rate law for this reaction, as follows: $$ [\mathrm{HI}] = -kt + [\mathrm{HI}]_0 $$ where [HI] is the concentration of HI at time t, k is the rate constant, t is time, and [HI]\(_0\) is the initial concentration of HI.

Calculate the concentration of HI at 25 minutes

We are given the initial concentration of HI, which is 0.250 mol/L, and the rate constant, which is 1.20 x 10^{-4} mol/L⋅s. We convert 25 minutes to seconds, and then we can plug these values into our integrated rate law and calculate the concentration of HI at 25 minutes. $$ t = 25\,\text{minutes} \times \frac{60\,\text{s}}{1\,\text{minute}} = 1500\,\text{s} $$ $$ [\mathrm{HI}] = -kt + [\mathrm{HI}]_0 \\ [\mathrm{HI}] = -(1.20 \times 10^{-4}\,\text{mol/L⋅s})(1500\,\text{s}) + (0.250\,\text{mol/L}) \\ [\mathrm{HI}] = 0.070\,\text{mol/L} $$ So, the concentration of HI after 25 minutes is 0.070 mol/L.

Calculate the time required for complete decomposition of HI

To find the time it takes for all of the 0.250 mol/L HI to decompose, we can set the HI concentration at time t to 0 and solve for t using the integrated rate law. $$ 0 = -kt + [\mathrm{HI}]_0 \\ t = \frac{[\mathrm{HI}]_0}{k} \\ t = \frac{0.250\,\text{mol/L}}{1.20 \times 10^{-4}\,\text{mol/L⋅s}} $$ $$ t = 2083.3\,\text{s} $$ Converting this to minutes: $$ t = \frac{2083.3\,\text{s}}{60\,\text{s/min}} \approx 34.7\,\text{minutes} $$ So, it takes approximately 34.7 minutes for all of the 0.250 mol/L HI to decompose.

Key concepts

Rate Constant

The rate constant (k) is a crucial factor in understanding reaction rates. It's a specific value for a given reaction at a certain temperature, influencing how quickly the reaction occurs. The units of the rate constant vary with the order of the reaction. For a zero-order reaction, such as the decomposition of hydrogen iodide, the unit is typically in \(\text{mol/L} \cdot \text{s}\).
The given reaction has a rate constant of \(1.20 \times 10^{-4} \text{ mol/L} \cdot \text{s}\), indicating the rate at which hydrogen iodide decomposes over time. The value remains constant throughout the reaction, regardless of changes in concentration, because zero-order reactions do not depend on reactant concentrations.

Integrated Rate Law

The integrated rate law for zero-order reactions provides a mathematical relationship between reactant concentration and time. It helps predict the concentration of reactants at any point in time. The equation for zero-order reactions is: \[ \[\mathrm{HI}\] = -kt + \[\mathrm{HI}\]_0 \]
Here, \([\mathrm{HI}]_0\) is the initial concentration, \(t\) is the elapsed time, and \(k\) is the rate constant. This linear equation means if you graph concentration versus time, the plot will be a straight line. Zero-order reactions are unique in that the rate of reaction is constant, and the concentration changes uniformly over time.
In the case of hydrogen iodide's decomposition, using this integrated rate law allows us to calculate concentrations at different times or the time required for complete decomposition.

Decomposition of Hydrogen Iodide

The decomposition of hydrogen iodide is an interesting chemical reaction where two molecules of HI break down to form \(\text{H}_2\) and \(\text{I}_2\). The reaction is facilitated by a gold catalyst at a temperature of \(150^{\circ} \text{C}\) and is classified as a zero-order reaction.
Because it is zero-order, the rate of decomposition remains constant throughout the process, independent of any fluctuations in \(\text{HI}\) concentration. This constant rate illustrates that the number of reactant molecules does not interfere with the reaction progress, which is a characteristic of reactions on surfaces or involving catalysts.
Understanding such reactions can be significant in industrial applications where consistent output is desired, and conditions like temperature and catalyst presence are controlled.

Reaction Kinetics

Reaction kinetics studies the speed and mechanism behind chemical reactions. It's crucial for predicting how long a reaction will take and what factors will affect its rate.
Key elements include:

• **Reaction Order**: Determines how the concentration of reactants affects the rate. Zero-order reactions have rates unaffected by reactant concentration.
• **Temperature**: Higher temperatures generally increase reaction rates.
• **Catalysts**: Substances that increase the reaction rate without being consumed, such as gold in the hydrogen iodide decomposition.

In the case of hydrogen iodide, the reaction kinetics reveal that the constant rate of decomposition is controlled by catalyst presence rather than the concentration of \(\text{HI}\). This is particularly pivotal in processes where precise control over reaction duration and rate is required, highlighting the importance of understanding and applying reaction kinetics.