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 \(1.20 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) $$\begin{array}{c} 2 \mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t}=k=1.20 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \end{array}$$ a. If the initial HI concentration was 0.250 mol/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 \(0.250 \mathrm{M}\) HI to decompose?

Short answer

The concentration of HI at 25 minutes after the start of the reaction is 0.070 M, and it will take approximately 34.72 minutes for all the 0.250 M HI to decompose.

Step-by-step solution

Recall the integrated rate law for zero-order reactions

For a zero-order reaction, the integrated rate law is given by the formula: \[ [A]_t = [A]_0 - kt \] where [A]_t is the concentration of A at time t, [A]_0 is the initial concentration of A, k is the rate constant, and t is the time.

Find the concentration of HI at 25 minutes

We are given the initial concentration of HI, [HI]_0 = 0.250 mol/L, the rate constant k = \(1.20 \times 10^{-4}\) mol/L·s, and the time t = 25 minutes. Convert minutes to seconds: \[ 25 \, \text{min} \times \frac{60 \, \text{s}}{1 \, \text{min}} = 1500 \, \text{s} \] Now, use the integrated rate law to calculate the concentration of HI at 25 minutes: \[ [\mathrm{HI}]_{t} = [\mathrm{HI}]_0 - k \cdot t \] \[ [\mathrm{HI}]_{t} = 0.250 \, \mathrm{M} - (1.20 \times 10^{-4} \, \mathrm{M}\, \mathrm{s}^{-1})(1500 \, \mathrm{s}) \] \[ [\mathrm{HI}]_{t} = 0.250 \, \mathrm{M} - 0.180 \, \mathrm{M} \] \[ [\mathrm{HI}]_{t} = 0.070 \, \mathrm{M} \] So, the concentration of HI at 25 minutes after the start of the reaction is 0.070 M.

Find the time for all 0.250 M HI to decompose

To find the time it takes for all the 0.250 M HI to decompose, we need to determine when the concentration of HI reaches zero. Using the integrated rate law formula, we set [HI]_t = 0 and solve for t: \[ 0 = [\mathrm{HI}]_0 - kt \] \[ kt = [\mathrm{HI}]_0 \] \[ t = \frac{[\mathrm{HI}]_0}{k} \] \[ t = \frac{0.250 \, \mathrm{M}}{1.20 \times 10^{-4} \, \mathrm{M} \, \mathrm{s}^{-1}} \] \[ t = 2083.33 \, \mathrm{s} \] Now, convert seconds to minutes: \[ t = 2083.33 \, \text{s} \times \frac{1 \, \text{min}}{60 \, \text{s}} \] \[ t = 34.72 \, \text{min} \] So, it will take approximately 34.72 minutes for all the 0.250 M HI to decompose.

Key concepts

Integrated rate law

The integrated rate law is a powerful tool in kinetics, allowing us to connect the concentration of reactants with time as a chemical reaction progresses. For zero-order reactions like the decomposition of hydrogen iodide, this rate law takes a particularly simple form. The formula \[ [A]_t = [A]_0 - kt \] describes how the concentration of a reactant, denoted as \([A]_t\) at any time \(t\), changes from its initial concentration \([A]_0\) over the course of the reaction. Here, \(k\) represents the rate constant, a measure of the reaction's speed, which is unique in zero-order reactions because it remains consistent over time.
Understanding the integrated rate law is crucial for calculating how long a reaction will take and predicting the concentration of reactants at any given moment. These calculations are foundational in fields like chemistry and pharmacology, where reaction time is critical.

Reaction kinetics

Reaction kinetics is the study of the rates of chemical processes and the steps through which they occur. By understanding the speed and mechanisms of reactions, scientists can better control and utilize these processes, from industrial chemistry to biological systems.
In simple terms, kinetics examines factors such as:

• The concentration of reactants
• Temperature
• The presence of a catalyst
• The nature of the reactants themselves

For zero-order reactions, as seen with hydrogen iodide, the rate is independent of the concentration of the reactants. This type of reaction often occurs under conditions where the reacting surfaces or catalysts are saturated.
Through the lens of kinetics, understanding zero-order reactions is essential, particularly in catalysis, as it uncovers why, even with more reactants, the product formation remains constant until the initial reactant is consumed.

Hydrogen iodide decomposition

Hydrogen iodide (HI) decomposition is a classic example of a zero-order reaction occurring on a catalyst. In this reaction, HI gas decomposes on the surface of finely divided gold to form hydrogen (H\(_2\)) and iodine (I\(_2\)) gases.
The balanced chemical equation for this decomposition is:\[ 2 \mathrm{HI}(g) \longrightarrow \mathrm{H}_2(g) + \mathrm{I}_2(g) \]Here, gold serves as a catalyst, providing a surface for the reaction. Despite the simplicity of the reaction products, the mechanism is interesting because it reflects the kinetics of zero-order reactions.
This means the decomposition rate remains constant under specific conditions, and doesn't change with varying concentrations of HI. Studying HI decomposition provides insight into surface reactions and catalysis, which are pivotal in developing efficient industrial processes and understanding natural phenomena.