Question

The decomposition of hydrogen iodide on finely divided gold at \(150^{\circ} \mathrm{C}\) is zero order with respect to \(\mathrm{HI}\). The rate defined below is constant at \(1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). $$ \begin{aligned} & 2 \mathrm{HI}(g) \stackrel{\text { ?u }}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t} &=k=1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ 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 \(0.250 \mathrm{M}\) HI to decompose?

Short answer

a. The concentration of HI at 25 minutes after the start of the reaction is \(0.070\, mol/L\). b. It will take approximately 34.72 minutes for all of the 0.250 M HI to decompose.

Step-by-step solution

Part a: Calculate the concentration of HI at 25 minutes

To calculate the concentration of HI at 25 minutes, we need to plug the given values into the zero-order rate formula: \([HI]_{25} = [HI]_0 - k \cdot t\) where \([HI]_{25}\) is the concentration of HI after 25 minutes, \([HI]_0\) is the initial concentration which is \(0.250\, mol/L\), \(k\) is the rate constant which is \(1.20 \times 10^{-4}\, mol/L \cdot s\), and \(t\) is the time which is 25 minutes. First, we need to convert 25 minutes to seconds: \(25\, min \times \frac{60\, s}{1\, min} = 1500\, s\) Now, we can plug the values into the formula: \([HI]_{25} = 0.250\, mol/L - (1.20 \times 10^{-4}\, mol/L \cdot s)(1500\, s)\) Now, calculate the concentration: \([HI]_{25} = 0.250 - (1.20 \times 10^{-4} \times 1500)\) \([HI]_{25} = 0.250 - 0.180\) \([HI]_{25} = 0.070\, mol/L\) The concentration of HI at 25 minutes after the start of the reaction is \(0.070\, mol/L\).

Part b: Calculate the time it will take for all the 0.250 M HI to decompose

To calculate the time it will take for all the 0.250 M HI to decompose, we can use the zero-order rate formula again: \([HI]_t = [HI]_0 - kt\) This time we want to find \(t\), and the final concentration \([HI]_t\) will be 0. Rearrange the formula to solve for t: \(t = \frac{[HI]_0 - [HI]_t}{k}\) Plug in the values: \(t = \frac{0.250 - 0}{1.20 \times 10^{-4}}\) \(t = \frac{0.250}{1.20 \times 10^{-4}}\) Now, calculate the time: \(t = 2083.33\, s\) For convenience, we can convert the time from seconds to minutes: \(2083.33\, s \times \frac{1\, min}{60\, s} = 34.72\, min\) It will take approximately 34.72 minutes (rounded) for all of the 0.250 M HI to decompose.

Key concepts

Rate Equation

Zero-order kinetics is a fascinating concept in chemistry. The rate equation for a zero-order reaction is unique because the rate of reaction does not depend on the concentration of the reactant. The rate is constant, which might sound slightly confusing initially. But it simply means that the reactant is consumed at a consistent rate over time.

For example, consider the rate equation for the decomposition of hydrogen iodide: \(-\frac{\Delta [\mathrm{HI}]}{\Delta t} = k\).

In this equation:- \(\Delta [\mathrm{HI}]\) represents the change in concentration of hydrogen iodide over time.- \(\Delta t\) is the time during which the concentration changes.- \(k\) is the rate constant with units of concentration per time, here specifically, mol/L·s.
The rate equation shows that irrespective of how much \(\mathrm{HI}\) is present initially, it breaks down at a steady rate denoted by \(1.20 \times 10^{-4} \, \mathrm{mol/L \cdot s}\). This peculiarity makes zero-order reactions easy to predict and calculate, as the rate of reaction remains constant.

Concentration Change

Concentration change in a zero-order reaction is straightforward to understand yet crucial. It describes how the reactant's amount decreases over time. The concentration after a certain period can be calculated using the formula:\([HI]_t = [HI]_0 - k \cdot t\)

Where:- \(\ [HI]_t \) is the concentration after time \( t \).- \(\ [HI]_0 \) is the initial concentration.- \(\ k \) is the rate constant.- \(\ t \) is the time interval.

For the hydrogen iodide example, if the initial concentration is \(0.250\, \mathrm{mol/L}\), after 25 minutes, the concentration reduces as calculated: \(0.070 \mathrm{mol/L}\). The units of time should match the rate constant's unit (usually seconds), so converting time is often necessary. Multiply minutes by 60 to convert into seconds, as seen here with the given 25 minutes \( = 1500 \, \mathrm{s}\).

This linear reduction means that every second, a fixed amount, determined by the rate constant, of \( \mathrm{HI} \) breaks down, leading to consistent concentration reduction rates.

Reaction Time Calculation

Calculating the time required for complete decomposition in zero-order kinetics is relatively simple. We determine the time taken for a reactant to fully break down with zero concentration remaining.

By rearranging the original concentration change equation, we find:\( t = \frac{[HI]_0 - [HI]_t}{k} \)

Here:- \( t \) = time needed for the reaction.- \(\ [HI]_0 \) = initial concentration.- \(\ [HI]_t \) = final concentration, which is zero for complete decomposition.- \(\ k \) = rate constant.

Plugging in our values for the complete decomposition of hydrogen iodide: \( t = \frac{0.250 - 0}{1.20 \times 10^{-4}} = 2083.33 \mathrm{s}\). Converting seconds to minutes, since experiments are easier to handle in common units, gives us approximately 34.72 minutes for full decomposition of the reactant. This simplicity and predictability of zero-order reactions facilitate straightforward planning and execution in chemical processes.