Problem 24 - Chemistry and Chemical Reactivity

Question

Hydrogen iodide decomposes when heated, forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g}) .\) The rate law for this reaction is \(-\Delta|\mathrm{HI}| / \Delta t=k|\mathrm{HI}|^{2} \cdot \mathrm{At} 443^{\circ} \mathrm{C}\) \(k=30 . \mathrm{L} / \mathrm{mol} \cdot\) min. If the initial HI(g) concentration is \(1.5 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) what concentration of \(\mathrm{HI}(\mathrm{g})\) will remain after \(10 .\) minutes?

Step-by-Step Solution

Answer

The concentration after 10 minutes is \(2.73 \times 10^{-3} \, \mathrm{mol/L}\).

Step 1: Understand the Rate Law

The rate law given is \(-\frac{\Delta[\mathrm{HI}]}{\Delta t} = k[\mathrm{HI}]^2\), meaning the reaction rate depends on \([\mathrm{HI}]^2\). Here, \(k\) is the rate constant and is provided as \(30 \, \mathrm{L} / \mathrm{mol} \cdot \mathrm{min}\). Initial concentration of \(\mathrm{HI}\) is \(1.5 \times 10^{-2} \, \mathrm{mol} / \mathrm{L}\). We need to find the concentration of \(\mathrm{HI}\) after 10 minutes.

Step 2: Write the Integrated Rate Law for Second Order Reactions

For a second-order reaction, the integrated rate law is given by \(\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt\), where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.

Step 3: Substitute the Given Values

Substitute the initial concentration \([\mathrm{HI}]_0 = 1.5 \times 10^{-2} \, \mathrm{mol/L}\), \(k = 30 \, \mathrm{L/mol} \cdot \mathrm{min}\), and \(t = 10\) minutes into the equation: \[\frac{1}{[\mathrm{HI}]_t} = \frac{1}{1.5 \times 10^{-2}} + 30 \times 10.\]

Step 4: Calculate \(\frac{1}{[A]_t}\)

Calculate \(\frac{1}{1.5 \times 10^{-2}} = 66.67\). Then calculate \(30 \times 10 = 300\). Thus, \[\frac{1}{[\mathrm{HI}]_t} = 66.67 + 300 = 366.67.\]

Step 5: Solve for \([A]_t\)

Take the reciprocal of \(366.67\) to find \([\mathrm{HI}]_t\): \([\mathrm{HI}]_t = \frac{1}{366.67} \approx 2.73 \times 10^{-3} \, \mathrm{mol/L}\).

Step 6: Conclusion

After 10 minutes, the concentration of \(\mathrm{HI}\) remaining is \(2.73 \times 10^{-3} \, \mathrm{mol/L}\).

Key Concepts

Second-Order Reaction

A second-order reaction is one where the reaction rate depends on the concentration of one reactant squared or the concentrations of two different reactants. In this case, the decomposition of hydrogen iodide (HI) is second-order with respect to \[\text{HI} \] which means the rate is proportional to the square of its concentration. This type of reaction often occurs when molecules collide and react together.
Understanding the order of a reaction helps us determine how the concentration of reactants affects the speed at which the reaction proceeds. For HI, being second-order means doubling the concentration of HI will increase the rate by a factor of four.

Rate Law

The rate law is an equation that links the reaction rate with the concentrations of reactants. For the decomposition of HI, the rate law is given by:
\[-\frac{\Delta[\text{HI}]}{\Delta t} = k[\text{HI}]^2\]
- Here, \(k\) is the rate constant, which is given as \(30 \ \text{L/mol} \cdot \text{min}\).
- The rate or speed of the reaction is dependent on \([\text{HI}]^2\), which means it's sensitive to changes in the concentration of HI.
Understanding the rate law helps predict how fast a reaction will occur under different initial conditions.

Reaction Rate

The reaction rate describes how quickly a reaction proceeds. For second-order reactions like the decomposition of HI, the rate (\(-\frac{\Delta[\text{HI}]}{\Delta t}\)) is related to the square of the reactant concentration. This indicates that reactions of this order can accelerate rapidly as the concentration of the reactant increases.
Because the rate depends on the concentration squared, small changes in concentration can lead to significant changes in the reaction speed. Therefore, managing concentration is crucial in experiments and industrial processes involving second-order reactions.

Integrated Rate Law

The integrated rate law provides a relationship between the concentrations of reactants at different times. For second-order reactions, it is expressed as:
\[\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt\]where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
Using this formula, we can solve for the concentration of HI after a certain period. For example, if we start with an initial concentration of \(1.5 \times 10^{-2} \ \text{mol/L}\) and a time of 10 minutes, substituting these values gives a remaining concentration of approximately \(2.73 \times 10^{-3} \ \text{mol/L}\).
Understanding the integrated rate law is crucial for calculating concentrations over time, which is important for planning and optimizing reactions.