Question

Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after \(5.12\) s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

Short answer

(a) The half-life of the reaction is \(2.56 \, seconds\). (b) After \(5.12 \, s\), the remaining concentration of molecular iodine is \(0.0109 \, M\), assuming that the iodine atoms do not recombine to form I₂.

Step-by-step solution

Determine Half-Life

To determine the half-life of a first-order reaction, we'll use the formula: \[t_{1/2} = \frac{0.693}{k}\] \[t_{1/2}\] represents the half-life, and \(k\) represents the first-order rate constant. We are given the rate constant, \(k = 0.271 s^{-1}\). Plug in the given value of \(k\) into the formula and calculate the half-life: \[t_{1/2} = \frac{0.693}{0.271 s^{-1}}\]

Calculate Half-Life

After plugging in the values, perform the calculation to get the half-life: \[t_{1/2} = 2.56 s\] So the half-life of the reaction is 2.56 seconds.

Determine the Remaining Concentration of Molecular Iodine

We have been asked to calculate the remaining concentration of molecular iodine after 5.12 seconds, given the initial concentration as 0.050 M. To do this, we use the formula for a first-order reaction: \[A_t = A_0 e^{-kt}\] Where: - \(A_t\) is the remaining concentration of molecular iodine at time t - \(A_0\) is the initial concentration of molecular iodine - \(k\) is the rate constant - \(t\) is the time (in seconds) We have the values for the initial concentration \((A_0 = 0.050 M)\), the rate constant \((k = 0.271 s^{-1})\), and the time \((t = 5.12 s)\). Plug these values into the equation and solve for the remaining concentration, \(A_t\).

Calculate the Remaining Concentration

Plug in the given values into the formula and calculate \(A_t\): \[A_t = (0.050 M) × e^{-(0.271 s^{-1})(5.12s)}\] \[A_t = (0.050 M) × e^{-1.39}\] \[A_t = 0.0109 M\] So after 5.12 seconds, the remaining concentration of molecular iodine is 0.0109 M, assuming that the iodine atoms do not recombine to form I₂.

Key concepts

Understanding Half-Life Calculation

The concept of half-life is an important part of first-order reaction kinetics. It represents the time required for the concentration of a reactant to decrease to half of its initial concentration. This is especially useful for understanding how quickly a substance undergoes a reaction.
The formula used to determine the half-life (\[t_{1/2}\]) of a first-order reaction is:

• \[t_{1/2} = \frac{0.693}{k}\]

In this equation, \[k\] is the first-order rate constant, which is a unique value that reflects how fast the reaction progresses.
In our exercise, we had a given rate constant of \[0.271 \text{ s}^{-1}\]. By plugging this value into the formula, we computed the half-life to be \[2.56\] seconds. This means every 2.56 seconds, the concentration of iodine molecules in the reaction is halved.
Understanding half-life calculations can help you predict how much of a reactant will remain after a certain period, which is vital in contexts like pharmacology and environmental science.

Rate Constant in First-Order Reactions

The rate constant \(k\) is integral in the study of chemical kinetics. For a first-order reaction, the rate of the reaction depends logarithmically on the concentration of one reactant. Essentially, it measures how fast or slow a reaction proceeds.
In mathematical terms, the rate of a first-order reaction can be expressed as:

• \( ext{Rate} = k imes [ ext{Reactant}]\)

Here, the rate constant \(k\) remains the same at a constant temperature and provides insights into the speed of the reaction.
In the context of our iodine reaction at 625 K, the rate constant was given as \(0.271 \, ext{s}^{-1}\). This relatively high value indicates that the reaction proceeds quite fast. Calculating \(k\) is critical to understanding how half-life and other kinetic expressions work.
It is noteworthy that the rate constant varies with temperature; an increase in temperature usually leads to an increase in the rate constant, making the reaction faster.

Dissociation of Iodine

The dissociation of iodine (\[ ext{I}_2\]) into iodine atoms is a compelling example of a first-order reaction. In this process, the molecule splits into two separate iodine atoms, represented by the equation:

• \[ ext{I}_2 (g) \rightarrow 2 ext{ I} (g)\]

At high temperatures, such as 625 K, this reaction becomes significant as the thermal energy allows the \[ ext{I}_2\] molecules to overcome forces holding the atoms together.
In our problem, after 5.12 seconds, with an initial concentration of 0.050 M, the calculated remaining concentration of molecular iodine was 0.0109 M. This was found using the formula for the concentration of a first-order reaction:

• \[A_t = A_0 e^{-kt}\]

This formula helps in understanding how much reactant is left after a given time, provided the iodine atoms don't recombine. This formula shows that even if a reaction is quick, as with iodine's dissociation, the concentration will decrease exponentially, rather than linearly.
Real-world applications of iodine dissociation include its role in atmospheric chemistry and understanding complex reaction networks in chemistry.