Question

The first order reaction, \(\mathrm{SO}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{SO}_{2}+\mathrm{Cl}_{2}\), has a half life of \(8.75\) hours at \(593 \mathrm{~K}\). How long will it take for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to fall to \(12.5 \%\) of its initial value? a. \(6.24\) hours b. \(26.2\) hours c. \(16.2\) hours d. \(22.6\) hours

Short answer

It takes approximately 26.2 hours for the concentration to fall to 12.5%.

Step-by-step solution

Understand the Half-Life Formula for First Order Reactions

For first order reactions, the relationship between the half-life \(t_{1/2}\) and the rate constant \(k\) is given by the formula \(t_{1/2} = \frac{0.693}{k}\). We will use this formula to find \(k\).

Calculate the Rate Constant k

Given the half-life \(t_{1/2} = 8.75\) hours, substitute it into the half-life formula: \[8.75 = \frac{0.693}{k}\].Solve for \(k\):\[k = \frac{0.693}{8.75} \approx 0.0792 \, \text{hr}^{-1}\].

Use the Integrated Rate Law for First Order Reactions

The integrated rate law for a first order reaction is \( [A] = [A]_0 e^{-kt} \), where \([A]\) is the concentration at time \(t\), and \([A]_0\) is the initial concentration. We want to find \(t\) when \([A] = 0.125[A]_0\) (i.e., 12.5% of the initial concentration).

Set Up the Equation to Solve for Time t

Use the relationship from Step 3:\[0.125[A]_0 = [A]_0 e^{-kt}\]Cancel \([A]_0\) from both sides:\[0.125 = e^{-kt}\].

Apply the Natural Logarithm to Solve for Time t

Take the natural logarithm on both sides to solve for \(t\):\[\ln(0.125) = -kt\]\[-2.079 = -0.0792t\].

Solve the Equation for Time t

Finally, solve for \(t\):\[t = \frac{2.079}{0.0792} \approx 26.2\, \text{hours}\].

Key concepts

half-life calculation

In a first order reaction, half-life is an essential concept that helps understand how long it takes for a reactant to decrease to half its original amount. For these reactions, an important feature is that the half-life is constant and does not depend on the initial concentration. It's determined using the simple formula:
\( t_{1/2} = \frac{0.693}{k} \)
where \( t_{1/2} \) is the half-life, and \( k \) is the rate constant.

• This relationship is specifically useful because, unlike zero-order or second-order reactions, knowing just the rate constant allows you to determine how long the reactant takes to reach half its original concentration regardless of starting amounts.
• In practical applications, the stability and predictability of the half-life measurement allow chemists to predict reaction progress over time.

rate constant determination

The rate constant \( k \) is a critical factor in first order reactions, defining the speed at which the reaction proceeds. Calculating it is straightforward with the half-life data:
First, use the known formula of half-life:
\[ 8.75 = \frac{0.693}{k} \]From this equation, solve for \( k \) to find:
\[ k = \frac{0.693}{8.75} \approx 0.0792 \text{ hr}^{-1} \]

• The value of \( k \) tells us that each hour, about 7.92% of the reactant decomposes, which is a direct translation of the reaction's speed.
• This constant remains the same throughout the reaction and is an important factor in calculating how the concentration of reactants will change over time.

integrated rate law

The integrated rate law for first order reactions is crucial when determining the concentration of a reactant at any point in time. It is expressed as:
\[ [A] = [A]_0 e^{-kt} \] where:

• \([A]\) is the concentration at time \( t \).
• \([A]_0\) is the initial concentration.
• \(e\) is the base of the natural logarithm.
This formula helps assess the progress and control of chemical reactions by connecting concentration, reaction time, and the rate constant.To find the time when a concentration drops to a specific percentage, like 12.5%, the equation is rearranged:
\[0.125[A]_0 = [A]_0 e^{-kt}\]Canceling \([A]_0\), you're left with:
\[0.125 = e^{-kt}\]This lays the foundation of solving for \( t \) by isolating it effectively.

natural logarithm application

Natural logarithms can simplify the equations involved in reaction kinetics, specifically for first order processes. To find the time \( t \) when the concentration is a certain fraction of the original, apply the natural logarithm:
Taking the natural log of both sides of our equation from the integrated rate law gives us:
\[ \ln(0.125) = -kt \]Simplifying this by applying our value for \( k \):
\[ -2.079 = -0.0792t \]Finally, solving for time \( t \):
\[ t = \frac{2.079}{0.0792} \approx 26.2\, \text{hours} \] The use of logarithms here allows you to easily solve equations involving exponential concepts. This simplification is valuable in kinetic studies for determining time frames to reach specific concentration levels.