Problem 53 - Chemistry and Chemical Reactivity

Question

Formic acid decomposes at \(550^{\circ} \mathrm{C}\) according to the equation $$\mathrm{HCO}_{2} \mathrm{H}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ The reaction follows first-order kinetics. In an experiment, it is determined that \(75 \%\) of a sample of \(\mathrm{HCO}_{2} \mathrm{H}\) has decomposed in 72 seconds. Determine \(t_{1 / 4}\) for this reaction.

Step-by-Step Solution

Answer

\( t_{1/4} \approx 18.7 \, \text{s} \).

Step 1: Understanding First-Order Kinetics

A first-order reaction is described by the formula \( k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \), where \( k \) is the rate constant, \([A]_0\) is the initial concentration, and \([A]\) is the concentration at time \( t \).

Step 2: Determine Rate Constant (k)

Given that 75% of the sample decomposed, 25% remains. Using the formula: \[ \ln \left( \frac{[A]_0}{[A]} \right) = \ln \left( \frac{1}{0.25} \right) = \ln(4) \]. With \( t = 72 \) seconds, calculate \( k \):\[ k = \frac{\ln(4)}{72} \approx 0.0193 \, \text{s}^{-1} \].

Step 3: Determine Time for 25% Decomposition (\( t_{1/4} \))

The time for a quarter decomposition \( t_{1/4} \) implies 75% of the sample remains. Use the rearranged formula:\[ t = \frac{1}{k} \ln \left( \frac{1}{0.75} \right) \]. Substitute \( k = 0.0193 \, \text{s}^{-1} \) and solve:\[ t_{1/4} = \frac{1}{0.0193} \ln(\frac{4}{3}) \approx 18.7 \, \text{s} \].

Key Concepts

Chemical Decomposition

Chemical decomposition is a process where a single compound breaks down into two or more simpler substances. In this exercise, formic acid (\(\text{HCO}_2\text{H}\)) decomposes into carbon dioxide (\(\text{CO}_2\)) and hydrogen gas (\(\text{H}_2\)). The reaction is represented by the equation:

\[ \text{HCO}_2\text{H} \rightarrow \text{CO}_2 + \text{H}_2 \]

This type of reaction is common in chemistry, where heat or another form of energy is often required to break the chemical bonds in the compound.
In our case, the reaction occurs at a high temperature of \(550^{\circ} \text{C}\).

Key points about chemical decomposition:

• It requires energy input to break bonds.
• It results in simpler products.
• The process follows specific reaction kinetics, such as first-order.

Rate Constant

The rate constant, represented by \(k\), is a fundamental value in reaction kinetics. It indicates how fast a reaction proceeds. For first-order reactions, it is determined using the equation:

\[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]} \right) \]

Here, \([A]_0\) is the initial concentration, and \([A]\) is the concentration at a given time \(t\).
In our problem, we calculated \(k\) for a reaction where 75% of \(\text{HCO}_2\text{H}\) decomposes in 72 seconds.

Substituting values:
\[ k = \frac{\ln(4)}{72} \approx 0.0193 \, \text{s}^{-1} \]

This value tells us about the speed and nature of the reaction. The larger the rate constant, the faster the reaction.

Keep in mind:

• The units of \(k\) vary with the order of the reaction.
• First-order reactions have \(k\) in units of \(\text{s}^{-1}\).

Reaction Time

Reaction time is a crucial aspect in studying chemical kinetics, reflecting how long it takes for a certain percentage of the reactants to transform into products. In the context of this exercise, we focus on the time needed for a specific fraction of decomposition.

For 25% decomposition (or 75% of the substance remaining), the reaction time \(t_{1/4}\) can be calculated using:
\[ t = \frac{1}{k} \ln \left( \frac{1}{0.75} \right) \]

Substituting in the given \(k = 0.0193 \, \text{s}^{-1}\):
\[ t_{1/4} \approx \frac{1}{0.0193} \ln\left(\frac{4}{3}\right) \approx 18.7 \, \text{s} \]

This shows that it takes approximately 18.7 seconds for 75% of the formic acid to remain.

Important points about reaction time:

• Provides insight into reaction speed.
• Varies with conditions like temperature and pressure.
• Is essential for determining suitable reaction durations in industrial applications.