Problem 35
Question
The decomposition of \(\mathrm{HI}(\mathrm{g})\) at \(700 \mathrm{K}\) is followed for \(400 \mathrm{s},\) yielding the following data: at \(t=0,[\mathrm{HI}]=\) \(1.00 \mathrm{M} ;\) at \(t=100 \mathrm{s},[\mathrm{HI}]=0.90 \mathrm{M} ;\) at \(t=200 \mathrm{s}, [\mathrm{HI}]=0.81 \mathrm{M} ; t=300 \mathrm{s},[\mathrm{HI}]=0.74 \mathrm{M} ;\) at \(t=400 \mathrm{s}, [\mathrm{HI}]=0.68 \mathrm{M} .\) What are the reaction order and the rate constant for the reaction: $$\mathrm{HI}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g}) ?$$ Write the rate law for the reaction at 700 K.
Step-by-Step Solution
Verified Answer
The reaction order is 1 (first order). The rate law for the reaction at 700 K is \[Rate = k[HI]\].
1Step 1: Determine the reaction order
To deduce the reaction order, we observe the behavior of the concentration of \(HI\) over time. We can see that the concentration of \(HI\) is decreasing by a factor of 0.9 for every 100 s time interval (from 1.00 M to 0.90 M, then 0.90 M to 0.81 M and so on), suggesting a first-order reaction.
2Step 2: Calculate the rate constant using the first-order reaction formula
For a first-order reaction, the rate constant \(k\) can be calculated using the formula: \[k = \frac{1}{t} ln\frac{[HI]_0} {[HI]_t}\] where \([HI]_0\) is the initial concentration of \(HI\), \([HI]_t\) is the concentration of \(HI\) at a certain time \(t\). Given that \([HI]_0\) = 1.00 M, \([HI]_t\) = 0.90 M, and \(t\) = 100 s, we can substitute these values into the formula: \[k = \frac{1}{100 s} ln\frac{1.00 M}{0.90 M}\]
3Step 3: Write the rate law of the reaction
Once we have determined the order of the reaction and the rate constant, we can write the rate law of the reaction. For a first-order reaction involving a reactant \(HI\), the rate law is: \[Rate = k[HI]\] where \(k\) is the rate constant we calculated and \([HI]\) is the concentration of \(HI\).
Key Concepts
Rate LawFirst-order ReactionRate Constant Calculation
Rate Law
Understanding the rate law of a chemical reaction is key to predicting how the concentration of reactants changes over time. For any reaction, the rate law expresses the rate of the reaction as a function of the concentration of reactants. It typically takes the form:
Knowing the rate law helps in understanding how altering concentrations or the conditions can impact the course of the reaction. For reactions involving gases like HI, the concentration is often measured in molarity (M). Recognizing this allows chemists to carefully manipulate and control reactions.
- Rate = k[A]^m[B]^n
Knowing the rate law helps in understanding how altering concentrations or the conditions can impact the course of the reaction. For reactions involving gases like HI, the concentration is often measured in molarity (M). Recognizing this allows chemists to carefully manipulate and control reactions.
First-order Reaction
A first-order reaction is a common type of reaction where the rate depends linearly on the concentration of one reactant. In our example with the decomposition of HI, the observed concentration changes over even time intervals suggest it follows first-order kinetics.
- Characteristic Behavior: In a first-order reaction, if you plot the natural logarithm of the concentration (ln[HI]) versus time, it yields a straight line.
- Equation: The integrated rate law for a first-order reaction is:\[ ln[HI]_t = ln[HI]_0 - kt \]where \([HI]_0\) is the initial concentration, and \([HI]_t\) is the concentration at time t.
Rate Constant Calculation
Once you establish that a reaction is first-order, the next step is to calculate the rate constant, k. This constant is vital as it quantifies the reaction rate.
- Using the Formula: For first-order reactions, the rate constant k can be calculated using:\[ k = \frac{1}{t} \ln \left(\frac{[HI]_0}{[HI]_t}\right) \]where t is the time elapsed from the start of the reaction.
- Example Calculation: Given \([HI]_0 = 1.00 \, \text{M}\), \([HI]_{100s} = 0.90 \, \text{M}\), and \(t = 100 \, \text{s}\), substituting these values gives:\[ k = \frac{1}{100} \ln \left(\frac{1.00}{0.90}\right) \]Calculating this yields a specific value for k, indicating how fast the HI concentration decreases at 700 K.
Other exercises in this chapter
Problem 33
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