Problem 239
Question
For the reaction \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}\left[\mathrm{H}^{+}\right]^{\mathrm{n}} .\) If \(\mathrm{pH}\) of the reaction medium changes from two to one, rate becomes 100 times that of the value at \(\mathrm{pH}=2\). The order of reaction is
Step-by-Step Solution
Verified Answer
The order of reaction is 2.
1Step 1: Understanding the Reaction Rate Equation
The given equation for the reaction rate is \( \frac{\mathrm{dx}}{\mathrm{dt}} = \mathrm{k} [\mathrm{H}^{+}]^n \), where \( k \) is the rate constant, \( [\mathrm{H}^{+}] \) is the concentration of hydrogen ions, and \( n \) is the order of the reaction with respect to \([\mathrm{H}^{+}]\).
2Step 2: Calculate Initial Hydrogen Ion Concentration
At \( \text{pH} = 2 \), the hydrogen ion concentration, \([\mathrm{H}^{+}]\), can be calculated using the formula \([\mathrm{H}^{+}] = 10^{-\text{pH}}\). Thus, \([\mathrm{H}^{+}] = 10^{-2} = 0.01 \text{M}\).
3Step 3: Calculate Hydrogen Ion Concentration at New pH
For \( \text{pH} = 1 \), the hydrogen ion concentration is \([\mathrm{H}^{+}] = 10^{-1} = 0.1 \text{M}\).
4Step 4: Relate Rate Changes to pH Changes
Let the initial rate at \( \text{pH} = 2 \) be \( R_1 = k (0.01)^n \) and the rate at \( \text{pH} = 1 \) be \( R_2 = k (0.1)^n \). According to the problem, \( R_2 = 100 \times R_1 \).
5Step 5: Set Up the Equation for Rate Change
We have \( k (0.1)^n = 100 \times k (0.01)^n \). Simplifying, we get \( (0.1)^n = 100 \times (0.01)^n \).
6Step 6: Simplify and Solve for n
Divide both sides by \( (0.01)^n \) to get \( (0.1)^n (0.01)^{-n} = 100 \), which simplifies to \( 10^n = 100 \). Therefore, \( n = 2 \).
7Step 7: Conclusion
The reaction order with respect to \([\mathrm{H}^{+}]\) is \( n = 2 \).
Key Concepts
Rate EquationpH and Hydrogen Ion ConcentrationReaction Rate Dependence on pHOrder of Reaction Determination
Rate Equation
In the study of chemical kinetics, a rate equation is a mathematical expression that describes how the rate of a chemical reaction depends on the concentration of reactants. It often takes the form \( \frac{\mathrm{dx}}{\mathrm{dt}} = \mathrm{k} [A]^n \), where \( \mathrm{k} \) is the rate constant, \([A]\) is the concentration of a reactant, and \( n \) is the reaction order.
- \( \frac{\mathrm{dx}}{\mathrm{dt}} \): This symbolizes the rate of reaction, or how quickly the reactant is converted into a product over time.
- \( \mathrm{k} \): The rate constant is a proportional factor specific to a given reaction at a certain temperature.
- \( n \): The order of reaction, denoting how the concentration of a reactant influences the rate of reaction.
pH and Hydrogen Ion Concentration
The concept of pH is fundamental in chemistry as it measures the acidity or basicity of a solution. It is a logarithmic scale:
- \( \text{pH} = -\log_{10} [\mathrm{H}^+] \), where \([\mathrm{H}^+] \) is the hydrogen ion concentration.
- A lower pH indicates a higher concentration of hydrogen ions and hence a more acidic solution.
- A change by one pH unit alters the hydrogen ion concentration by a factor of 10.
Reaction Rate Dependence on pH
The rate of a reaction can be highly sensitive to changes in pH, particularly for reactions involving hydrogen ions. In the provided exercise, the reaction rate increases dramatically, becoming 100 times faster when the pH changes from 2 to 1. This rate increase can be attributed to:
- Increased concentration of hydrogen ions, \( [\mathrm{H}^+] \), boosting the rate of reaction.
- Greater accessibility of reactants to participate effectively in the reaction.
Order of Reaction Determination
The order of reaction is a key concept in chemical kinetics. It is determined by how the rate of reaction changes with the concentration of reactants:
- Experimentally, by observing how rate varies when concentrations change.
- In our scenario, if changing the pH from 2 to 1 boosts the rate by a factor of 100, we deduce this represents two powers of 10. Thus, the order \( n \) must be 2.
- This implies the rate depends quadratically on \([\mathrm{H}^+]\).
Other exercises in this chapter
Problem 237
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