A + 2B C + 2D →>> The reaction is known to be first order in [A] and second order in [B]. When 0.30 moles A and 0.30 moles of B are placed in a 1.00 L container at 25 °C, 1.8 x 10-4 moles of C are formed per second with no change in volume. What is the rate constant at 25 °C? k = [?] Do not include units in your answer. Magnitude of k Enter
A + 2B C + 2D →>> The reaction is known to be first order in [A] and second order in [B]. When 0.30 moles A and 0.30 moles of B are placed in a 1.00 L container at 25 °C, 1.8 x 10-4 moles of C are formed per second with no change in volume. What is the rate constant at 25 °C? k = [?] Do not include units in your answer. Magnitude of k Enter
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Reaction Rate Calculation
**Reaction:**
\[ A + 2B \rightarrow C + 2D \]
**Given:**
- The reaction is first order in [A] and second order in [B].
- Initial concentrations: 0.30 moles of A and 0.30 moles of B.
- Volume of container: 1.00 L.
- Temperature: 25 °C.
- Rate of formation of C: \( 1.8 \times 10^{-4} \) moles per second.
**Task:**
Determine the rate constant, \( k \), at 25 °C.
### Calculation:
1. **Determine the Concentrations:**
- \([A] = \frac{0.30 \text{ moles}}{1.00 \text{ L}} = 0.30 \text{ M}\)
- \([B] = \frac{0.30 \text{ moles}}{1.00 \text{ L}} = 0.30 \text{ M}\)
2. **Rate Law:**
Since the reaction is first order in [A] and second order in [B], the rate law can be expressed as:
\[ \text{Rate} = k [A][B]^2 \]
3. **Plug in the Given Values:**
- Rate: \( 1.8 \times 10^{-4} \) M/s
- \([A] = 0.30 \text{ M}\)
- \([B] = 0.30 \text{ M}\)
\[ 1.8 \times 10^{-4} = k (0.30)(0.30)^2 \]
4. **Solve for \( k \):**
\[ k = \frac{1.8 \times 10^{-4}}{(0.30)(0.09)} \]
\[ k = \frac{1.8 \times 10^{-4}}{0.027} \]
\[ k = 6.67 \times 10^{-3} \]
### Answer:
- Enter the magnitude of \( k \): **\[ 6.67 \times 10^{-3} \]**
Please ensure not to include units in the answer text box. The rate constant, \( k \), is calculated based on the provided reaction conditions at 25 °C.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6e1cbf0-df3a-4970-a570-9594e2279a44%2F0b5d6a18-f392-4456-acf8-a9b86415ff53%2Fh408iv8c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Reaction Rate Calculation
**Reaction:**
\[ A + 2B \rightarrow C + 2D \]
**Given:**
- The reaction is first order in [A] and second order in [B].
- Initial concentrations: 0.30 moles of A and 0.30 moles of B.
- Volume of container: 1.00 L.
- Temperature: 25 °C.
- Rate of formation of C: \( 1.8 \times 10^{-4} \) moles per second.
**Task:**
Determine the rate constant, \( k \), at 25 °C.
### Calculation:
1. **Determine the Concentrations:**
- \([A] = \frac{0.30 \text{ moles}}{1.00 \text{ L}} = 0.30 \text{ M}\)
- \([B] = \frac{0.30 \text{ moles}}{1.00 \text{ L}} = 0.30 \text{ M}\)
2. **Rate Law:**
Since the reaction is first order in [A] and second order in [B], the rate law can be expressed as:
\[ \text{Rate} = k [A][B]^2 \]
3. **Plug in the Given Values:**
- Rate: \( 1.8 \times 10^{-4} \) M/s
- \([A] = 0.30 \text{ M}\)
- \([B] = 0.30 \text{ M}\)
\[ 1.8 \times 10^{-4} = k (0.30)(0.30)^2 \]
4. **Solve for \( k \):**
\[ k = \frac{1.8 \times 10^{-4}}{(0.30)(0.09)} \]
\[ k = \frac{1.8 \times 10^{-4}}{0.027} \]
\[ k = 6.67 \times 10^{-3} \]
### Answer:
- Enter the magnitude of \( k \): **\[ 6.67 \times 10^{-3} \]**
Please ensure not to include units in the answer text box. The rate constant, \( k \), is calculated based on the provided reaction conditions at 25 °C.
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