Consider the first-order reaction described by the equation H,C– CH, H-C- -с —н H H Cyclopropane Propene At a certain temperature, the rate constant for this reaction is 5.84 × 104s-!. Calculate the half-life of cyclopropane at this temperature. - מו4 Given an initial cyclopropane concentration of 0.00700 M, calculate the concentration of cyclopropane that remains after 1.70 hours. concentration:

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### First-Order Reaction: Cyclopropane to Propene

**Consider the first-order reaction described by the equation:**

![Cyclopropane to Propene Reaction Diagram](image-url)

**Cyclopropane** (C₄H₆) transforms into **Propene** (C₃H₆) as shown in the chemical equation below:

\[ \text{Cyclopropane} \rightarrow \text{Propene} \]

At a certain temperature, the rate constant for this reaction is given as \( 5.84 \times 10^{-4} \, \text{s}^{-1} \).

#### Calculate the half-life of cyclopropane at this temperature:

The half-life ( \( t_{1/2} \) ) of a first-order reaction is calculated using the formula:

\[ t_{1/2} = \frac{0.693}{k} \]

where:
- \( t_{1/2} \) is the half-life,
- \( k \) is the rate constant.

\[ t_{1/2} = \]

[**Input box for the half-life calculation in seconds**]

---

#### Given an initial cyclopropane concentration of 0.00700 M, calculate the concentration of cyclopropane that remains after 1.70 hours.

For a first-order reaction, the relationship between the concentration of the reactant at a given time \( t \) and the rate constant \( k \) is given by:

\[ [A]_t = [A]_0 e^{-kt} \]

where:
- \([A]_t\) is the concentration at time \( t \),
- \([A]_0\) is the initial concentration,
- \( k \) is the rate constant,
- \( t \) is the time.

Convert 1.70 hours to seconds (since \( k \) is in \( \text{s}^{-1} \)):

\[ t = 1.70 \, \text{hours} \times 3600 \, \text{s/hour} \]

Then calculate:

[**Input box for the concentration calculation in Molarity (M)**]

#### Explanation of Diagrams:
- **Cyclopropane (structure on the left)**: Three carbon atoms form a triangle with each carbon bonded to two hydrogen atoms (\( C_3H_6 \)).
  
- **Reaction Arrow**
Transcribed Image Text:### First-Order Reaction: Cyclopropane to Propene **Consider the first-order reaction described by the equation:** ![Cyclopropane to Propene Reaction Diagram](image-url) **Cyclopropane** (C₄H₆) transforms into **Propene** (C₃H₆) as shown in the chemical equation below: \[ \text{Cyclopropane} \rightarrow \text{Propene} \] At a certain temperature, the rate constant for this reaction is given as \( 5.84 \times 10^{-4} \, \text{s}^{-1} \). #### Calculate the half-life of cyclopropane at this temperature: The half-life ( \( t_{1/2} \) ) of a first-order reaction is calculated using the formula: \[ t_{1/2} = \frac{0.693}{k} \] where: - \( t_{1/2} \) is the half-life, - \( k \) is the rate constant. \[ t_{1/2} = \] [**Input box for the half-life calculation in seconds**] --- #### Given an initial cyclopropane concentration of 0.00700 M, calculate the concentration of cyclopropane that remains after 1.70 hours. For a first-order reaction, the relationship between the concentration of the reactant at a given time \( t \) and the rate constant \( k \) is given by: \[ [A]_t = [A]_0 e^{-kt} \] where: - \([A]_t\) is the concentration at time \( t \), - \([A]_0\) is the initial concentration, - \( k \) is the rate constant, - \( t \) is the time. Convert 1.70 hours to seconds (since \( k \) is in \( \text{s}^{-1} \)): \[ t = 1.70 \, \text{hours} \times 3600 \, \text{s/hour} \] Then calculate: [**Input box for the concentration calculation in Molarity (M)**] #### Explanation of Diagrams: - **Cyclopropane (structure on the left)**: Three carbon atoms form a triangle with each carbon bonded to two hydrogen atoms (\( C_3H_6 \)). - **Reaction Arrow**
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