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Answer :
Sure! Let's solve this problem step-by-step.
Step 1: Understand the Reaction and Given Information
- The chemical reaction is: [tex]\(2A \rightarrow B + 2C\)[/tex].
- We have a first-order reaction with respect to A, which means the rate of reaction depends on the concentration (or pressure, in this case) of A.
- We are given the initial pressure of A, [tex]\(P_0 = 40.0\)[/tex] torr, and the final pressure, [tex]\(P_t = 35.9\)[/tex] torr.
- The rate constant, [tex]\(k\)[/tex], is [tex]\(5.41 \times 10^{-3} \, \text{s}^{-1}\)[/tex].
Step 2: Use the First-Order Kinetics Formula
For a first-order reaction, the relationship between the initial and final pressures is expressed by the formula:
[tex]\[ \ln\left(\frac{P_0}{P_t}\right) = k \cdot t \][/tex]
Where:
- [tex]\(\ln\)[/tex] is the natural logarithm,
- [tex]\(P_0\)[/tex] is the initial pressure,
- [tex]\(P_t\)[/tex] is the final pressure,
- [tex]\(k\)[/tex] is the rate constant,
- [tex]\(t\)[/tex] is the time.
Step 3: Rearrange the Formula to Solve for Time (t)
To find the time [tex]\(t\)[/tex], rearrange the equation:
[tex]\[ t = \frac{\ln(P_0 / P_t)}{k} \][/tex]
Step 4: Plug in the Known Values
Substitute the given values into the rearranged formula:
[tex]\[ t = \frac{\ln(40.0 / 35.9)}{5.41 \times 10^{-3}} \][/tex]
Step 5: Calculate the Result
- First, compute the ratio [tex]\(\frac{40.0}{35.9}\)[/tex].
- Next, take the natural logarithm of that ratio.
- Finally, divide the result by the rate constant, [tex]\(5.41 \times 10^{-3}\)[/tex].
Step 6: Conclusion
The calculated time is approximately [tex]\(19.99\)[/tex] seconds. This is the time it will take for the pressure of [tex]\(A\)[/tex] to decrease from [tex]\(40.0\)[/tex] torr to [tex]\(35.9\)[/tex] torr.
Step 1: Understand the Reaction and Given Information
- The chemical reaction is: [tex]\(2A \rightarrow B + 2C\)[/tex].
- We have a first-order reaction with respect to A, which means the rate of reaction depends on the concentration (or pressure, in this case) of A.
- We are given the initial pressure of A, [tex]\(P_0 = 40.0\)[/tex] torr, and the final pressure, [tex]\(P_t = 35.9\)[/tex] torr.
- The rate constant, [tex]\(k\)[/tex], is [tex]\(5.41 \times 10^{-3} \, \text{s}^{-1}\)[/tex].
Step 2: Use the First-Order Kinetics Formula
For a first-order reaction, the relationship between the initial and final pressures is expressed by the formula:
[tex]\[ \ln\left(\frac{P_0}{P_t}\right) = k \cdot t \][/tex]
Where:
- [tex]\(\ln\)[/tex] is the natural logarithm,
- [tex]\(P_0\)[/tex] is the initial pressure,
- [tex]\(P_t\)[/tex] is the final pressure,
- [tex]\(k\)[/tex] is the rate constant,
- [tex]\(t\)[/tex] is the time.
Step 3: Rearrange the Formula to Solve for Time (t)
To find the time [tex]\(t\)[/tex], rearrange the equation:
[tex]\[ t = \frac{\ln(P_0 / P_t)}{k} \][/tex]
Step 4: Plug in the Known Values
Substitute the given values into the rearranged formula:
[tex]\[ t = \frac{\ln(40.0 / 35.9)}{5.41 \times 10^{-3}} \][/tex]
Step 5: Calculate the Result
- First, compute the ratio [tex]\(\frac{40.0}{35.9}\)[/tex].
- Next, take the natural logarithm of that ratio.
- Finally, divide the result by the rate constant, [tex]\(5.41 \times 10^{-3}\)[/tex].
Step 6: Conclusion
The calculated time is approximately [tex]\(19.99\)[/tex] seconds. This is the time it will take for the pressure of [tex]\(A\)[/tex] to decrease from [tex]\(40.0\)[/tex] torr to [tex]\(35.9\)[/tex] torr.
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