For all your questions, big or small, IDNLearn.com has the answers you need. Discover detailed answers to your questions with our extensive database of expert knowledge.
Sagot :
To determine how long it will take for the concentration of [tex]\( SO_2Cl_2 \)[/tex] to be reduced to 1% of its initial value, we can use the concept of half-life in a first-order reaction. The half-life of a substance is the time it takes for the concentration to decrease to half of its initial value.
Given:
- Half-life ([tex]\( t_{1/2} \)[/tex]) = 8.0 minutes
- Initial concentration ([tex]\( [SO_2Cl_2]_0 \)[/tex]) = 100% (for simplicity)
- Final concentration ([tex]\( [SO_2Cl_2]_t \)[/tex]) = 1% of initial value = 1%
We use the relationship for first-order kinetics:
[tex]\[ [SO_2Cl_2]_t = [SO_2Cl_2]_0 \cdot \left(\frac{1}{2}\right)^{t/t_{1/2}} \][/tex]
where:
- [tex]\( [SO_2Cl_2]_t \)[/tex] is the concentration at time [tex]\( t \)[/tex]
- [tex]\( [SO_2Cl_2]_0 \)[/tex] is the initial concentration
- [tex]\( t \)[/tex] is the time
- [tex]\( t_{1/2} \)[/tex] is the half-life
Substituting the given values into the formula:
[tex]\[ 1 = 100 \cdot \left(\frac{1}{2}\right)^{t / 8} \][/tex]
To isolate [tex]\( t \)[/tex], we divide both sides of the equation by 100:
[tex]\[ \frac{1}{100} = \left(\frac{1}{2}\right)^{t / 8} \][/tex]
or equivalently:
[tex]\[ 0.01 = \left(\frac{1}{2}\right)^{t / 8} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides of the equation:
[tex]\[ \ln(0.01) = \ln\left(\left(\frac{1}{2}\right)^{t / 8}\right) \][/tex]
Using the property of logarithms:
[tex]\[ \ln(0.01) = \left(\frac{t}{8}\right) \ln\left(\frac{1}{2}\right) \][/tex]
We know that [tex]\( \ln\left(\frac{1}{2}\right) = -\ln(2) \)[/tex], so the equation becomes:
[tex]\[ \ln(0.01) = \left(\frac{t}{8}\right) \cdot (-\ln(2)) \][/tex]
Rearranging to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.01)}{-\ln(2)} \cdot 8 \][/tex]
Finally, we substitute the known values of [tex]\( \ln(0.01) \)[/tex] and [tex]\( \ln(2) \)[/tex]:
[tex]\[ \ln(0.01) \approx -4.60517 \][/tex]
[tex]\[ \ln(2) \approx 0.69315 \][/tex]
Thus:
[tex]\[ t = \frac{-4.60517}{-0.69315} \cdot 8 \][/tex]
[tex]\[ t \approx 53.150849518197795 \][/tex]
Therefore, it will take approximately 53.15 minutes for the concentration of [tex]\( SO_2Cl_2 \)[/tex] to be reduced to 1% of its initial value.
Given:
- Half-life ([tex]\( t_{1/2} \)[/tex]) = 8.0 minutes
- Initial concentration ([tex]\( [SO_2Cl_2]_0 \)[/tex]) = 100% (for simplicity)
- Final concentration ([tex]\( [SO_2Cl_2]_t \)[/tex]) = 1% of initial value = 1%
We use the relationship for first-order kinetics:
[tex]\[ [SO_2Cl_2]_t = [SO_2Cl_2]_0 \cdot \left(\frac{1}{2}\right)^{t/t_{1/2}} \][/tex]
where:
- [tex]\( [SO_2Cl_2]_t \)[/tex] is the concentration at time [tex]\( t \)[/tex]
- [tex]\( [SO_2Cl_2]_0 \)[/tex] is the initial concentration
- [tex]\( t \)[/tex] is the time
- [tex]\( t_{1/2} \)[/tex] is the half-life
Substituting the given values into the formula:
[tex]\[ 1 = 100 \cdot \left(\frac{1}{2}\right)^{t / 8} \][/tex]
To isolate [tex]\( t \)[/tex], we divide both sides of the equation by 100:
[tex]\[ \frac{1}{100} = \left(\frac{1}{2}\right)^{t / 8} \][/tex]
or equivalently:
[tex]\[ 0.01 = \left(\frac{1}{2}\right)^{t / 8} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides of the equation:
[tex]\[ \ln(0.01) = \ln\left(\left(\frac{1}{2}\right)^{t / 8}\right) \][/tex]
Using the property of logarithms:
[tex]\[ \ln(0.01) = \left(\frac{t}{8}\right) \ln\left(\frac{1}{2}\right) \][/tex]
We know that [tex]\( \ln\left(\frac{1}{2}\right) = -\ln(2) \)[/tex], so the equation becomes:
[tex]\[ \ln(0.01) = \left(\frac{t}{8}\right) \cdot (-\ln(2)) \][/tex]
Rearranging to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.01)}{-\ln(2)} \cdot 8 \][/tex]
Finally, we substitute the known values of [tex]\( \ln(0.01) \)[/tex] and [tex]\( \ln(2) \)[/tex]:
[tex]\[ \ln(0.01) \approx -4.60517 \][/tex]
[tex]\[ \ln(2) \approx 0.69315 \][/tex]
Thus:
[tex]\[ t = \frac{-4.60517}{-0.69315} \cdot 8 \][/tex]
[tex]\[ t \approx 53.150849518197795 \][/tex]
Therefore, it will take approximately 53.15 minutes for the concentration of [tex]\( SO_2Cl_2 \)[/tex] to be reduced to 1% of its initial value.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
