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.
