To determine the half-life of a first-order reaction given the rate constant [tex]\( k = 4.2 \times 10^{-4} \, \text{s}^{-1} \)[/tex], we can use the half-life formula for a first-order reaction, which is:
[tex]\[ t_{1/2} = \frac{\ln(2)}{k} \][/tex]
Here, [tex]\( \ln(2) \)[/tex] is the natural logarithm of 2, which is approximately 0.693.
1. Calculate the half-life:
[tex]\[ t_{1/2} = \frac{0.693}{4.2 \times 10^{-4} \, \text{s}^{-1}} \][/tex]
2. Simplify the division:
[tex]\[ t_{1/2} = \frac{0.693}{4.2 \times 10^{-4}} \][/tex]
3. Perform the calculation:
[tex]\[ t_{1/2} \approx 1650.3504299046315 \, \text{s} \][/tex]
So the half-life of the reaction is approximately [tex]\( 1650.35 \, \text{s} \)[/tex].
4. Compare this value with the given options:
- [tex]\( 2.9 \times 10^{-4} \, \text{s} \)[/tex] is way too small.
- [tex]\( 1.2 \times 10^3 \, \text{s} \)[/tex] is 1200 seconds, which is smaller but closer.
- [tex]\( 2.4 \times 10^3 \, \text{s} \)[/tex] is 2400 seconds, which is larger.
- [tex]\( 1.7 \times 10^3 \, \text{s} \)[/tex] is 1700 seconds, which is very close.
Therefore, the closest given option to our calculated half-life is [tex]\( 1.7 \times 10^3 \, \text{s} \)[/tex].
So, the half-life for the reaction is approximately [tex]\( 1.7 \times 10^3 \, \text{s} \)[/tex].
