Answer:
41.4 s
Explanation:
Given data
- Rate constant (k): 8.42 × 10⁻² s⁻¹ at 800 °C
- Initial concentration of A ([A]₀): 5.00 M
- Concentration of A at a time t ([A]): 0.153 M
Let's consider the following reaction of first order with respect to A.
2 A ⇒ B
We can find the time that it will take for A to decrease from 5.00 M to 0.153 M using the following expression.
[tex]ln([A]/[A]_0)=-k.t\\ln(0.153M/5.00M)=-8.42 \times 10^{-2}s^{-1} .t\\t = 41.4 s[/tex]
