Answer:
[HI] = 0.097 M
Explanation:
Let's consider the following reaction.
2 HI(g) ⇄ H₂(g) + I₂(g)
The order of reaction for HI is 1. Thus, we can calculate the concentration of HI ([HI]) at certain time using the following expression:
ln [HI] = ln [HI]₀ - k. t
where,
[HI]₀ is the initial concentration of HI
k is the rate constant
t is the time elapsed
When [HI]₀ = 0.440 M and t = 0.210 s, the concentration of HI is
ln [HI] = ln (0.440) - 7.21 s⁻¹ × 0.210 s
ln [HI] = -2.33
[HI] = 0.097 M