Complete question is;
Consider this reaction: 2 HI(g) ⇄ H₂(g) + I₂(g)
At a certain temperature, this reaction follows first - order kinetics with a rate constant of 0.684 s^(-1). Suppose a vessel contains HI at a concentration of 0.32 M. Calculate the concentration of HI in the vessel 0.88 seconds later.
You may assume no other reaction is important.
Round your answer to 2 significant digits
Answer:
0.18 M
Step-by-step explanation:
We are given;
Initial concentration of HI = 0.32 M
Time = 0.88 s
Rate is given as 0.684 s^(-1)
Now, we want to find the concentration of HI in the vessel 0.88 seconds later. It can be gotten from the formula;
ln (HI) = ln (HI)₀ - (kt)
Where;
(HI)₀ is the initial concentration
k is the reaction rate constant
t is time
Plugging in the relevant values, we have;
ln (HI) = In (0.32) - (0.684 × 0.88)
ln (HI) = -1.1394 - 0.6019
ln (HI) = -1.7413
(HI) = e^(-1.7413)
(HI) = 0.175 M ≈ 0.18 M