Final answer:
The concentration of HI remaining after 45.6 seconds is calculated using the integrated rate law for second-order reactions with a rate constant of 1.80 x 10^-3 M^-1 s^-1 and the initial concentration of 3.81 M, resulting in 2.90 M.
Step-by-step explanation:
To determine the concentration of HI remaining after 45.6 seconds given the second-order rate constant (rate constant), we will use the integrated rate law for second-order reactions:
\[\frac{1}{[HI]} - \frac{1}{[HI]_0} = kt\]
Where \([HI]\) is the final concentration of HI, \([HI]_0\) is the initial concentration, k is the rate constant, and t is the time in seconds.
Given:\
k = 1.80 \(\times\) 10^\(-3\) M^-1 s^-1
t = 45.6 s
\([HI]_0\) = 3.81 M
Plugging these values into the equation:
\[\frac{1}{[HI]} = \frac{1}{3.81} + (1.80 \(\times\) 10^\(-3\) \(\times\) 45.6)\]
Solving for \([HI]\), we get:
\[\frac{1}{[HI]} = \frac{1}{3.81} + (0.08208)\]
\[\frac{1}{[HI]} = 0.26247 + 0.08208\]
\[\frac{1}{[HI]} = 0.34455\]
\[[HI] = \frac{1}{0.34455}\]
\[[HI] = 2.90 M\]
So, the concentration of HI remaining after 45.6 seconds is 2.90 M (Option c).