The reaction: 2 HI==> H2 + I2, is second order and the rate constant at 800 K is 9.70 x 10-2 M-1 s-1. How long will it take for 8.00 x 10-2 mol/L of HI to decrease to one-fourth of its initial concentration?
can someone please show me how to set this up?
To solve this problem, we can use the integrated rate law for a second-order reaction:
1/[A]t - 1/[A]0 = kt
Where:
[A]t is the concentration of HI at time t
[A]0 is the initial concentration of HI
k is the rate constant
t is the time
We are given:
Initial concentration, [A]0 = 8.00 x 10^-2 mol/L
Final concentration, [A]t = 1/4 * [A]0 = 1/4 * 8.00 x 10^-2 mol/L = 2.00 x 10^-2 mol/L
Rate constant, k = 9.70 x 10^-2 M^-1 s^-1
Now, we can substitute these values into the equation and solve for t:
1/[A]t - 1/[A]0 = kt
1/(2.00 x 10^-2) - 1/(8.00 x 10^-2) = (9.70 x 10^-2)(t)
Simplifying this equation:
50 - 12.5 = (9.70 x 10^-2)(t)
37.5 = (9.70 x 10^-2)(t)
Now, we can solve for t:
t = 37.5 / (9.70 x 10^-2)
t ≈ 386.6 seconds
Therefore, it will take approximately 386.6 seconds for the concentration of HI to decrease to one-fourth of its initial concentration.