Hydrogen iodide decomposition via a second-order process to produce hydrogen and iodine according to the following chemical equation:
2 HI(g) -> H2 (g) + I2 (g)
If it takes 489 seconds for the initial concentration of HI to decrease from 0.0693 M to 0.0174 M, what is the rate constant for the reaction under these conditions? Units are in M-1 *s-1
So far i have done the following:
Ln of (concentration A divided by Concentration A at time 0) = -k * T
natural log (0.0174 M / 0.0693 M) / -489 sec
which equals to 2.826 x10^-3 s-1
1/A - 1/Ao = kt
A = 0.0174 M
Ao = 0.0693 M
t = 489 s
Solve for k.
To determine the rate constant for the reaction, we can use the integrated rate law for a second-order reaction.
The integrated rate law for a second-order reaction is given by the equation:
1/[A]t = kt + 1/[A]0
Where:
[A]t is the concentration of the reactant at time t
[A]0 is the initial concentration of the reactant
k is the rate constant
Given that the initial concentration [A]0 is 0.0693 M and the concentration [A]t at time t is 0.0174 M, we can plug these values into the equation and solve for k.
1/[0.0174 M] = k * 489 s + 1/[0.0693 M]
Simplifying:
57.47 M-1 = k * 489 s + 14.44 M-1
Rearranging the equation:
k * 489 s = 57.47 M-1 - 14.44 M-1
k * 489 s = 43.03 M-1
Finally, solving for k:
k = (43.03 M-1) / (489 s)
k ≈ 0.088 M-1 * s-1
Therefore, the rate constant for the reaction under these conditions is approximately 0.088 M-1 * s-1.