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?
To find the time it takes for the concentration of HI to decrease to one-fourth of its initial concentration, we can use the integrated rate law for a second-order reaction.
The integrated rate law for a second-order reaction is given by:
1/[A]t - 1/[A]0 = kt
Where:
[A]t = concentration of HI at time t
[A]0 = initial concentration of HI
k = rate constant
t = time
We can rearrange this equation to solve for t:
t = 1/[(1/[A]0) - (1/[A]t)] / k
Given:
[A]t = 1/4 * [A]0
[A]0 = 8.00 x 10^-2 mol/L
k = 9.70 x 10^-2 M^-1 s^-1
Substituting the given values into the equation:
t = 1 / [(1/8.00 x 10^-2) - (1/(1/4 * 8.00 x 10^-2))] / 9.70 x 10^-2
t = 1 / [(1/8.00 x 10^-2) - (1/(1/4 * 8.00 x 10^-2))] / 9.70 x 10^-2
Simplifying this equation will give you the time it takes for the concentration of HI to decrease to one-fourth of its initial concentration.
To find out how long it will take for the concentration of HI to decrease to one-fourth of its initial concentration, you can use the integrated rate law for a second-order reaction. The integrated rate law equation for a second-order reaction is:
1/[A]t = kt + 1/[A]0
where [A]t is the concentration of HI at time t, [A]0 is the initial concentration of HI, k is the rate constant, and t is the time.
In this case, we want to solve for t, so we need to rearrange the equation to isolate t. Rearranging the equation, we get:
t = (1/[A]t - 1/[A]0) / k
Substituting the given values:
[A]t = 1/4 * [A]0 = 1/4 * 8.00 x 10^(-2) mol/L = 2.00 x 10^(-2) mol/L
[A]0 = 8.00 x 10^(-2) mol/L
k = 9.70 x 10^(-2) M^(-1) s^(-1)
Now, plug these values into the equation:
t = (1/(2.00 x 10^(-2)) - 1/(8.00 x 10^(-2))) / (9.70 x 10^(-2))
Calculating this expression will give you the time it takes for the concentration of HI to decrease to one-fourth of its initial concentration.
(1/A) - (1/Ao) = akt
Just make up a number for Ao, then put in 1/4 of that number for A..