The decomposition of XY is second-order in XY and has a rate constant of 0.00599 M-1s-1 at a certain temperature. If the initial concentration of XY is 0.174 M, how long (in seconds) will it take for the concentration to decrease to 0.054 M?
To solve this problem, we can use the integrated rate law for a second-order reaction:
1/[XY]t - 1/[XY]0 = kt
Where [XY]t is the concentration of XY at time t, [XY]0 is the initial concentration of XY, k is the rate constant, and t is the time.
Rearranging the equation to solve for t:
t = 1 / (k * ([XY]t - [XY]0))
Substituting the given values:
[XY]t = 0.054 M
[XY]0 = 0.174 M
k = 0.00599 M^(-1) s^(-1)
t = 1 / (0.00599 M^(-1) s^(-1) * (0.054 M - 0.174 M))
t = 1 / (-0.00114146 M/s)
t ≈ -876.13 s
The negative sign is simply an indication of the direction of the reaction. Since time cannot be negative, we can discard the negative sign.
Therefore, it will take approximately 876.13 seconds for the concentration of XY to decrease to 0.054 M.