The decomposition of hydrogen iodide on a gold surface at 150 oC
HI(g) ½ H2(g) + ½ I2(g)
is zero order in HI with a rate constant of 1.20E-4 Ms-1.
If the initial concentration of HI is 0.575 M, the concentration of HI will be 0.116 M after how many seconds have passed?
See your post below.
To solve this problem, we can use the zero-order rate equation, which is:
Rate = k[HI]^0 = k
Where:
- Rate is the rate of the reaction,
- k is the rate constant,
- [HI] is the concentration of hydrogen iodide.
In this case, the rate constant (k) is given as 1.20E-4 Ms^(-1).
We are asked to find the time it takes for the concentration of HI to decrease from 0.575 M to 0.116 M.
To do this, we'll use the integrated rate law for a zero-order reaction:
[HI]t = [HI]0 - kt
Where:
- [HI]t is the concentration of HI at time t,
- [HI]0 is the initial concentration of HI,
- k is the rate constant,
- t is the time.
We are given [HI]0 = 0.575 M and [HI]t = 0.116 M. We can plug these values into the equation and solve for t:
0.116 M = 0.575 M - (1.20E-4 Ms^(-1)) * t
Rearranging the equation to isolate t:
(1.20E-4 Ms^(-1)) * t = 0.575 M - 0.116 M
(1.20E-4 Ms^(-1)) * t = 0.459 M
t = (0.459 M) / (1.20E-4 Ms^(-1))
Calculating this expression will give us the time in seconds.