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.
In one experiment, when the initial concentration of HI was 0.206 M, the concentration of HI dropped to 3.71E-2 M after 898 seconds had passed.
Based on these data, the rate constant for the reaction is?
I need and answer asap please anyone
A = Ao - kt
To determine the rate constant for the reaction, we can use the integrated rate law for a zero-order reaction, which is given by:
[Reactant] = [Reactant]₀ - kt
Where:
[Reactant] = concentration of the reactant at a given time
[Reactant]₀ = initial concentration of the reactant
k = rate constant
t = time
In this case, we are given the initial concentration of HI ([HI]₀ = 0.206 M) and the concentration of HI after a certain time ([HI] = 3.71E-2 M) as well as the time elapsed (t = 898 seconds).
We can substitute these values into the integrated rate law and solve for the rate constant (k).
[HI] = [HI]₀ - kt
3.71E-2 M = 0.206 M - k * 898 s
To make the equation easier to work with, we can rearrange it:
k * 898 s = 0.206 M - 3.71E-2 M
k * 898 s = 0.169 M
Now, we can solve for k by dividing both sides of the equation by 898 s:
k = 0.169 M / 898 s
k ≈ 0.000187 M/s
Therefore, the rate constant for the reaction is approximately 0.000187 M/s.