00:01
In this question we've been told that the reaction follows first order decay kinetics.
00:07
So if that's the case, then lean the final concentration at a time t is equal to lynn the initial concentration minus kt.
00:16
We also know that k is equal to lin 2 divided by the half -life for that process.
00:23
So using this information, we are seeing the value of k if we are making the substitutions here, we are saying lean 0 .0 let's just have lean r not it is equal to lean the final concentration plus k2 so bearing this in mind we are saying lean the initial concentration it is equal to lean 0 .0451 plus 30 .5 this is our first equation.
01:05
We also know that lynn, the initial concentration, it is equal to lynn 0 .0321 plus 45k, which is our equation two.
01:19
So if we equate these two, we are going to have a k that is equal to 0 .34 divided by 1 .4 .5, which gives us a rate constant that is equal to 0 .0234 in units of per second.
01:34
Now that we have the value of k, we can then substitute the value of k to then make t half the half -life subject of the formula...