a certain first order decomposition reaction has a half-life of 15.0 minutes. What is the rate constant for this reaction?
what is the time it will take for the reaction to reach the 10% completion point?
k = 0.693/t1/2
ln(No/N) = kt
No = 100
N = 90
k from above.
Solve for t.
To find the rate constant for a first-order decomposition reaction, you can use the formula:
t1/2 = (0.693 / k)
where t1/2 is the half-life and k is the rate constant.
Given that the half-life (t1/2) is 15.0 minutes, we can substitute this value into the formula:
15.0 = (0.693 / k)
To isolate the rate constant, we can rearrange the equation as follows:
k = 0.693 / 15.0
Solving this equation gives us the rate constant:
k ≈ 0.0462 min^-1
Now let's proceed to find the time it will take for the reaction to reach the 10% completion point.
The amount remaining after a certain time in a first-order reaction can be calculated using the equation:
A(t) = A₀ * e^(-kt)
where A(t) is the amount remaining at time t, A₀ is the initial amount, k is the rate constant, and e is the base of the natural logarithm (approximately equal to 2.718).
In this case, we are interested in finding the time it will take for the reaction to reach 10% completion. So we can rewrite the formula as:
0.10 = e^(-kt)
Taking the natural logarithm (ln) of both sides, we get:
ln(0.10) = -kt
Rearranging the equation to solve for time (t), we have:
t = -ln(0.10) / k
Substituting the known values of ln(0.10) and the rate constant (k ≈ 0.0462 min^-1), we can calculate the time it will take for the reaction to reach 10% completion:
t ≈ -ln(0.10) / 0.0462
Calculating this, we find:
t ≈ 9.51 minutes
Therefore, it will take approximately 9.51 minutes for the reaction to reach the 10% completion point.