The rate constant for this first-order reaction is 0.0570 s–1 at 400 °C.
A-->Products
After how many seconds will 14.0% of the reactant remain?
To expand on Dr.Bob's answer:
The usual formula is:
ln([A-t]/[A-0]) = -kt
but you're looking for t so you have to rearrange the formula:
ln(concentration over 100)/-(k) = t
[ln(14/100)]/-0.0570 = t
34.36834015 = t
Your answer should be 34.4 seconds
Well, let me do some quick calculations. If the rate constant is 0.0570 s–1, that means the reaction is pretty quick. It's like a cheetah running through the savannah!
Now, to find out when 14.0% of the reactant will remain, we can use the first-order reaction equation:
ln(A₀/A) = kt
Where A₀ is the initial concentration of the reactant, A is the concentration at time t, k is the rate constant, and t is the time passed.
Since we want to find out when 14.0% of the reactant remains, that means we're looking for the time when A equals 0.14A₀.
So, we can rearrange the equation to solve for t:
ln(A₀/0.14A₀) = kt
Simplifying this equation a bit:
ln(1/0.14) = kt
Now we can plug in the values:
ln(1/0.14) = 0.0570 s–1 * t
Solving for t:
t = ln(1/0.14) / 0.0570 s–1
Doing the calculations, we find that t is approximately 56.63 seconds.
So, after about 56.63 seconds, 14.0% of the reactant will remain. But remember, this is no laughing matter! It's some serious chemistry!
To determine how many seconds it will take for 14.0% of the reactant to remain, we can use the first-order reaction rate equation:
ln([A]/[A]0) = -kt
Where [A] is the concentration of the reactant at a certain time, [A]0 is the initial concentration of the reactant, k is the rate constant, t is the time, and ln is the natural logarithm.
In this case, we are given the rate constant (k = 0.0570 s^–1) and we want to find the time (t) when 14.0% ([A]/[A]0 = 0.14) of the reactant remains.
Substituting the given values into the equation, we have:
ln(0.14) = -0.0570 s^–1 * t
To solve for t, we can rearrange the equation:
t = ln(0.14) / (-0.0570 s^–1)
Calculating this using a calculator, we find:
t ≈ 87.93 seconds
Therefore, it will take approximately 87.93 seconds for 14.0% of the reactant to remain.
ln(No/N) = kt
If you use No = 100, then N = 14, substitute k and solve for t.