A first-order reaction is 40% complete at the end of 50 minutes. What is the value of the rate constant (in min^-1)?
A. 1.8 x 10^-2
B. 1.0 x 10^-2
C. 1.2 x 10^-2
D. 8.0 x 10^-3
E. none of these
how would i approach this problem?
i know that to find the rate constant, i could use the formula k=rate/reactants, but i don't have the reactants or anything...so i don't know what to do..
i would say it option (D)
if u chang 40% to .40
.40/50
=.008= 8.0x10^-3
Could you do something like this?
Reactants ==> Products
100 = reactants to 0 products initially.
Then if it is 40% complete, then
100-40 = 60 reactants and 40 products. And this has occurred in 50 minutes.
To approach this problem, you can use the concept of a first-order reaction and the given information to determine the rate constant.
In a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant raised to the power of 1. Since the reaction is 40% complete at the end of 50 minutes, it means that 40% of the reactant has been consumed, leaving 60% of the reactant remaining.
Let's assume the initial concentration of the reactant is A₀. After 50 minutes, the remaining concentration of the reactant is 0.60A₀ (60% of A₀).
Now, we can use the formula for a first-order reaction:
ln(A/A₀) = -kt
In this case, A is the remaining concentration of the reactant after 50 minutes (0.60A₀), and A₀ is the initial concentration of the reactant. We can rearrange the equation to solve for k:
k = -ln(A/A₀) / t
Plugging in the values:
k = -ln(0.60) / 50
Using a calculator, you can evaluate the natural logarithm of 0.60, and then divide the result by 50 to find the value of k.
Let's calculate it:
k ≈ 0.01813 min^-1
Therefore, the value of the rate constant (in min^-1) is approximately 0.01813. Comparing it with the given options, we can see that it matches option A: 1.8 x 10^-2.