The reaction 2NO2--->2NO+O2 has the rate constant k= 0.68 m-1s-1.
If the initial concentration of NO2 is 9.30×10−2 M, how long it would take for the concentration to decrease to 2.00×10−2 M ?
please answer and show the steps
To solve this problem, we can use the first-order rate equation:
Rate = k * [NO2]
We can rearrange this equation to solve for time (t):
t = ln([NO2]₀ / [NO2]) / k
Where [NO2]₀ is the initial concentration of NO2 and [NO2] is the final concentration of NO2.
Given:
Initial concentration [NO2]₀ = 9.30×10^(-2) M
Final concentration [NO2] = 2.00×10^(-2) M
Rate constant k = 0.68 m^(-1)s^(-1)
Substituting these values into the equation, we have:
t = ln(9.30×10^(-2) / 2.00×10^(-2)) / 0.68
t = ln(4.65) / 0.68
Using a calculator,
t ≈ 2.29 s
Therefore, it would take approximately 2.29 seconds for the concentration of NO2 to decrease from 9.30×10^(-2) M to 2.00×10^(-2) M.
To determine how long it would take for the concentration of NO2 to decrease to 2.00×10−2 M, we can use the integrated rate equation for a first-order reaction:
ln([NO2]t/[NO2]0) = -kt
Where [NO2]t is the concentration of NO2 at time t, [NO2]0 is the initial concentration of NO2, k is the rate constant, and t is the time.
In this case, we are given the initial concentration [NO2]0 = 9.30×10−2 M, the final concentration [NO2]t = 2.00×10−2 M, and the rate constant k = 0.68 m-1s-1.
Plugging in these values into the equation, we get:
ln(2.00×10−2/9.30×10−2) = (-0.68 m-1s-1) * t
Simplifying the left side of the equation:
ln(2.00×10−2/9.30×10−2) = ln(2.00/9.30)
Now, we can solve for t by rearranging the equation:
(-0.68 m-1s-1) * t = ln(2.00/9.30)
Dividing both sides by -0.68 m-1s-1:
t = ln(2.00/9.30) / -0.68
Using a calculator, we can find the value of t:
t ≈ 4.46 seconds (rounded to two decimal places)
Therefore, it would take approximately 4.46 seconds for the concentration of NO2 to decrease to 2.00×10−2 M.
ln (A)t - ln(A)o = -kt
ln2.00x10-2 - ln(9.30x10-2) = -.68 t
t = 2.26s
Hope this is right!