The decomposotion ofN2O5(g)-->NO3(g) proceeeds as a first order reaction with a half life of 30.0 seconds at a certain temperature. If the initial concentration [N2O5]0=0.400 M ,what is the concentration after 120 seconds

Answer 1

First, we need to find the rate constant k for this reaction. We can do this using the half-life formula for a first order reaction:

t1/2 = 0.693 / k

30.0 s = 0.693 / k

k = 0.693 / 30 = 0.0231 s⁻¹

Next, we want to find the concentration of N2O5 at a given time (t = 120s). We can use the integrated rate law for a first order reaction:

[N2O5]t = [N2O5]0 * e^(-kt)

[N2O5]t = 0.400 M * e^(-0.0231 s⁻¹ × 120 s)

[N2O5]t = 0.400 M * e^(-2.772)

[N2O5]t = 0.400 M * 0.0625 = 0.025 M

Therefore, the concentration of N2O5 after 120 seconds is approximately 0.025 M.