Question

Consider the following reaction 2 N2O(g) => 2 N2(g) + O2(g) rate = k[N2O]. For an initial concentration of N2O of 0.50 M, calculate the concentration of N2O remaining after 2.0 min if k = 3.4 x 10-3 s-1.

Answer 1

Answer:

After 2.0 minutes the concentration of N2O is 0.3325 M

Explanation:

Step 1: Data given

rate = k[N2O]

initial concentration of N2O of 0.50 M

k = 3.4 * 10^-3/s

Step 2: The balanced equation

2N2O(g) → 2 N2(g) + O2(g)  

Step 3: Calculate the concentration of N2O after 2.0 minutes

We use the rate law to derive a time dependent equation.

-d[N2O]/dt = k[N2O]

ln[N2O] = -kt + ln[N2O]i

 ⇒ with k = 3.4 *10^-3 /s

⇒ with t = 2.0 minutes = 120s

⇒ with [N2O]i = initial conc of N2O = 0.50 M

ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)

ln[N2O] = -1.101

e^(ln[N2O]) = e^(-1.1011)

[N2O} = 0.3325 M

After 2.0 minutes the concentration of N2O is 0.3325 M