Question

For the first-order reaction, 2 N2O(g) → 2 N2(g) + O2(g), what is the concentration of N2O after 3 half-lives if 0.25 mol of N2O is initially placed into a 1.00-L reaction vessel?

Answer 1

Answer: The concentration of $N_2O$ after 3 half-lives is 0.03125 M.

Explanation:

• To calculate the concentration of $N_2O$, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Moles of $N_2O$ = 0.25 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Initial concentration of }N_2O=\frac{0.25mol}{1L}=0.25M$

• To calculate the concentration of $N_2O$ after given half-lives, we use the equation:

$a=\frac{a_o}{2^n}$

where,

a = concentration of reactant left after n-half lives = ?

$a_o$ = Initial concentration of the reactant =  0.25 mol/L

n = number of half lives  = 3

Putting values in above equation, we get:

$a=\frac{0.25}{2^3}$

$a=0.03125M$

Hence, the concentration of $N_2O$ after 3 half-lives is 0.03125 M.