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

Question

Quizplus · Accepted Answer

First, we need to determine the rate constant (k) of the reaction using the half-life (t1/2) given. t1/2 = 24.6 min We can use the following equation: ln(2)/k = t1/2 ln(2)/k = 24.6 min k = 0.0282 min^-1 Next, we can use the integrated rate law for a first-order reaction: ln[N₂O]t = ln[N₂O]0 - kt Where [N₂O]t is the concentration of N₂O at time t and [N₂O]0 is the initial concentration of N₂O. After 3 half-lives, the concentration will be: [N₂O]t = [N₂O]0/2^3 [N₂O]t = 11ea888e_e2cb_ecb2_a96f_f1746b76a365_TB6107_11 mol / 2^3 L = 1.375 mol/L Now we can use the equation above to solve for the concentration of O₂: ln[O₂]t = ln[O₂]0 - kt [O₂]t = e^(ln[O₂]0 - kt) We know that the initial concentration of O₂ is 0, so: [O₂]t = e^(-kt) [O₂]t = e^(-0.0282 min^-1 * 3 * 24.6 min) = 3.8 × 10^-3 M Finally, we can use the stoichiometry of the reaction to determine the concentration of N₂: 2 mol N₂O produces 2 mol N₂ So the concentration of N₂ is also 3.8 × 10^-3 M. Therefore, the answer is B.

For the First-Order Reaction,2 N2O(g)→ 2 N2(g)+ O2(g),what Is the Concentration

For the First-Order Reaction,2 N₂O(g)→ 2 N₂(g)+ O₂(g),what Is the Concentration