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? 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? 3.1 × 10-2 m 1.2 × 10-1 m 6.2 × 10-2 m 1.6 × 10-2 m

Answer 1

The concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ after three half-lives is $\boxed{3.1 \times {{10}^{ - 2}}{\text{ M}}}$.

Further Explanation:

Order of reaction:

The order of any reaction describes the rate dependence on power of concentration of reactants involved in the reaction. On the basis of order of reaction, reactions can be first, second and third order and so on.

A reaction is said to be first-order reaction if its rate varies directly with the concentration of reactants.

The time period in which half of the initial amount of radioactive species is consumed is known as half-life period. It is represented by ${t_{{\text{1/2}}}}$.

Given reaction occurs as follows:

${\text{2}}{{\text{N}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right) \to 2{{\text{N}}_2}\left( {\text{g}} \right) + {{\text{O}}_2}\left( {\text{g}} \right)$  

Initial concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$  is 0.25 mol/L or 0.25 M.

Since half-life is the time after which concentration of substance becomes half, concentration of  ${{\text{N}}_{\text{2}}}{\text{O}}$ after one half-life can be calculated as follows:

 $\begin{aligned}\left[ {{{\text{N}}_{\text{2}}}{\text{O}}} \right] &= \frac{{0.25{\text{ M}}}}{2} \\ &= 12.5 \times {10^{ - 2}}{\text{ M}} \\\end{aligned}$

Similarly the concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ after two half-lives can be calculated as follows:

$\begin{aligned} \left[ {{{\text{N}}_{\text{2}}}{\text{O}}} \right] &= \frac{{12.5 \times {{10}^{ - 2}}{\text{ M}}}}{2} \\ &= 6.25 \times {10^{ - 2}}{\text{ M}} \\ \end{aligned}$  

Therefore the concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ after three half-lives can be calculated as follows:

 $\begin{aligned} \left[ {{{\text{N}}_{\text{2}}}{\text{O}}} \right] &= \frac{{6.25 \times {{10}^{ - 2}}{\text{ M}}}}{2} \\ &= 3.125 \times {10^{ - 2}}{\text{ M}} \\ & \approx 3.1 \times {10^{ - 2}}{\text{ M}} \\ \end{aligned}$

Therefore the concentration of ${{\text{N}}_{\text{2}}}{\text{O}}$ after three half-lives is $3.1 \times {10^{ - 2}}{\text{ M}}$.

Learn more:

1. Rate of chemical reaction: brainly.com/question/1569924
2. The main purpose of conducting experiments: brainly.com/question/5096428

Answer Details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical Kinetics

Keywords: N2O, N2, O2, first-order, half-life, one, two, three, 2N2O, 2N2, 3.1*10^-2 M, 6.25*10^-2 M, 12.5*10^-2 M, half, 0.25 mol/L, 0.25 M.

Answer 2

According to that each half-life we lose half of the concentration of N2O so,

Because we start with a concentration of N2O = 0.25 mol 
so after one half-life the concentration of N2O decrease to the half 0.25/2
= 12.5 x 10^-2 M
and after two half-lives the concentration of N2O of the one half-life decrease to the half (12.5 x 10^-2) / 2 
=6.25 x 10^-2 M
and after three half-lives the concentration of N2O of the two half-lives decrease to the half (6.25x10^-2) / 2
= 3.1 x 10^-2 M
∴ your correct answer is 3.1 x 10^-2 M