Question

Be sure to answer all parts. for the decomposition of gaseous dinitrogen pentaoxide, 2 n2o5(g)→4 no2(g) o2(g) the rate constant is k = 2.8 × 10−3 s−1 at 60°c. the initial concentration of n2o5 is 2.10 mol/l. (a) what is [n2o5] after 5.00 min? .91 mol l (b) what fraction of the n2o5 has decomposed after 5.00 min?

Answer 1

The concentration of N2O5 after 5.00 min is 0.91 M and 0.57 fraction of N2O5 has decomposed after 5.00 min.

Step-by-step explanation:

Given the rate constant (k) for the decomposition of gaseous dinitrogen pentaoxide, 2 N2O5(g) → 4 NO2(g) + O2(g), at 60°C, we can determine the concentration of N2O5 after 5.00 min.

(a) To find the concentration of N2O5 after 5.00 min, we can use the first-order rate equation:

ln([N2O5]t/[N2O5]0) = -kt

where [N2O5]t is the concentration at time t, [N2O5]0 is the initial concentration, k is the rate constant, and t is the time.

Substituting the given values, we have:

ln([N2O5]t/2.10) = -2.8 x 10^-3 x 300

[N2O5]t = 0.91 M

(b)To find the fraction of N2O5 that has decomposed after 5.00 min, we can use the equation:

Fraction decomposed = (initial concentration - concentration at time t) / initial concentration

Substituting the given values, we have:

Fraction decomposed = (2.10 - 0.91) / 2.10 = 0.57