Section 1: The first-order rate constant for the decomposition of N2O5, 2N2O5( g
ID: 879285 • Letter: S
Question
Section 1:
The first-order rate constant for the decomposition of N2O5,
2N2O5(g)4NO2(g)+O2(g),
at 70C is 6.82×103s1. Suppose we start with 2.80×102 mol of N2O5(g) in a volume of 1.5 L .
a) How many moles of N2O5 will remain after 7.0 min?
b) How many minutes will it take for the quantity of N2O5 to drop to 2.0×102 mol?
Section 2:
Consider the following hypothetical aqueous reaction: A(aq)B(aq). A flask is charged with 0.065 mol of A in a total volume of 100.0 mL. The following data are collected:
What is the half-life for the reaction?
Time (min) 0 10 20 30 40 Moles of A 0.065 0.051 0.042 0.036 0.031Explanation / Answer
section 1
Rate constant = 6.82×103s1
We know that
ln[N2O5]t=--kt + ln[N2O5]0
K is constant = - 6.82*10^-3 s^-1,
t is time = 7min =7 min * 60 ses/min = 420sec
[N2O5]0 is the intial mol we start with,
we know that Molarity = mol/liter of soln = .0282/1.5L=0.0188M
now ln[N2O5]t=--kt + ln[N2O5]0 = - 6.82*10^-3 X 420 + ln (0.0188) = - 2.864 + (-3.973) = -6.837
[N2O5]t = 0.00107 M.
b) again
ln[N2O5]t=--kt + ln[N2O5]0
ln(0.02/1.5) = -6.82*10^-3 x time + ln [0.028/1.5]
ln (0.0133) - ln (0.0188) = -6.82*10^-3 x time
-4.31 -(-3.973) = -6.82*10^-3 x time
Time = -8.283 / -6.82*10^-3 = 1318 seconds
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.