Answer:
After 7 min, 9.75 x 10⁻⁵ mol of N2O5 will remain
It will take 0.7 min for the quantity of N2O5 to drop to 1.7 x 10⁻² mol.
The half-life of N2O5 at 70 ºC is 50 s or 0.8 min.
Explanation:
Parta A
It is a first order reaction, so the velocity is given by the equation:
v = -k[N2O5]
where:
v = velocity
k = rate constant
[N2O5] = concentration of N2O5 in molarity.
the velocity of the reaction can also be written as a function of time:
v = 1/2(d[N2O5] / dt)
where:
d[N2O5] = infinitesimal variation of the concentration of N2O5
dt = infinitesimal variation of time.
(Note that everything is divided by two, the stoichiometric coefficient of N2O5)
then:
1/2(d[N2O5] / dt) = -k[N2O5]
separating variables:
d[N2O5]/[N2O5] = -2kdt
Integrating both sides from the initial concentration, [N2O5]₀, to the final concentration, [N2O5], and form t=0 to t=t:
ln ([N2O5]) - ln ([N2O5]₀) = -2kt
ln ([N2O5]) = -2kt + ln ([N2O5]₀) applying e to both sides:
[N2O5] = e^(-2kt + ln([N2O5]₀)
Now we have an equation that relates time with the concentration of the gas.
The initial concentration [N2O5]₀ = moles N2O5 / V
[N2O5]₀ = 3 .00x 10⁻² mol / 2.5 l = 0.012 M
the concentration at time t= 7.0 min (420 s) will be
[N2O5] =e^( -2 * 6.82 x 10⁻³ s⁻¹ * 420 s + ln 0.012)
[N2O5] = 3.9 x 10⁻⁵ M
The amount of moles of N2O5 present after 7 min is ( 3.9 x 10⁻⁵ mol/l * 2.5 l) 9.75 x 10⁻⁵ mol
Part B
Now we have the final amount of moles and have to find the time at which that quantity of N2O5 remains.
First, we need to know the concentration:
[N2O5] = 1.7 x 10⁻² mol / 2.5 l = 6.8 x 10⁻³ M
then using this equation:
ln [N2O5] = -2 * 6.82 x 10⁻³ s⁻¹ * t + ln ([N2O5]₀)
we have to solve for t:
t = ln [N2O5] - ln ([N2O5]₀ / (-2 * 6.82 x 10⁻³ s⁻¹)
t = ln (6.8 x 10⁻³ M / 0.012 M )/ (-2 * 6.82 x 10⁻³ s⁻¹) (logarithmic property applied)
t = 42 s or (42 s * 1 min / 60 s) 0.7 min.
Part C
The half-life is the time at wich the concentration of N2O5 is half the initial concentration, that is [N2O5] = 1/2 [N2O5]₀
Using this same equation:
ln ([N2O5]) = -2kt + ln ([N2O5]₀)
we replace [N2O5] by 1/2 [N2O5]₀ and solve for t:
ln (1/2 [N2O5]₀) = -2kt + ln ([N2O5]₀) applying logarithmic property
ln 1/2 + ln([N2O5]₀) = -2kt + ln ([N2O5]₀) solving for t and applying logarithmic property (ln 1/2 = ln 1 - ln 2 = 0 - ln 2 = - ln 2)
t = ln2 / 2k
t = 50 s or 0.8 min
