The first-order rate constant for the decomposition of N2O5,

2N2O5 (g) ?4NO2 (g) + O2 (g)

at 70? C is 6.82*10?3 s?1. Suppose we start with 3.00*10?2 mol of N2O5 (g) in a volume of 2.5 L.

Part A

How many moles of N2O5 will remain after 7.0 min?

Part B

How many minutes will it take for the quantity of N2O5 to drop to 1.7*10?2 mol?

Part C

What is the half-life of N2O5 at 70? C?

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