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

the product of the pressure and volume is also constant

the answer is option a.