1) To calculate the number of moles of N2O5 that will remain after 7.0 min, we can use the first-order rate law equation:
ln([N2O5]t/[N2O5]0) = -kt
where [N2O5]t is the concentration of N2O5 at time t, [N2O5]0 is the initial concentration of N2O5, k is the rate constant, and t is the time. Rearranging this equation, we get:
[N2O5]t = [N2O5]0 * e^(-kt)
Plugging in the given values, we get:
[N2O5]t = (2.70×10^−2 mol) * e^(-6.82×10^−3 s^-1 * 7.0 min * 60 s/min) = 1.20×10^−2 mol
Therefore, 1.20×10^−2 mol of N2O5 will remain after 7.0 min.
2) To calculate the time it takes for the quantity of N2O5 to drop to 1.7×10−2 mol, we can rearrange the equation from part 1 and solve for t:
t = -ln([N2O5]t/[N2O5]0)/k
Plugging in the given values, we get:
t = -ln(1.7×10^-2 mol/2.70×10^-2 mol)/(6.82×10^-3 s^-1) = 9.5 min
Therefore, it will take 9.5 min for the quantity of N2O5 to drop to 1.7×10^-2 mol.
3) The half-life of a first-order reaction is given by:
t1/2 = ln(2)/k
Plugging in the given rate constant, we get:
t1/2 = ln(2)/(6.82×10^-3 s^-1) = 101.6 s
Therefore, the half-life of N2O5 at 70°C is 101.6 s.