# How many moles remain? first order/second order

The first-order rate constant for the decomposition of N2O5, given below, at 70°C is 6.82 10-3 s-1. Suppose we start with 0.0500 mol of N2O5(g) in a volume of 2.0 L.

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

(a) How many moles of N2O5 will remain after 3.5 min?

(b) How many minutes will it take for the quantity of N2O5 to drop to 0.005 mol?

(c) What is the half-life of N2O5 at 70°C?

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If initial quantity of N2O5 = 0.0500 mol and the volume = 2.0L, the initial concentration, Ci = 0.0250 mol/L

ln (Cf/Ci) = kt where Cf is final concentration and t = (3.5 x 60) seconds

ln (Cf/0.0250) = 6.82 x 10

^{-3}x (3.5 x 60)So this should give you the final concentration in moles /L

As you have 2 L you can find the number of moles left

For part b, plug back into the same equation remembering that to end up to 0.005 moles you want a final concentration of 0.025 mole/L as this will be 0.005 mole in the 2 L that we have.

part c) plug into half life period = ln(1/2)/k

kingchemist

Thu, 2009-12-24 06:21