Question

The first order rate constant for the decomposition of N2O5 to NO2 and O2 at 700C is 6.82 x 10-3 s-1. Suppose we start with 0.300 mol of N2O5 in a 0.500 L container. How many moles of N2O5 will remain after 1.5 min?
0.162 mol
0.081 mol
0.205 mol
0.555 mol
0.297 mol

Answer 1

2N2O5(g) → 4NO2(g) + O2(g)

If initial quantity of N2O5 = 0.300 mol and the volume = 0.500 L, the initial concentration Ci = 0.3mol/0.5L = 0.6 mol/L

kt = log(Ci/Cf)

6.82·10-3*1.5*60 = log(0.6/Cf)

0.6138 = -0.22185 - logCf

logCf = -0.6138 – 0.22185

Cf = 0.1485 mol/L

Hence, 0.1485/0.5 = 0.297 mol of N2O5 will remain after 1.5 minutes.