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
1
Expert's answer
2018-10-03T04:59:59-0400
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.
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