Calculating an composition after a prior equillbrlum d... Nitrogen dioxide is on

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Calculating an composition after a prior equillbrlum d... Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric chemistry. It can react with itself to form another form of NOx, dinitrogen tetroxide A chemical engineer studying this reaction fills a 100. I tank at 16. °C with 43, mol of nitrogen dioxide gas. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 17. mol of nitrogen dioxide gas. The engineer then adds another 14. mol of nitrogen dioxide, and allows the mixture to come to equilibrium again. Calculate the moles of dinitrogen tetroxide after equilibrium is reached the second time. Round your answer to 2 significant digits 0-10 Explanation D P 0

Explanation / Answer

The reaction is:

2NO2 ------> N2O4

In first case:

ICE table:

[NO2] = mole/volume = 17/100 = 0.17

[N2O4] = 13/100 = 0.13 M

So, equilibrium constant = K = [N2O4]/[NO2]2 = 0.13/(0.17)2 = 4.50

K = 4.50

For second case:

ICE table:

[NO2] = (31-2x)/100

[N2O4] = (13+x)/100

K = [N2O4]/[NO2]2

4.50 = [(13+x)/100]/[(31-2x)/100]2

4.50 = 100(13+x)/(31-2x)2

0.045 = (13+x)/(31-2x)2

0.045(31-2x)2 = 13+x

0.045(961+4x2-124x) = 13+x

43.245+0.18x2-5.58x = 13+x

0.18x2-6.58x+30.245=0

on solving, we get x = 5.4

So, moles of N2O4 at equilibrium = 13+x = 13+5.4 = 18.4 mol

Answer is 18*100 mol

[NO2] [N2O4] Initial 43 mol 0 Equilibrium 17 mol 1/2(43-17) = 13 mol

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