1) The two most abundant gases in the atmosphere, N 2 and O 2 , react to a tiny
ID: 504676 • Letter: 1
Question
1) The two most abundant gases in the atmosphere, N2 and O2, react to a tiny extent at 25oC in the presence of a catalyst. If the partial pressure of O2 is 0.210atm and the partial pressure of N2 is 0.780atm, will be the equilibrium concentration of NO in a 1.00L container at 25oC in the presence of a catalyst?
N2(g) + O2(g) ---> 2 NO(g) Kc = 4.35 x 10-31 at 25oC
a) 1.1 x 10-17M b) 3.33 x 10-9M c) 5.7 x 10-5M d) 7.1 x 10-3M e) 0.087M
Okay so i know that this is a equlibruim problem and i think a dalton's law of partial pressures. The steps to follow are 1.write equation, 2. determine K (right over left no solids or liquids) which i solved and got [(NO)2/(N2)1(O2)1], 3. then you do data tables, 4. find direction K=Q K<Q going left K>Q going right. 5. complete data table, 6.do K caluclation.
Im unsure what to do for the data tables, I know that it has to be in Molarity and the direction is going towards zeros. Dalton's law PV=NRT to find moles? molar mass? I appreciate all the help.
Explanation / Answer
1) answer : a) 1.1 x 10-17M
concentration of O2 = 0.210 x 1.00 / 0.0821 x 298
= 8.58 x 10^-3 M
concentration of N2 = 0.780 x 1 / 0.0821 x 298 = 0.03188 M
N2 + O2 -------------------> 2 NO
0.03188 8.58 x 10^-3 0
0.03188 - x 8.58 x 10^-3 - x 2x
Kc = 4.35 x 10-31
Kc = (2x)^2 / (0.03188 - x)( 8.58 x 10^-3 - x )
4.35 x 10-31 = 4x^2 / x^2 - 0.04046 x + 2.735 x 10^-4
x = 5.5 x 10^-18
concentration of NO = 1.1 x 10^-17 M
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