When one-half mole of oxygen and one-half mole of nitrogen are heated to 3000 K
ID: 897607 • Letter: W
Question
When one-half mole of oxygen and one-half mole of nitrogen are heated to 3000 K at a pressure of 5000 kPa, some of the mixture will react to form NO by the reaction equation! O^2 + ! N2 ~ NO. Assume these are the only components which react. Calculate: Chemical equilibrium constant for this reaction at these conditions using Table A-3. Number of moles of NO at equilibrium. Number of moles of O_2 at equilibrium. Number of moles of NO at equilibrium if total pressure is doubled. Number of moles of NO at equilibrium if there was originally one-half mole of oxygen, one-half mole of nitrogen, and one mole of argon at 5000 kPa total pressure.Explanation / Answer
(a): The given chemical reaction in equilibrium is
N2(g) + O2(g) <---> 2NO(g)
The standard entropy change for the above reaction is DeltaG0 = 173.14 KJ/mol = 173140 J/mol
Temperature, T = 3000K
Now equilibrium constant at temperature T can be calculated from the following formulae
DeltaG0 = - R*T*lnK
=> ln K = - DeltaG0 / RT = - 173140 Jmol-1 / (8.314 J.K-1mol-1 * 3000K) = - 6.9417
=> K = (exp)-6.9417 = 9.667*10-4
(b) : Goben the total initial pressure, P = 5000 KPa = 5 MPa
Initial moles of N2 = 0.5 mol
Initial moles of O2 = 0.5 mol
N2(g) + O2(g) <---> 2NO(g)
Initial moles : 0.5 0.5 0
Eqm.moles: (0.5 - x) (0.5 - x) 2x
Total pressure at equilibrium = (0.5 - x)+(0.5 - x)+2x = 1 mol
K = 9.667x10-4 = (2x)2 / (0.5 - x)(0.5 - x) = (2x)2 / (0.5 - x)2
=> 2x / (0.5 - x) = underroot(K) = 3.11 x 10-2
=> x = 0.007656 mol
Hence moles of NO at equilibrium = 2x = 0.0153 mol
(c) moles of O2 at equilibrium = 0.5 - x = 0.5 - 0.007656 = 0.492 mol
(d) Since equilibrium moles is independent of pressure, total moles of NO at equilibrium = 0.0153 mol
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