 #1
i_love_science
 80
 2
 Homework Statement:

Consider the reaction 3 O2(g) > 2 O3(g).
At 175°C and a pressure of 128 torr, an equilibrium mixture of O2 and O3 has a density of 0.168 g/L. Calculate Kp for the above reaction at 175°C.
 Relevant Equations:
 PV = nRT
My solution:
PV = nRT
M = dRT / P
M is the weighted average molar mass of o2 and o3 > M = 32x + 16(1x), where x is the % by mass of o2 and 1x is the % by mass of o3.
M = 32x + 16(1x) = (0.168 g/L) (62.36 L*torr/mol*K) (448 K) / (128 torr) = 36.668 g/mol
solving the equation gives x = 0.708, 1x = 0.282
Concentration of o2: 0.708 * 0.168 g/L * 1 mol / 32 g = 0.00372 mol/L
Concentration of o3: 0.282 * 0.168 g/L * 1 mol/48 g = 0.00102 mol/L
Kc = (0.00102 mol/L)^3 / (0.00372 mol/L)^2
Kp = Kc / (RT)^(delta n) = (0.00102 mol/L)^3 / (0.00372 mol/L)^2 / (0.08206 L*mol/atm*K)^(1) = 0.550
My answer was wrong, the correct answer is 1.5. Could anyone explain where I went wrong? Thanks.
PV = nRT
M = dRT / P
M is the weighted average molar mass of o2 and o3 > M = 32x + 16(1x), where x is the % by mass of o2 and 1x is the % by mass of o3.
M = 32x + 16(1x) = (0.168 g/L) (62.36 L*torr/mol*K) (448 K) / (128 torr) = 36.668 g/mol
solving the equation gives x = 0.708, 1x = 0.282
Concentration of o2: 0.708 * 0.168 g/L * 1 mol / 32 g = 0.00372 mol/L
Concentration of o3: 0.282 * 0.168 g/L * 1 mol/48 g = 0.00102 mol/L
Kc = (0.00102 mol/L)^3 / (0.00372 mol/L)^2
Kp = Kc / (RT)^(delta n) = (0.00102 mol/L)^3 / (0.00372 mol/L)^2 / (0.08206 L*mol/atm*K)^(1) = 0.550
My answer was wrong, the correct answer is 1.5. Could anyone explain where I went wrong? Thanks.