- #1

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- 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(1-x), where x is the % by mass of o2 and 1-x is the % by mass of o3.

M = 32x + 16(1-x) = (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, 1-x = 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(1-x), where x is the % by mass of o2 and 1-x is the % by mass of o3.

M = 32x + 16(1-x) = (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, 1-x = 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.