Question #190402

A 15.0 L closed vessel contains sulfur trioxide, sulfur dioxide, and oxygen gases that are in equilibrium at 350°C, with Kequal to 1.8x10-5.

2SO3 (g) ⇌ 2SO2 (g) + O2 (g)

If the equilibrium partial pressures of SO2 and O2 are 0.024 atm and 0.057atm, respectively, how many grams of SO3 (MM=80 g/mol) are present in the container? (Hint: Use the ideal gas equation.) 


1
Expert's answer
2021-05-09T00:00:25-0400

2SO3 (g) ⇌ 2SO2 (g) + O2 (g)


Let Partial pressure of SO2SO_2 is x atm.


So

kp=PO2.Pso22Pso32=0.057×0.0242x2x2=3.28×1051.8×105x=1.82=1.35 atmk_p=\dfrac{P_{O_2}.P_{so_2}^2}{P_{so_3}^2}=\dfrac{0.057\times 0.024^2}{x^2}\\[9pt]x^2=\dfrac{3.28\times 10^{-5}}{1.8\times 10^{-5}}\\[9pt]\Rightarrow x=\sqrt{1.82}=1.35\text{ atm}



Now using the ideal gas equation-


PV=nRT1.35×15=n×0.08314×623Kn=20.2551.796=0.39PV=nRT\\[9pt] \Rightarrow 1.35\times 15=n\times 0.08314\times 623K \\[9pt]\Rightarrow n=\dfrac{20.25}{51.796}=0.39


Grams of SO3SO_3 = moles of SO3×SO_3 \times Molar mass of SO3SO_3

=0.39×80=31.27g=0.39\times 80=31.27g


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Comments

Assignment Expert
19.05.21, 10:08

Dear Anime By multiplying 0.24^2 and 0.057


Anime
11.05.21, 12:44

Where did u get that 3.28x10^-5

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