Answer to Question #190402 in General Chemistry for ganst

Question

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

2SO_{3 (g)} ⇌ 2SO_{2 (g)} + O_{2 (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.) 

Answer 1

2SO_{3 (g)} ⇌ 2SO_{2 (g)} + O_{2 (g)}

Let Partial pressure of $SO_2$ is x atm.

So

$k_p=\dfrac{P_{O_2}.P_{so_2}^2}{P_{so_3}^2}=\dfrac{0.057\times 0.024^2}{x^2}\\[9pt]x^2=\dfrac{3.28\times 10^{-5}}{1.8\times 10^{-5}}\\[9pt]\Rightarrow x=\sqrt{1.82}=1.35\text{ atm}$

Now using the ideal gas equation-

$PV=nRT\\[9pt] \Rightarrow 1.35\times 15=n\times 0.08314\times 623K \\[9pt]\Rightarrow n=\dfrac{20.25}{51.796}=0.39$

Grams of $SO_3$ = moles of $SO_3 \times$ Molar mass of $SO_3$

$=0.39\times 80=31.27g$