Answer :
Answer:
The answer is "[tex]\bold{4.97 \times 10^{-2}}[/tex]"
Explanation:
Please find the complete question in the attached file.
Equation:
[tex]2SO_2+O_2 \leftrightharpoons 2SO_3[/tex]
at [tex]t=0 3.3 \ \ \ \ \ \ \ \ \ \ 0.79[/tex]
at equilibrium [tex]3.3-p \ \ \ \ \ \ \ \ \ \ 0.79 - \frac{P}{2} \ \ \ \ \ \ \ \ \ \ \ \ P[/tex]
[tex]p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\[/tex]
[tex]=\frac{0.47^2}{2.83^2\times 0.555}\\\\=4.97 \times 10^{-2}[/tex]
