Question

The value of kp for the reaction is 5 2so2(g)+o2(g)←−→2so3(g) what is po2 when mones of so2 and so3 are equal

Answer 1

For given equilibrium reaction,

                                       $2SO_{2} + O_{2} = 2SO_{3}$

Initially moles                         1                  1              0

At equil.                              (1-2x)            (1-x)           2x

According to question,

1-2x = 2x ⇒x= 0.25

Totals moles at equilibrium = 1-2x+1-x+2x = 2-x or (2-0.25)= 1.75

$P_{SO_{3[tex]P_{SO_{2} } = \frac{0.5}{1.75} = \frac{1}{35}$} } = \frac{2*0.25}{1.75} = \frac{1}{35}[/tex]

$P_{O_{2} } = \frac{0.75}{1.75} = \frac{3}{7}$

$K_{p} = \frac{(P_{SO_{3} })^2 }{P_{O_{2} }*P_{SO_{2} }^2  }$

On substituting all values in above equation

$5 = \frac{7}{3}*P_{O_{2} }$

$P_{O_{2} } = \frac{15}{7} =  2.14 atm$

Thus the pressure of oxygen at equilibrium = 2.14 atm  

Answer 2

the given reaction  

2SO2 + O2→2SO  3

Kp  can be calculated as

Kp  =  [PSO2 ] [ PO2 ]

           ------------------

           [P  SO  3  ]

​as given  [PSO3]=[PSO  2]

so ,

1/5=  [O2]

⟹[P  O  2 ]=  0.2 ATM

HOPE YOU ARE SATISFIED WITH MY ANS

IF YES PLZ MARK BRAINLIEST

​