2 SO3 (g) 2 SO2 (g) O2 (g) A sample of 2 0 mole of SO3 (g)is placed...

a initial concentration of SO3 number of moles volume 2 mol 2L 1 molL 2 SO3 g 2 SO2 g O2 ....

2 SO3 (g) â†’ 2 SO2 (g) + O2 (g) A sample of 2.0 mole...

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Chemistry

2 SO3 (g) â†’ 2 SO2 (g) + O2 (g) A sample of 2.0 mole of SO3 (g)is placed into an evacuated 2.0 L container and heated to 700K. Thereaction occurs and the concentration of SO3 (g) as a function oftime is shown in the graph below. Sorry the graph will not copy butis shows that SO3 at equilibrium is 0.4 mol/L
a) Determine the equilibrium concentrations of
SO2 (g) and O2 (g)
b) Calculate the value of the Kc and Kp equilibriumconstants forthe reaction

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4.1 Ratings (697 Votes)

a)
initial concentration of SO3 = number of moles/ volume
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 mol / 2L
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1 mol/L

2 SO3 (g) â†’ 2 SO2 (g) + O2 (g)
1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â (initial concentration)
1-2xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â (at equilibrium)

But given that equilibrium concentration of SO3 = 0.4 mol/L
so,
1-2x = 0.4
x= 0.3 mol/L

equilibrium concentration of SO2 = 2x = 2*0.3 = 0.6 mol/L
equilibrium concentration of O2 = x = 0.3 mol/L

b)
Kc = [SO2]^2 [O2] / [SO3]^2
Â Â Â Â Â = 0.6^2 *0.3/ {0.4^2}
Â Â Â Â Â = 0.675

use:
Kp = Kc* (RT)^ (delta n)
Â Â Â Â Â = 0.675* (0.0821*700)^ (2+1-2)
Â Â Â Â Â = 0.675* (0.0821*700)^1
Â Â Â Â Â = 0.675* (0.0821*700)
Â Â Â Â Â Â =38.8

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