What Are the New Equilibrium Concentrations After Removing 1.0 mol/L of SO2?

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SUMMARY

The discussion focuses on calculating new equilibrium concentrations after the removal of 1.0 mol/L of SO2 from a reaction involving O2 and SO2 forming SO3. Initially, the system contained 4.0 M O2 and 5.0 M SO2, with 3.0 moles/L of SO3 at equilibrium. Upon removing SO2, the reaction shifts to the right, necessitating the use of a rice table to determine the new equilibrium concentrations of O2, SO2, and SO3.

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  • Understanding of chemical equilibrium principles
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  • Ability to construct and interpret rice tables
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  • Calculate the equilibrium constant (K) for the reaction 3O2(g) + SO2(g) <--> 2SO3(g)
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dnartS
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8.0 mole of O2(g) is mixed with 10.0 moles of SO2(g) in a 2.0L container forming SO3(g). If 3.0 moles/L of SO3(g) remains at equilibrium. 1.0 mol/L of SO2(g) is removed from the equilibrium, calculate the new eq'm [ ]'s.So I make the rice table

3O2(g) + SO2(g) <--> 2SO3(g)
r 3 ... : ... 1 ... : 2
4.0M ... 5.0M ... 0M (shift right)
c -3x ... -3x ... +2x
e 4.0 -3x ... 5.0-3x ... 2xwhat do I do now?
 
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Guise I need help on this homework question please, it's really bothering me and I am stuck.
 
What were the concentrations of all gases before SO2 was removed? Can you calculate them from the stoichiometry? After they are calculated - can you use them to calculate equilibrium constant?

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