How does adding oxygen (o2) to this reaction change the equilibrium? 2so2(g) + o2(g) ⇌ 2so3(g) a. the equilibrium shifts right to produce more so3 molecules. b. the equilibrium shifts left to produce more o2 molecules. c. the equilibrium shifts right because of decreased collisions between so2 and o2 molecules. d. the equilibrium shifts left with an increase in so2 and o2 molecules. e. the equilibrium shifts left because of increased collisions between so2 and o2 molecules.

Based on Le Chatelier's principle, if the equilibrium of a given reaction is disturbed by external factors like temperature, pressure or concentration, then it will shift in a direction to undo the effects of the change produced.

The given reaction is:-

2SO2 (g) + O2(g)↔ 2SO3(g)

Thus, adding O2 will shift the equilibrium in a direction in which more of it will be consumed i.e. in the forward direction.

Ans A) The equilibrium will shift to the right to produce more SO3 molecules

hey there!:

2HgO (s) =>  2Hg (l) + O2 (g)

2 moles of HgO decompose to form 2 moles of Hg and 1 mole of O2 according to the reaction mentioned in the question.

So 4.00 moles of HgO must give 4 moles of Hg and 2 moles of O2 theoretically.

603 g of Hg = 603 / 200.6 = 3 moles

Percent yield = ( actual yield / theoretical yield) * 100

= ( 3/4) * 100

= 75 %

Hope this helps!