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Consider the following reaction: 2 SO2(g)O2(g)2 SO3(g) If 1.55 moles of SO2 and 1.55 moles of O2 are added to an empty sealedPlease answer all four of these questions and explain WHY. I don't understand any of this. Thank you!

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Answer #1

Ans :

1) the rate of reverse reaction is zero

When the reaction has just started , there are only reactants in the system and no products , so the reaction will only move forward. SO3 hasn't formed yet , so the rate of reverse reaction is zero.

2) Greater than zero, but less than the rate of forward reaction

Now when some of SO3 has been formed in the system ,the reaction will also proceed in reverse direction , but with the same rate as forward but less because the system hasn't reached equillibrium yet.

3) Greater than zero , equal to the rate of forward reaction

At equillibrium the rate of forward reaction becomes equal to the reverse reaction.

4) Some , but less than 1.55 moles

Here SO2 amount is less so it is limiting. The amount of SO3 will be formed but less than SO3 , as at the equillibrium there is still some SO2 present in the system.

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Please answer all four of these questions and explain WHY. I don't understand any of this. Thank you! Consider the following reaction: 2 SO2(g)O2(g)2 SO3(g) If 1.55 moles of SO2 and 1.55 moles of...
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