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Trava [24]
3 years ago
13

Consider the following system at equilibrium:

Chemistry
1 answer:
Vika [28.1K]3 years ago
4 0

Answer:

1) Rightward shift

2) Rightward shift

3) Leftward shift

4) Leftward shift

5) Leftward shift

6) Rightward shift

7) No shift

8) No shift                                                              

   

Explanation:

To evaluate each case we need to consider Le Chatelier's Principle, which states that the adding of additional reactants or products to a system will shift the equilibrium in the opposite direction, to maintain the equilibrium of the system. On the contrary, if we remove a reactant or a product in the system, the equilibrium will be shifted in the direction of the reactant or product reduced, to produce more of it (and thus maintain balance).        

Taking into account the above, let's see each statement, in the following equation:

A(aq) + B(aq)  ⇄  2C(aq)    (1)

1) Increase A. This will cause a rightward shift in equation 1 in order to consume the reactant added.

2) Increase B. Same as 1), this will cause a rightward in equation 1.

3) Increase C. This will cause a leftward shift in order to consume the excess of product in the system.  

4) Decrease A. This will produce a leftward shift to produce the reactant that is being reduced.

5) Decrease B. Same as 4), a leftward shift.

6) Decrease C. This will produce a rightward shift to produce the product that is being reduced.

7) Double A, half B. The double A will cause a rightward shift and the half B will produce a leftward shift, which results in no shift.

8) Double both B and C. Double B will produce a rightward shift and double C will produce the contrary, a leftward shift, so the final result is no shift.

               

I hope it helps you!

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First let's balance the C.

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Next let's balance the H.

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Lastly, let's balance the O.

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