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why isnt it A for q13 (1 Viewer)

Luukas.2

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For the second one, shifting to the right means the rate of the forward reaction (A + B ---> C) is faster / higher than that of the reverse reaction. Prior to returning to equilibrium, graph A has rate forward > rate reverse (moving right) while (B) has rate reverse > rate forward.
 

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For the second one, shifting to the right means the rate of the forward reaction (A + B ---> C) is faster / higher than that of the reverse reaction. Prior to returning to equilibrium, graph A has rate forward > rate reverse (moving right) while (B) has rate reverse > rate forward.
Are those js the initial rates tho? Isnt the forward rxn technically increasing in B?
 

Luukas.2

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For Q 13, adding aqueous NaOH to a saturated Mg(OH)2 solution will lead to further precipitation of Mg(OH)2 (s) and with a decrease in the concentration of magnesium ions and an increase in the concentration of hydroxide ions such that [Mg2+][OH-]2 remains constant. This assumes that the sodium hydroxide solution was sufficiently concentrated that the effect of its addition was an increase in the hydroxide ion concentration that more than compensated for the dilution effect on the magnesium ion concentration. If the NaOH was extremely dilute then the result could be a solution of magnesium hydroxide is no longer saturated... but that's unlikely. Option (B) appears based on this possibility, but it's not literally true because (even in that case) the solubility hasn't changed, it's just that the solution has been diluted sufficiently to no longer be saturated. To me, none of the answers are good but (A) is the best choice available.
 
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Luukas.2

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The net effect in (A) has forward faster than reverse, though decreasing as it approaches equilibrium. The forward rate is increasing in (B) but the reverse rate is higher / faster so the net movement is to the left.
 

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