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The reaction [tex]$A(aq) \rightleftharpoons B(aq)$[/tex] is at equilibrium. [tex]B[/tex] forms at a rate of [tex]0.5 \, M/s[/tex]. At what rate does [tex]A[/tex] form?

A. [tex]1 \, M/s[/tex]
B. Impossible to know
C. [tex]A[/tex] does not form because [tex]A[/tex] is a reactant, not a product
D. [tex]0.5 \, M/s[/tex]

Sagot :

GinnyAnsweridn
To determine the rate at which [tex]\( A \)[/tex] forms in the given reaction [tex]\( A(aq) \rightleftharpoons B(aq) \)[/tex], we need to understand the concept of chemical equilibrium. At equilibrium, the forward and reverse reactions occur at the same rate. This means that the rate of formation of [tex]\( B \)[/tex] from [tex]\( A \)[/tex] is equal to the rate of formation of [tex]\( A \)[/tex] from [tex]\( B \)[/tex].

Given that [tex]\( B \)[/tex] forms at a rate of [tex]\( 0.5 \)[/tex] M/s, we can conclude that [tex]\( A \)[/tex] also forms at the same rate because the system is in equilibrium, and the rates of the forward and reverse reactions must be equal.

Thus, the rate at which [tex]\( A \)[/tex] forms is [tex]\( 0.5 \)[/tex] M/s.

So, the correct answer is:
[tex]\[ \boxed{0.5 \, M/s} \][/tex]
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