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Sagot :
Certainly! Let's go through the steps to write the equilibrium expression for the given reaction.
### Step 1: Identify the balanced chemical equation
The given reaction for the [tex]$pH$[/tex] indicator bromophenol blue is:
[tex]\[ (C_{19}H_8Br_4O_5S)H^{-} \leftrightarrow (C_{19}H_8Br_4O_5S)^{2-} + H^{+} \][/tex]
### Step 2: Write the general form of the equilibrium constant expression
The general formula for the equilibrium constant [tex]\(K_{eq}\)[/tex] for a reaction of the type:
[tex]\[ aA + bB \leftrightarrow cC + dD \][/tex]
is given by:
[tex]\[ K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \][/tex]
Where:
- [tex]\( [C] \)[/tex] and [tex]\( [D] \)[/tex] are the molar concentrations of the products,
- [tex]\( [A] \)[/tex] and [tex]\( [B] \)[/tex] are the molar concentrations of the reactants,
- [tex]\( a, b, c, d \)[/tex] are the stoichiometric coefficients of the reactants and products.
### Step 3: Apply the specific reaction to the general form
For the reaction:
[tex]\[ (C_{19}H_8Br_4O_5S)H^{-} \leftrightarrow (C_{19}H_8Br_4O_5S)^{2-} + H^{+} \][/tex]
The products are:
- [tex]\((C_{19}H_8Br_4O_5S)^{2-}\)[/tex]
- [tex]\(H^+\)[/tex]
The reactant is:
- [tex]\((C_{19}H_8Br_4O_5S)H^{-}\)[/tex]
### Step 4: Substitute the concentrations and coefficients into the equilibrium expression
In this reaction, the stoichiometric coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] are all equal to 1, so the equilibrium constant expression becomes:
[tex]\[ K_{eq} = \frac{[(C_{19}H_8Br_4O_5S)^{2-}][H^+]}{[(C_{19}H_8Br_4O_5S)H^{-}]} \][/tex]
### Conclusion
The equilibrium expression for the given reaction is:
[tex]\[ K_{eq} = \frac{[(C_{19}H_8Br_4O_5S)^{2-}][H^+]}{[(C_{19}H_8Br_4O_5S)H^{-}]} \][/tex]
### Step 1: Identify the balanced chemical equation
The given reaction for the [tex]$pH$[/tex] indicator bromophenol blue is:
[tex]\[ (C_{19}H_8Br_4O_5S)H^{-} \leftrightarrow (C_{19}H_8Br_4O_5S)^{2-} + H^{+} \][/tex]
### Step 2: Write the general form of the equilibrium constant expression
The general formula for the equilibrium constant [tex]\(K_{eq}\)[/tex] for a reaction of the type:
[tex]\[ aA + bB \leftrightarrow cC + dD \][/tex]
is given by:
[tex]\[ K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b} \][/tex]
Where:
- [tex]\( [C] \)[/tex] and [tex]\( [D] \)[/tex] are the molar concentrations of the products,
- [tex]\( [A] \)[/tex] and [tex]\( [B] \)[/tex] are the molar concentrations of the reactants,
- [tex]\( a, b, c, d \)[/tex] are the stoichiometric coefficients of the reactants and products.
### Step 3: Apply the specific reaction to the general form
For the reaction:
[tex]\[ (C_{19}H_8Br_4O_5S)H^{-} \leftrightarrow (C_{19}H_8Br_4O_5S)^{2-} + H^{+} \][/tex]
The products are:
- [tex]\((C_{19}H_8Br_4O_5S)^{2-}\)[/tex]
- [tex]\(H^+\)[/tex]
The reactant is:
- [tex]\((C_{19}H_8Br_4O_5S)H^{-}\)[/tex]
### Step 4: Substitute the concentrations and coefficients into the equilibrium expression
In this reaction, the stoichiometric coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] are all equal to 1, so the equilibrium constant expression becomes:
[tex]\[ K_{eq} = \frac{[(C_{19}H_8Br_4O_5S)^{2-}][H^+]}{[(C_{19}H_8Br_4O_5S)H^{-}]} \][/tex]
### Conclusion
The equilibrium expression for the given reaction is:
[tex]\[ K_{eq} = \frac{[(C_{19}H_8Br_4O_5S)^{2-}][H^+]}{[(C_{19}H_8Br_4O_5S)H^{-}]} \][/tex]
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