Respuesta :
Answer:
EQUILIBRIUM FAVOURS THE PRODUCTS.
Explanation:
The thing we will be considering to specify whether the equilibrium favors starting materials or products is whether our EQUILIBRIUM CONSTANT is less or higher than 1. If the equilibrium constant is less than one, then the equilibrium favours the starting material and vice versa.
Check the ionic chemical equation for the reaction of acetic acid with sodium bromide below in equation (1);
CH3COOH + Na^+ --------------> CH3COO^- + NaH.-----------------------------------------------------------------------(1).
Note: the Na^+ is from Sodium bromide(NaBr) and it is acting as the base in this Reaction, while the acetic acid is acting as the acid. The NaH is the conjugate acid.
The pKa of the conjugate acid,NaH is 35 approximately and the pKa of acetic acid is 4.7. Therefore, the PK(equivalent) =pKa acid - pKa conjugate acid.
That is, PK(eq)= 4.7 - 35.
PK(equivalent)= - 30.3.
Then, the equilibrium constant= 10^-pK(equ).
That is; 10^-(-30.3)= 1.9953 × 10^30.
Here, we can see that the equilibrium constant is 1.9953 × 10^30, that is greater than one, so, the EQUILIBRIUM FAVOURS THE PRODUCTS.
The equilibrium favors starting products.
The Calculation for Equibilrium constant:
We need to check whether EQUILIBRIUM CONSTANT is less or higher than 1. If the equilibrium constant is less than one, then the equilibrium favors the starting material and vice versa.
The ionic chemical equation:
[tex]CH_3COOH + Na^+ ---- > CH_3COO^- + NaH[/tex]
The pKa of the conjugate acid, NaH is 35 approximately and the pKa of acetic acid is 4.7.
Therefore, the PK(equivalent) =pKa acid - pKa conjugate acid.
That is, PK(eq)= 4.7 - 35
pk(equivalent)= - 30.3
Then, the equilibrium constant= [tex]10^{-pK(equ)}[/tex]
That is; [tex]10^{-(-30.3)}= 1.9953 *10^{30}[/tex].
Here, we can see that the equilibrium constant is [tex]1.9953 * 10^{30}[/tex] that is greater than one, so, the equilibrium favors starting products.
Find more information about the equilibrium constant here:
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