Describe two ways you could make this equilibrium shift toward reactants that haven't already been discussed in this problem. Explain your reasoning.
Electronic Effects
The effect of electrons that are located in the chemical bonds within the atoms of the molecule is termed an electronic effect. The electronic effect is also explained as the effect through which the reactivity of the compound in one portion is controlled by the electron repulsion or attraction producing in another portion of the molecule.
Drawing Resonance Forms
In organic chemistry, resonance may be a mental exercise that illustrates the delocalization of electrons inside molecules within the valence bond theory of octet bonding. It entails creating several Lewis structures that, when combined, reflect the molecule's entire electronic structure. One Lewis diagram cannot explain the bonding (lone pair, double bond, octet) elaborately. A hybrid describes a combination of possible resonance structures that represents the entire delocalization of electrons within the molecule.
Using Molecular Structure To Predict Equilibrium
Equilibrium does not always imply an equal presence of reactants and products. This signifies that the reaction reaches a point when reactant and product quantities remain constant as the rate of forward and backward reaction is the same. Molecular structures of various compounds can help in predicting equilibrium.
Image # 2 is the question.
![Check Your Understanding
Describe two ways you could make this equilibrium shift toward reactants that haven't already
been discussed in this problem. Explain your reasoning.
Answer
Possibilities include removing HCN from the system or adding an acid (other than HCN),
which would increase [H+].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F783eec75-1af8-40ca-a14b-856d041c9d32%2F95050b67-be28-4999-95da-fdef2b62e02c%2F9a6x0ky_processed.png&w=3840&q=75)
![Solution
The equilibrium is
CH3 COOH(aq) 2H*(aq)+CH3COO (aq)
(a) Sodium acetate is added and will dissociate into Na* and CH3 COO¯. The sodium ion is not
involved in this equilibrium, but by adding CH; COO¯,we increase the concentration of a
product. This in turn increases the value of Q, so Q > K. Equilibrium must shift toward
reactants, so additional CH3 COOH will form.
(b) Additional acetic acid is added. Because CH3 COOH is a reactant, the increase in
concentration of the acid decreases the value of Q. Q becomes less than K, so equilibrium
shifts toward products. Some of the added CH3 COOH will dissociate, and the concentration of
both Ht and CH3 COO will increase.
(c) Sodium hydroxide is added. Again the sodium ion does not play a significant role in this
perturbation, and at first glance, it might seem like the hydroxide ion doesn't either. But if you
use a little bit of the chemical intuition you've developed by this point in the course, you
should realize that OH- and H+ ions will react with one another to form water. This will
decrease the concentration of H+. Because the concentration of a product is reduced, the size
of the numerator decreases, making Q < K. Thus the equilibrium shifts to form more
products.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F783eec75-1af8-40ca-a14b-856d041c9d32%2F95050b67-be28-4999-95da-fdef2b62e02c%2Fqpwpwrg_processed.png&w=3840&q=75)
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