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When phosphine reacts with ammonium, will the equilibrium shift to right or left?​

2. when CH2COO- reacts with hydrochloric acid, will the equilibrium shift to right or left?

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When phosphine (PH3) reacts with ammonium (NH4+), the reaction can be represented as follows:

PH3 + NH4+ ⇌ NH3 + H2P-

In this reaction, the equilibrium will shift to the right. The formation of ammonia (NH3) and the phosphide ion (H2P-) is favored, as they are the products of the reaction. This can be understood by Le Chatelier's principle, which states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift to counteract that change. In this case, as ammonia and the phosphide ion are on the right side of the equation, an increase in their concentrations will cause the equilibrium to shift to the right to consume some of the reactants (phosphine and ammonium) and produce more products.

When CH2COO- (acetate ion) reacts with hydrochloric acid (HCl), the reaction can be represented as follows:

CH2COO- + HCl ⇌ CH3COOH + Cl-

In this reaction, the equilibrium will shift to the left. The formation of acetic acid (CH3COOH) and chloride ion (Cl-) is favored. This is because hydrochloric acid is a strong acid, and when it donates a proton (H+) to the acetate ion, it forms acetic acid and chloride ion. Acetic acid is a weaker acid than HCl, so the equilibrium will shift to the left to minimize the formation of more acetic acid. This is in accordance with Le Chatelier's principle, as an increase in the concentration of acetic acid will cause the equilibrium to shift to the left to consume some of the products and favor the reactants.

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User Lotz
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