Question

In a system \(A(s) \rightleftharpoons 2 B(g)+3 C(g)\), if the concentration of \(C\) at equilibrium is increased by a factor of 2 , it will cause the equilibrium concentration of \(B\) to change to : (a) two times the original value (b) one half of its original value (c) \(2 \sqrt{2}\) times to the original value (d) \(\frac{1}{2 \sqrt{2}}\) times the original value

Short answer

The equilibrium concentration of \(B\) will change to \(\frac{1}{2 \sqrt{2}}\) times the original value.

Step-by-step solution

Understanding the Equilibrium Expression

Establish the equilibrium expression for the reaction using the law of mass action. For the reaction, the equilibrium constant expression (K) is written based on the stoichiometry of the reaction as follows: \(K = \frac{{[B]^2[C]^3}}{{[A]}}\)

Analyzing the Effect of Changing Concentration

When the concentration of \(C\) is increased by a factor of 2, the new concentration of \(C\) at equilibrium becomes \(2[C]\). However, because the value of K must remain the same at constant temperature, increasing \([C]\) by a factor of 2 will require adjusting \([B]\) to maintain the equilibrium constant value.

Applying the Change to the Equilibrium Expression

Because the equilibrium constant must remain constant, if we represent the original concentration of \(B\) as \([B]\), the new equilibrium concentration of \(B\) can be represented as \(x[B]\). Plug in the values into the equilibrium expression to solve for \(x\): \(K = \frac{{(x[B])^2(2[C])^3}}{{[A]}}\). Since \(K\) and \([A]\) are constant, and the original expression was \(K = \frac{{[B]^2[C]^3}}{{[A]}}\), by the coefficients of the reaction, the change in \([C]\) to the power of 3 will be compensated by a change in \([B]\) to the power of 2 equating \(x^2 = \frac{1}{2^3} \Rightarrow x = \frac{1}{2 \sqrt{2}}\).

Key concepts

Understanding the Equilibrium Constant

The equilibrium constant, represented by the symbol K, is a number that provides us with a ratio of the concentration of the products to the concentration of the reactants for a reaction at equilibrium, each raised to the power of their coefficients. The formula for a reaction \( A(s) \rightleftharpoons 2 B(g)+3 C(g) \) would be \( K = \frac{{[B]^2[C]^3}}{{[A]}} \).

This number is vital because it tells us the extent to which a reaction will proceed before reaching equilibrium. A high value indicates a greater concentration of products, while a low value suggests a reaction that favors the reactants. In the classroom, it's crucial to highlight that different reactions have their own unique equilibrium constants which depend on the temperature of the system.

Applying Le Chatelier's Principle

Le Chatelier's principle is a rule of thumb that helps us predict the effect of a change in conditions on a chemical equilibrium. The principle states that if an external condition is imposed on a system at equilibrium, the system adjusts to partially oppose the effect of the disturbance. It's like a chemical tug-of-war.

For example, increasing the concentration of a reactant or product will shift the equilibrium in the direction that tends to decrease that concentration. If we focus on our classroom example where the concentration of C is doubled, Le Chatelier's principle would predict a shift in the equilibrium to counteract this change, affecting the concentrations of other substances in the system, such as B.

Stoichiometry's Role in Equilibrium

Stoichiometry is the calculation of reactants and products in chemical reactions. It is fundamental for understanding chemical equilibrium because it relates the proportions in which reactants combine and products form. When dealing with equilibrium, as we have in our exercise, the stoichiometric coefficients come into play in the equilibrium constant expression.

In our reaction, the coefficients are 1 for A, 2 for B, and 3 for C. This means that an equilibrium shift affecting C will have a more significant effect due to its higher coefficient. Students must appreciate that stoichiometry isn't just numbers; it's about the relative amounts and how they dictate the behavior of a chemical system.

The Mass Action Law Fundamentals

The law of mass action is the foundation upon which the concept of chemical equilibrium rests. It states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised to the power corresponding to the coefficient in the balanced equation. In essence, the concentrations of the chemicals at equilibrium are governed by this law.

It's important in our subject exercise that an increase in the concentration of \(C\) must be offset by a decrease in the concentration of \(B\) to maintain the K value according to the law of mass action. When the concentration of \(C\) is doubled, our equation requires solving for the new concentration of \(B\) while keeping K constant, which is directly influenced by the stoichiometry of the reacting substances. The mass action law is the key that allows us to systematically approach the change in equilibrium and solve for the unknown concentrations.