Question

According to the law of mass action, which of these conditions will drive the reaction \(\mathrm{A}+\mathrm{B} \rightleftarrows \mathrm{C}\) to the right? a. an increase in the concentration of \(\mathrm{A}\) and \(\mathrm{B}\) b. a decrease in the concentration of \(C\) c. an increase in the concentration of enzyme d. both \(a\) and \(b\) e. both \(b\) and \(c\)

Short answer

The correct answer is option d (both a and b). Increasing the concentrations of \(A\) and \(B\) and decreasing the concentration of \(C\) will drive the reaction \(A+B \rightleftarrows C\) to the right, favoring the formation of \(C\) according to the law of mass action.

Step-by-step solution

Law of Mass Action

The law of mass action states that the rate of a reaction is directly proportional to the product of the concentrations of the reacting substances raised to the powers of their stoichiometric coefficients. For the given reaction \(A+B \rightleftarrows C\), we can write the following equilibrium expression: \[K = \frac{[C]}{[A][B]}\]

Determining the Effects of Changing Concentrations

We need to figure out what will happen when we change the concentrations of \(A\), \(B\), and \(C\), or introduce an enzyme, which of these options will result in the reaction proceeding to the right? a. An increase in the concentration of \(A\) and \(B\) will cause the forward reaction to accelerate, pushing the equilibrium towards product \(C\). b. A decrease in the concentration of \(C\) will also push the equilibrium towards product \(C\), because the concentration of \(C\) in the denominator of the equation and a decrease would make the ratio less than K. It would need to reverse the equilibrium. c. An increase in the concentration of the enzyme does not affect the equilibrium itself; it only increases the rate at which equilibrium is reached. It does not favor any side, either reactants or products.

Combining the Results from the Options

Based on the information above, we can combine the potential conditions that will drive the reaction to the right: Both increasing the concentrations of \(A\) and \(B\) (option a) and decreasing the concentration of \(C\) (option b) will drive the reaction to the right, favoring the formation of \(C\). The correct answer is option d (both a and b).

Key concepts

Chemical Equilibrium

Chemical equilibrium plays a pivotal role in understanding how chemical reactions occur within a closed system. When a reaction reaches equilibrium, the rate of the forward reaction, where reactants convert to products, equals the rate of the reverse reaction, where products revert to reactants. This balance does not imply that the reactants and products are present in equal amounts; rather, their concentrations remain constant over time. An important aspect of equilibrium involves Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change.

For the equilibrium of the reaction \(A + B \rightleftarrows C\), if we altered the system by increasing the concentration of reactants \(A\) and \(B\), or by decreasing the concentration of product \(C\), according to Le Chatelier’s Principle, the reaction would shift towards the products to restore equilibrium. This explains why options a and b in the exercise would drive the reaction to the right, towards the formation of more \(C\).

Reaction Rate

Reaction rate refers to the speed at which reactants turn into products. It is usually expressed as the change in concentration of a reactant or product per unit time. This rate can be influenced by several factors such as temperature, pressure, concentration of the reactants, and the presence of a catalyst or enzyme.

For the reaction discussed, \(A + B \rightleftarrows C\), increasing the concentrations of the reactants \(A\) and \(B\) would generally result in an increased reaction rate due to more frequent collisions between the reacting molecules. Conversely, the concentration of an enzyme typically affects only the speed at which equilibrium is reached, not the position of the equilibrium itself. Enzymes lower the activation energy required for the reaction to proceed, facilitating a faster reaction without shifting the balance of reactants and products at equilibrium.

Equilibrium Constant

The equilibrium constant, denoted by \(K\), provides a quantitative measure of the position of equilibrium in a chemical reaction. For a generic reaction \(aA + bB \rightleftarrows cC + dD\), the equilibrium constant can be written as \[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]. This expression includes the concentrations of the products and reactants raised to the power of their stoichiometric coefficients.

If the value of \(K\) is known for a certain temperature, it can be used to predict the direction of the reaction when the concentration of one or more components is changed. A higher concentration of products relative to reactants will result in a value greater than \(K\), indicating that the reaction will shift towards the reactants to achieve equilibrium. Conversely, if this ratio is less than \(K\), the reaction will shift towards the products to restore equilibrium.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. By using balanced chemical equations, stoichiometry provides a means to predict the amounts of substances consumed and produced in a given reaction.

In the context of the law of mass action, stoichiometry is used to determine the powers to which the concentration of each reactant is raised in the equilibrium expression. In the equation \(A+B \rightleftarrows C\), each reactant has a stoichiometric coefficient of 1, which is why the concentrations of \(A\) and \(B\) are to the first power in the equilibrium expression. In more complex reactions, understanding stoichiometry allows chemists to manipulate conditions (such as concentrations) to favor the production of desirable products while minimizing unwanted byproducts.