Problem 15 - Chemistry in 30 days

Question

For the reaction \(2 A(g)+B(g) \rightleftharpoons 3 C(g)+D(g)\) Two moles each of \(A\) and \(B\) were taken into a 11 flask. The following must always be true when the system attained equilibrium (a) \([A]=[B]\) (b) \(|A|<[B \mid\) (c) \([B]=[C]\) (d) \(\\{A \mid+[B]<[C]+[D]\)

Step-by-Step Solution

Answer

Choice (d) is always possible when enough reaction occurs, making \([A] + [B] < [C] + [D]\).

Step 1: Understand the Initial Setup

We start with 2 moles each of gases A and B in a flask at equilibrium. The balanced equation for the reaction is \(2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)\). With these as reactants, the change that occurs at equilibrium needs to be evaluated.

Step 2: Define Changes and Equilibrium Expression

Let the change in moles of \(B\) be \(x\). Therefore, \([A] = 2 - 2x\), \([B] = 2 - x\), \([C] = 3x\), \([D] = x\) at equilibrium, because \(x\) moles of \(B\) are used, reducing \(A\) by \(2x\), increasing \(C\) by \(3x\), and \(D\) by \(x\).

Step 3: Compare the Given Choices

Given these equilibrium expressions, we compare the choices: (a) \([A] = 2 - 2x\) and \([B] = 2 - x\), so generally \([A] eq [B]\). (b) \([A] < [B]\) would be true as \(2 - 2x < 2 - x\). (c) \([B] eq [C]\) since \(2 - x eq 3x\) generally. (d) \([A] + [B] = 4 - 3x\) and \([C] + [D] = 4x\); hence, \(4 - 3x < 4x\) might be true when \(x > 0\).

Step 4: Evaluate Inequality in Choice (d)

For \([A] + [B] < [C] + [D]\), solve \(4 - 3x < 4x\). This simplifies to \(4 < 7x\) or \(x > \frac{4}{7}\). This represents the scenario where choice (d) can be true upon a sufficient forward reaction shift.

Key Concepts

Equilibrium Expression

The equilibrium expression is a fundamental aspect of chemical equilibrium, describing the relationship between the concentrations of reactants and products at equilibrium. In a reversible chemical reaction like \(2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)\), you can use the equilibrium constant, \(K_{eq}\), to express this balance.

The equilibrium expression for the reaction is formulated based on the balanced equation and is written as:

• \(K_{eq} = \frac{[C]^3[D]}{[A]^2[B]}\)

Here, the concentrations of the products \(C\) and \(D\) are in the numerator, raised to the power of their coefficients (3 and 1, respectively), while the concentrations of the reactants \(A\) and \(B\) are in the denominator, reflecting their respective stoichiometric coefficients from the balanced equation.

This expression helps to predict how the concentrations of various species will change as the system reaches equilibrium, providing insight into the direction of reaction shifts under different conditions.

Reaction Stoichiometry

Stoichiometry encompasses the quantitative relationships between the reactants and products in a chemical reaction. For the reaction \(2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)\), stoichiometry tells us how the amounts of substances relate through their coefficients.

In this equation:

• 2 moles of \(A\) react with 1 mole of \(B\) to produce 3 moles of \(C\) and 1 mole of \(D\).

These coefficients guide how we calculate the changes in moles as the reaction proceeds towards equilibrium. If the change in moles of \(B\) is \(x\), accordingly:

• \([A] = 2 - 2x\)
• \([B] = 2 - x\)
• \([C] = 3x\)
• \([D] = x\)

This shows that for every unit change in \(B\), there's a proportional change in \(A\), \(C\), and \(D\), determined by their respective stoichiometric proportions.

Mole Concept

The mole concept is crucial in understanding how quantities of substances are measured and related in chemistry. It allows us to convert between atoms, molecules, or ions and moles, the SI unit for amount of substance.

In the context of our reaction \(2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)\), it helps us understand how much of each substance is involved:

• Initially, we have 2 moles each of \(A\) and \(B\).
• As the reaction proceeds and reaches equilibrium, the changes in moles are tracked using the stoichiometric coefficients.
• This means as \(x\) moles of \(B\) react, \(2x\) moles of \(A\) are used, and \(3x\) moles of \(C\) and \(x\) moles of \(D\) are formed.

The mole concept essentially helps in predicting and measuring the transformations occurring in the chemical reaction until equilibrium is achieved.

Dynamic Equilibrium

Dynamic equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction, leading to no net change in the concentrations of reactants and products. Despite no apparent changes on the macroscopic level, continuous microscopic changes occur.

In the reaction \(2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)\), dynamic equilibrium is achieved when:

• The formation of products from reactants is balanced by the reverse process.
• Concentrations of \(A\), \(B\), \(C\), and \(D\) remain constant over time.

The key characteristic of dynamic equilibrium is this balance in change as seen in step-by-step reactions on both sides of the equation.

This state highlights why \([A] + [B]\) can be less than \([C] + [D]\) when equilibrium shifts due to changes like pressure, concentration, or temperature, in agreement with Le Chatelier's principle. Dynamic equilibrium emphasizes that while the chemical reactions continue to occur, the overall concentrations no longer change.