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Chapter 7: Problem 67

\(1.1\) mole of \(\Lambda\) is mixed with \(2.2\) molc of \(B\) and the mixture is then kept in \(1-\mathrm{L}\) flask till the cquilibrium is attained \(\Lambda+2 \mathrm{~B} \rightleftharpoons 2 \mathrm{C}+\mathrm{D} \cdot \Lambda \mathrm{t}\) the equilibrium \(0.2\) mole of \(C\) are formed. The equilibrium constant of the reaction is (1) \(0.001\) (2) \(0.222\) (3) \(0.003\) (4) \(0.004\)

Short Answer

Expert verified
The equilibrium constant is 0.001, which corresponds to option (1).

Step by step solution

01

Write the balanced chemical equation

The given chemical reaction is: \[ \Lambda + 2B \rightleftharpoons 2C + D \]
02

Determine initial moles

Initially, we have 1.1 moles of \( \Lambda \) and 2.2 moles of \( B \). The initial moles of \( C \) and \( D \) are zero.
03

Define the change in moles at equilibrium

At equilibrium, 0.2 moles of \( C \) are formed. Since the stoichiometric coefficient of \( C \) is 2, the change in moles for \( C \) will be twice the change in moles for \( \Lambda \). If 0.2 moles of \( C \) are formed, the changes are as follows: \[ \Delta [C] = +0.2 \text{ moles} \] Thus, the change for \( \Lambda \) is \[ \Delta [\Lambda] = -0.1 \text{ moles} \] And for \( B \), the change is \[ \Delta [B] = -2 \times 0.1 = -0.2 \text{ moles} \] For \( D \), which has a stoichiometric coefficient of 1, the change will be half that of \( C \): \[ \Delta [D] = +0.1 \text{ moles} \]
04

Calculate equilibrium moles

Equilibrium moles for each substance are: \[ [\Lambda]_{eq} = 1.1 - 0.1 = 1.0 \text{ moles} \] \[ [B]_{eq} = 2.2 - 0.2 = 2.0 \text{ moles} \] \[ [C]_{eq} = 0 + 0.2 = 0.2 \text{ moles} \] \[ [D]_{eq} = 0 + 0.1 = 0.1 \text{ moles} \]
05

Calculate equilibrium concentrations

The volume of the flask is 1 L, so the concentrations are equal to the moles: \[ [\Lambda] = 1.0 \text{ M} \] \[ [B] = 2.0 \text{ M} \] \[ [C] = 0.2 \text{ M} \] \[ [D] = 0.1 \text{ M} \]
06

Write the expression for the equilibrium constant

The expression for the equilibrium constant for the given reaction is: \[ K_c = \frac{[C]^2 [D]}{[\Lambda] [B]^2} \]
07

Substitute the equilibrium concentrations into Kc expression

Substitute the values: \[ K_c = \frac{(0.2)^2 \times 0.1}{1.0 \times (2.0)^2} = \frac{0.04 \times 0.1}{1 \times 4} = \frac{0.004}{4} = 0.001 \]
08

Identify the correct option

The equilibrium constant is 0.001, which corresponds to option (1).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
Chemical equilibrium refers to the state in which the concentrations of reactants and products remain constant over time. The equilibrium constant, symbolized as \( K_c \), quantifies the ratio of product concentrations to reactant concentrations at equilibrium. For a balanced reaction
Stoichiometry
Stoichiometry deals with the quantitative relationship between reactants and products in a chemical reaction. This relationship is essential for solving equilibrium problems. In the given exercise, the stoichiometric coefficients from the balanced equation \( \Lambda + 2B \rightleftharpoons 2C + D \) dictate the mole ratios: 1 mole of \( \Lambda \) reacts with 2 moles of \( B \) to produce 2 moles of \( C \) and 1 mole of \( D \).
Mole Calculations
In chemical reactions, mole calculations involve tracking the number of moles of each substance as they change from reactants to products. In the provided solution, you start with initial moles of \( \Lambda \) and \( B \) and use stoichiometry to find the moles of \( C \) formed and the corresponding changes for all reactants and products.
Chemical Reactions
Chemical reactions describe the process where reactants convert into products. Each reaction follows a specific pathway dictated by the balanced chemical equation and the conditions set by stoichiometry and kinetics. To analyze and predict the outcomes of these reactions, especially at equilibrium, it's crucial to comprehend the balanced equation and the behaviors of involved substances.

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