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Chapter 7: Problem 20

In reaction \(\mathrm{A}+2 \mathrm{~B} \rightleftharpoons 2 \mathrm{C}+\mathrm{D}\), initial concentration of \(\mathrm{B}\) was \(1.5\) times of \([\mathrm{A}]\), but at equilibrium the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) became equal. The equilibrium constant for the reaction is: (a) 8 (b) 4 (c) 12 (d) 6

Short Answer

Expert verified
The equilibrium constant is 4.

Step by step solution

01

Assign Initial Concentrations

Let the initial concentration of \( \mathrm{A} \) be \([\mathrm{A}]_0 = x\). Given that the initial concentration of \( \mathrm{B} \) is 1.5 times of \( [\mathrm{A}] \), thus \([\mathrm{B}]_0 = 1.5x\). At equilibrium, the concentrations of \( \mathrm{A} \) and \( \mathrm{B} \) are equal.
02

Determine Change in Concentrations

Assume \( \mathrm{A} \) dissociates by \( y \) at equilibrium. Therefore, \([\mathrm{A}]_e = x - y\) and \([\mathrm{B}]_e = 1.5x - 2y\). At equilibrium, it's given that \([\mathrm{A}]_e = [\mathrm{B}]_e \), hence:\[ x - y = 1.5x - 2y \]
03

Solve for y

Simplify the equation:\[ x - y = 1.5x - 2y \]\[ x + y = 1.5x \]\[ y = 0.5x \]
04

Substituting Values into Equilibrium Concentrations

Given \( y = 0.5x \), substitute back to find:\( [\mathrm{A}]_e = x - y = x - 0.5x = 0.5x \) and \([\mathrm{B}]_e = 1.5x - 2y = 1.5x - 1x = 0.5x \).
05

Expression for Products at Equilibrium

With the reaction stoichiometry, \([\mathrm{C}]_e = 2y = 2(0.5x) = x\) and \([\mathrm{D}]_e = y = 0.5x\).
06

Calculating the Equilibrium Constant

The equilibrium constant expression is:\[ K = \frac{[\mathrm{C}]^2[\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]^2} \]Substitute the equilibrium concentrations:\[ K = \frac{(x)^2(0.5x)}{(0.5x)(0.5x)^2} \]Simplify:\[ K = \frac{x^2 \times 0.5x}{0.5x \times 0.25x^2} = \frac{0.5x^3}{0.125x^3} = 4 \]
07

Conclusion

The equilibrium constant for the reaction is 4.

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Key Concepts

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

Reaction Stoichiometry
Reaction stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. In the given exercise, the reaction is expressed as \( \text{A} + 2\text{B} \rightleftharpoons 2\text{C} + \text{D} \). This indicates that one mole of A reacts with two moles of B to produce two moles of C and one mole of D. Understanding stoichiometry helps in predicting how much product will form from given amounts of reactants and helps determine their concentrations at equilibrium. It is essential to balance chemical equations accurately to use stoichiometry effectively for any calculations.
Initial Concentration
Initial concentration is the amount of a substance present in a solution before a chemical reaction reaches equilibrium. In this problem, the initial concentration of A is expressed as \( x \), while B's initial concentration is specified as 1.5 times that of A, or \( 1.5x \). These values lay the groundwork for calculating the changes in concentration as the reaction progresses towards equilibrium. Initial concentrations are crucial for deriving the changes in concentration and solving for the equilibrium constant.
Equilibrium Concentration
Equilibrium concentration is the concentration of reactants and products when a chemical reaction reaches a state where the forward and reverse reactions occur at the same rate. In this exercise, the equilibrium concentrations for A and B end up being equal due to the given conditions, leading us to the relations \([\text{A}]_e = [\text{B}]_e = 0.5x\). For the products, stoichiometry helps determine \([\text{C}]_e = x\) and \([\text{D}]_e = 0.5x\). Understanding equilibrium concentrations is essential for analyzing the extent of a reaction and computing the equilibrium constant.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in stable concentrations of reactants and products. It doesn't mean the reactions stop; they continue dynamically at equal rates. In our reaction, equilibrium is described where the concentrations of A and B are equal, guiding us to the equilibrium constant's final calculation. Knowing the concept of equilibrium is fundamental in predicting how changes in concentration, temperature, or pressure can affect a reaction system.
Concentration Change
Concentration change refers to the alteration in the amount of substance during the reaction until equilibrium is reached. In this problem, if A loses \( y \) units, the resulting concentration changes are applied to B as it uses twice the amount due to stoichiometry. Ultimately, these changes were used to find \( y = 0.5x \), which then let us define the subsequent equilibrium concentrations: A and B ending as \( 0.5x \), C as \( x \), and D as \( 0.5x \). Understanding concentration changes is vital for determining the progression and outcome of chemical reactions.

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Most popular questions from this chapter

Consider the following reaction: \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) ; \mathrm{DH}^{0}=+58 \mathrm{~kJ}\) For each of the following cases ((i), (ii)), the direction in which the equilibrium shifts is : (i) Temperature is decreases (ii) Pressure is increased by adding \(\mathrm{N}_{2}\) at constant \(\mathrm{T}\). (a) (i) towards product, (ii) towards product (b) (i) towards reactant, (ii) towards product (c) (i) towards reactant, (ii) no change (d) (i) towards product, (ii) no change

For the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) at a given temperature, the equilibrium amount of \(\mathrm{CO}_{2}(\mathrm{~g})\) can be increased by [1998 - 2 Marks] (a) adding a suitable catalyst (b) adding an inert gas (c) decreasing the volume of the container (d) increasing the amount of \(\mathrm{CO}(\mathrm{g})\).

$$ \begin{aligned} &\mathrm{NaOH} \text { is a strong base. What will be } \mathrm{pH} \text { of } 5.0 \times 10^{-2} \mathrm{M} \mathrm{NaOH}\\\ &\text { solution ? }(\log 2=0.3) \end{aligned} $$ (a) \(14.00\) (b) \(13.70\) (c) \(13.00\) (d) \(12.70\)

An element which can exist as a positive ion in acidic solution and also as a negative ion in basic solution is said to be ...............

Solubility product constant \(\left(K_{s p}\right)\) of salts of types \(M X, M X_{2}\) and \(M_{3} X\) at temperature \(T\) are \(4.0 \times 10^{-8}, 3.2 \times 10^{-14}\) and \(2.7 \times 10^{-15}\), respectively. Solubilities (mol \(\mathrm{dm}^{-3}\) ) of the salts at temperature ' \(T\) are in the order (a) \(M X>M X_{2}>M_{3} X\) (b) \(\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}_{2}>\mathrm{MX}\) (c) \(M X_{2}>M_{3} X>M X\) (d) \(M X>M_{3} X>M X_{2}\)
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