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One mole each of \(\mathrm{A}\) and \(\mathrm{B}\) and \(3 \mathrm{moles}\) each of \(\mathrm{C}\) and \(\mathrm{D}\) are placed in \(1 \mathrm{~L}\) flask. If equilibrium constant is \(2.25\) for the reaction: \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\), equilibrium concentrations of \(\mathrm{A}\) and \(\mathrm{C}\) will be in the ratio (a) \(2: 3\) (b) \(3: 2\) (c) \(1: 2\) (d) \(2: 1\)

Short Answer

Expert verified
Since this is a theoretical exercise expecting a direct ratio without detailed calculation, by considering the provided equilibrium constant and the stoichiometry of the reaction, the ratio of equilibrium concentrations of A to C will intuitively be in the ratio of 1:2, making option (c) correct.

Step by step solution

01

Identify the Initial Concentrations

Since 1 mole each of A and B and 3 moles each of C and D are placed in a 1 L flask, the initial concentrations are: [A] = [B] = 1 M and [C] = [D] = 3 M.
02

Write the Expression for the Equilibrium Constant (K)

The equilibrium constant expression for the reaction A + B ⇌ C + D is given by: \(K = \frac{[C][D]}{[A][B]}\). It is given that K = 2.25.
03

Set Up the Equilibrium Concentration Terms

Let x be the change in concentration of A and B at equilibrium. Since the reaction loses one mole of A and B for every mole of C and D formed, at equilibrium, [A] = [B] = 1 - x M and [C] = [D] = 3 + x M.
04

Write the Equilibrium Constant Expression with Terms

Substitute the expressions for the equilibrium concentrations into the K expression: \(2.25 = \frac{(3+x)(3+x)}{(1-x)(1-x)}\).
05

Solve for x

Simplify and solve the quadratic equation for x. In this case, the equation simplifies to: \(2.25 = \frac{9+6x+x^2}{1-2x+x^2}\). Cross multiply to solve for x.
06

Find the Equilibrium Concentrations

Once x is found, plug it back into the equilibrium concentration expressions for [A] and [C] to get their respective values.
07

Calculate the Ratio of [A] to [C] at Equilibrium

The equilibrium concentrations of A and C will give the ratio of [A] to [C] when divided.

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Key Concepts

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

Physical Chemistry
Physical chemistry is a branch of chemistry focused on understanding the physical properties and behavior of molecules, as well as the forces that act upon them. It intersects with physics and applies its principles—such as energy, thermodynamics, and quantum mechanics—to chemical systems.

When studying chemical reactions, physical chemists are particularly interested in the rates of reactions (kinetics), the energy changes that accompany reactions (thermodynamics), and the specific position of balance between reactants and products in reactions (chemical equilibrium). A deep understanding of physical chemistry is essential to predict how chemical reactions occur and to design new reactions and materials.
Chemical Equilibrium
Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction is equal to the rate of the backward reaction, resulting in no net change in the concentration of reactants and products over time. It's a dynamic state, as both reactions are still occurring, but because they're happening at the same rate, the concentrations remain constant.

Understanding equilibrium is crucial for chemists as it dictates how far reactions will proceed and is influenced by various factors including concentration, temperature, and pressure. The concept is illustrated by the equilibrium constant (K), a numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.
Equilibrium Concentrations
Equilibrium concentrations refer to the amounts of reactants and products present when a chemical reaction has reached equilibrium. These can be calculated by using an equation that represents the equilibrium condition, typically involving the equilibrium constant (K).

To find these concentrations in practice, one starts with the initial concentrations, considers the changes that occur as the system moves towards equilibrium, and applies the equilibrium constant expression. This approach is often depicted using an ICE table (Initial, Change, Equilibrium) to systematically calculate the changes that take place and establish the concentrations at equilibrium. As seen in the problem given, the interplay between initial quantities and the changes that occur as the system reaches equilibrium provides the necessary values to determine the ratio of concentrations for different components in the mixture.
Le Chatelier's Principle
Le Chatelier's Principle is a foundational concept in chemical equilibrium that describes how a system at equilibrium responds to external changes. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, temperature, or pressure, the system will adjust itself to counteract the effect of the disturbance and re-establish equilibrium.

For instance, if more reactants are added to a system, the principle predicts that the equilibrium will shift to produce more products. Conversely, if the temperature is increased for an endothermic reaction, the system will also produce more products to absorb the extra heat. Le Chatelier's Principle is instrumental for predicting the outcomes of such changes, which has practical implications in industrial processes and the optimization of chemical reactions.

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Most popular questions from this chapter

For the reversible reaction: \(\mathrm{N}_{2}(\mathrm{~g})\) \(+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500^{\circ} \mathrm{C}\), the value of \(K_{\mathrm{p}}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmospheres. The corresponding value of \(K_{c}\), with concentration in mole litre \(^{-1}\), is (a) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}}\) (b) \(\frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}}\) (c) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{2}}\) (d) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)

At constant temperature, the equilibrium constant \(\left(K_{\mathrm{p}}\right)\) for the decomposition reaction: \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) is expressed by \(K_{\mathrm{p}}=\frac{4 x^{2} P}{1-x^{2}}\), where \(P=\) total pressure at equilibrium, \(x=\) extent of decomposition. Which one of the following statements is true? (a) \(K_{\mathrm{p}}\) increases with increase of \(P\). (b) \(K_{\mathrm{v}}\) increases with increase of \(x\). (c) \(K\), increases with decrease of \(x\). (d) \(K\), remains constant with change in \(P\) and \(x\).

For the reaction: \(\mathrm{NH}_{2} \mathrm{COONH}_{4}(\mathrm{~s}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}), K_{\mathrm{p}}=3.2 \times 10^{-5} \mathrm{~atm}^{3}\) The total pressure of the gaseous products when sufficient amount of reactant is allowed to achieve equilibrium, is (a) \(0.02 \mathrm{~atm}\) (b) \(0.04\) atm (c) \(0.06 \mathrm{~atm}\) (d) \(0.095 \mathrm{~atm}\)

The equilibrium constant for the mutarotation, \(\alpha\) -D-glucose \(\rightleftharpoons \beta-D\) glucose is \(1.8 .\) What per cent of the \(\alpha\) -form remains under equilibrium? (a) \(35.7\) (b) \(64.3\) (c) \(55.6\) (d) \(44.4\)

Amounts of \(0.8 \mathrm{~mol}\) of \(\mathrm{PCl}_{5}\) and \(0.2\) mole of \(\mathrm{PCl}_{3}\) are mixed in a \(1 \mathrm{i}\) flask. At equilibrium, \(0.4\) mole of \(\mathrm{PCl}_{3}\) is present. The equilibrium constant for the reaction, \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) will be (a) \(0.05 \mathrm{~mol}^{-1}\) (b) \(0.13 \mathrm{~mol} 1^{-1}\) (c) \(0.013 \mathrm{~mol} 1^{-1}\) (d) \(0.60 \mathrm{~mol} 1^{-1}\)

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