For the reaction \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). If the initial concentration of \(\left[\mathrm{H}_{2}\right]=\left[\mathrm{CO}_{2}\right]\) and \(x\) moles/litre of hydrogen is consumed at equilibrium, the correct expression of \(K_{p}\) is : (a) \(\frac{x^{2}}{(1-x)^{2}}\) (b) \(\frac{(1+x)^{2}}{(1-x)^{2}}\) (c) \(\frac{x^{2}}{(2+x)^{2}}\) (d) \(\frac{x^{2}}{1-x^{2}}\)

Short Answer

Expert verified
The correct expression for \(K_{p}\) is \(\frac{x^{2}}{(1-x)^{2}}\), which is option (a).

Step by step solution

01

Write Down The Equilibrium Expression

For the given reaction \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\), the equilibrium constant \(K_p\) is expressed in terms of the partial pressures of the gases. \(K_p = \frac{P_{CO} \cdot P_{H2O}}{P_{H2} \cdot P_{CO2}}\)
02

Determine The Changes In Concentration

Initially, the concentrations of \(H_2\) and \(CO_2\) are the same. Let's call this initial concentration '1'. When 'x' moles of \(H_2\) are consumed, 'x' moles of \(CO_2\), \(CO\), and \(H_2O\) will react/form. The equilibrium concentrations will be: \[P_{H2} = P_{CO2} = 1 - x\] \[P_{CO} = P_{H2O} = x\]
03

Plug The Equilibrium Concentrations Into The Equilibrium Expression

Substitute the equilibrium concentrations into the \(K_p\) expression: \(K_p = \frac{x \cdot x}{(1 - x) \cdot (1 - x)} = \frac{x^{2}}{(1 - x)^{2}}\)
04

Choose The Correct Answer

The expression for \(K_p\) that we have derived is \(\frac{x^{2}}{(1 - x)^{2}}\), which corresponds to option (a).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in the concentrations of the reactants and products remaining constant over time. It's essential to understand that reaching equilibrium does not mean the reactants and products are present in equal amounts, but rather that their ratios don't change. At equilibrium, the system's macroscopic properties become stationary, yet at the molecular level, reactions still occur, just at equal rates in both directions.

It's crucial when evaluating equilibrium to recognize that while the macroscopic properties do not change, the forward and reverse reactions have reached a balance, not necessarily stopped. This caveat is often an area where misunderstanders arise, and it's important for clarity in explaining the concept to students.
Partial Pressures
Partial pressures refer to the pressure contributed by a single gas in a mixture of gases. Each gas in a mixture exerts a pressure independently as if the other gases weren't present. Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas. For gas-phase equilibria, like the one mentioned in the given exercise, partial pressures are directly proportional to their molar concentration when the temperature remains constant.

It's important for students to grasp that partial pressures reflect the mole fractions of the gases in a confined space and can be used to describe the concentration of gases in equilibrium calculations. Understanding how to measure and calculate partial pressures is key to solving problems involving gas equilibria.
Reaction Quotient
The reaction quotient, denoted as Q, is a value that changes as a reaction proceeds and reflects the concentrations of the reactants and products at a given instant, as opposed to the equilibrium constant K, which is fixed for a reaction at a specific temperature. The reaction quotient is calculated using the same expression as the equilibrium constant but with the current concentrations or partial pressures of the reactants and products.

By comparing Q to K, predictions can be made about the direction in which the reaction will proceed. If Q < K, the forward reaction will be favored; if Q > K, the reaction will shift towards the reactants; and if Q = K, the system is at equilibrium. Familiarity with Q is very useful for students as it allows them to gauge the progress of a reaction in real-time.
Equilibrium Calculations
The heart of solving equilibrium problems involves balance and proportions. Equilibrium calculations frequently require the setting up of an equation that incorporates the equilibrium constant and the expression for that constant based on concentrations or partial pressures of reactants and products. These calculations are central to understanding how systems respond to changes in conditions.

In the case of the exercise provided, the correct equilibrium constant expression is derived by substituting the equilibrium concentrations into the equilibrium expression. Helping students master equilibrium calculations often involves practice with setting up and solving algebraic equations and understanding how these solutions reflect the behavior of real chemical systems at equilibrium. Knowing how to manipulate these expressions and calculate the equilibrium constant from given data is an essential chemistry skill.

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Most popular questions from this chapter

In a system \(A(s) \rightleftharpoons 2 B(g)+3 C(g)\), if the concentration of \(C\) at equilibrium is increased by a factor of 2 , it will cause the equilibrium concentration of \(B\) to change to : (a) two times the original value (b) one half of its original value (c) \(2 \sqrt{2}\) times to the original value (d) \(\frac{1}{2 \sqrt{2}}\) times the original value

For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \quad \Delta H=-93.6 \mathrm{~kJ} \mathrm{~mol}^{-1} $$ the number of moles of \(\mathrm{H}_{2}\) at equilibrium will increase if : (a) volume is increased (b) volume is decreased (c) argon gas is added at constant volume (d) \(\mathrm{NH}_{3}\) is removed

The equilibrium constant \(K_{c}\) for the reaction \(\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)\) is \(16 .\) If 1 mole of each of all the four gases is taken in \(1 \mathrm{dm}^{3}\) vessel, the equilibrium concentration of NO would be : (a) \(0.4 \mathrm{M}\) (b) \(0.6 \mathrm{M}\) (c) \(1.4 \mathrm{M}\) (d) \(1.6 \mathrm{M}\)

In the presence of excess of anhydrous \(\mathrm{SrCl}_{2}\), the amount of water taken up is governed by \(K_{p}=10^{12} \mathrm{~atm}^{-4}\) for the following reaction at \(273 \mathrm{~K}\) $$ \mathrm{SrCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{SrCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s) $$ What is equilibrium vapour pressure (in torr) of water in a closed vessel that contains \(\mathrm{SrCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) ?\) (a) \(0.001\) torr (b) \(10^{3}\) torr (c) \(0.76\) torr (d) \(1.31\) torr

At a certain temperature, only \(50 \% \mathrm{HI}\) is dissociated at equilibrium in the following reaction: $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ The equilibrium constant for this reaction is : (a) \(0.25\) (b) \(1.0\) (c) \(3.0\) (d) \(0.5\)

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