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The equilibrium constant \(\left(K_{c}\right)\) for the reaction $$ 2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is \(4 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). What is the equilibrium constant for the reaction? $$ \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{HCl}(g) $$ (a) \(2 \times 10^{-17}\) (b) \(2.5 \times 10^{33}\) (c) \(5 \times 10^{6}\) (d) None of these

Short Answer

Expert verified
The equilibrium constant for the reaction 0.5 H2(g) + 0.5 Cl2(g) ⇌ HCl(g) is (b) 2.5 × 10^{33}.

Step by step solution

01

Identify the Relationship Between the Two Reactions

The original equilibrium reaction 2 HCl(g) ⇌ H2(g) + Cl2(g) has been given an equilibrium constant (Kc). The new equilibrium reaction 0.5 H2(g) + 0.5 Cl2(g) ⇌ HCl(g) can be viewed as the reverse of the given reaction with each of the coefficients divided by 2.
02

Calculate the Inverse of the Equilibrium Constant

To find the equilibrium constant for the reverse reaction, take the inverse of the given equilibrium constant. Thus, the equilibrium constant for the reverse reaction (the desired reaction) is Kc' = 1 / (4 × 10^(-34)).
03

Adjust for the Change in Coefficients

Given that the stoichiometric coefficients for the desired reaction are halved, raise the equilibrium constant for the reverse reaction to the power corresponding to the fraction of the change in coefficients. Thus, the equilibrium constant for the desired reaction will be (Kc')^(1/2).
04

Compute the New Equilibrium Constant

Perform the calculations to find Kc'^(1/2) = (1 / (4 × 10^(-34)))^(1/2).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium represents a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. In this balanced state, the concentrations of the reactants and products remain constant over time, even though the reactions continue to occur. Equilibrium can be reached in both physical processes, like the evaporation of water, and chemical processes, such as the synthesis and decomposition of hydrogen chloride gas.

It's crucial to note that being at equilibrium does not mean the reactants and products are present in equal amounts; rather, it means their ratios do not change. The equilibrium constant (\(K_{c}\)), fundamental to this concept, quantifies the equilibrium state in terms of the concentrations of reactants and products. Each reaction has its specific equilibrium constant that depends on temperature and reflects the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients.
Le Chatelier's Principle
Le Chatelier's principle is central to predicting how a system at equilibrium responds to changes in concentration, temperature, or pressure. It posits that when a stress is applied to a system in dynamic equilibrium, the system adjusts to minimize the effect of that disturbance.

For example, if more reactant is added to a system, the equilibrium will shift to produce more product; conversely, removing a product will result in the production of more of that product to counteract the change. Temperature changes can also alter equilibrium: increasing the temperature favors the endothermic direction of the reaction, while decreasing it favors the exothermic reaction. Understanding this principle helps chemists control product formation and optimize industrial processes.
Stoichiometry in Equilibrium Reactions
Stoichiometry in equilibrium reactions refers to the quantitative relationship between reactants and products. In the example given, the stoichiometric coefficients are crucial for calculating the equilibrium constant for a related reaction. If a balanced chemical equation is altered, the equilibrium constant must be adjusted to reflect these changes.

When we manipulate the coefficients of a balanced equation, we need to adjust the equilibrium constant to a power that corresponds to the changes in stoichiometry. If we halve all the coefficients in the reaction, as in the exercise, the equilibrium constant for the new reaction will be the square root of the equilibrium constant for the original reaction. This stoichiometric adjustment is an example of how equilibrium constants are sensitively linked to the reaction equation.
Gaseous Reactions
Gaseous reactions are chemical reactions involving gases, which are subject to the principles of chemical equilibrium and the laws of gases. In the case of the equilibrium constant expression for gaseous reactions, the concentrations are often expressed in terms of partial pressures, denoted as \(K_p\). However, for our example, we use concentrations and the equilibrium constant is given as \(K_c\).

The behavior of gaseous reactions can also be influenced by changes in pressure. According to Le Chatelier's principle, increasing the pressure on a gaseous system will shift the equilibrium towards the side with fewer moles of gas. This is a key consideration when studying the equilibrium of reactions involving gases, as varying the pressure can alter the equilibrium position.

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Most popular questions from this chapter

A plot of Gibbs energy of a reaction mixture against the extent of the reaction is : (a) minimum at equilibrium (b) zero at equilibrium (c) maximum at equilibrium (d) None of these

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) \(0.5\)

The equilibrium constant for the reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) is \(5 .\) How many moles of \(\mathrm{CO}_{2}\) must be added to 1 litre container already containing 3 moles each of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) to make \(2 \mathrm{M}\) equilibrium concentration of \(\mathrm{CO}\) ? (a) 15 (b) 19 (c) 5 (d) 20

The equilibrium constant for a reaction is \(K\), and the reaction quotient is \(Q .\) For a particular reaction mixture, the ratio \(\frac{K}{Q}\) is \(0.33 .\) This means that : (a) the reaction mixture will equilibrate to form more reactant species (b) the reaction mixture will equilibrate to form more product species (c) the equilibrium ratio of reactant to product concentrations will be 3 (d) the equilibrium ratio of reactant to product concentrations will be \(0.33\)

At \(1000 \mathrm{~K}\), a sample of pure \(\mathrm{NO}_{2}\) gas decomposes as: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{p}\) is \(156.25\) atm. Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is \(0.25\) atm at equilibrium. The partial pressure of \(\mathrm{NO}_{2}\) at equilibrium is : (a) \(0.01\) (b) \(0.02\) (c) \(0.04\) (d) None of these

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