Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write the expressions for the equilibrium constants \(K_{P}\) of these thermal decompositions: (a) \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{CaSO}_{4}(s) \rightleftharpoons 2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\)

Short Answer

Expert verified
The equilibrium constant expression for decomposition (a) is \(K_{P} = [CO_{2}] [H_{2} O]\) and for decomposition (b) is \(K_{P} = [SO_{2}]^{2} [O_{2}]\)

Step by step solution

01

Expression for Kp of Decomposition (a)

Starting with the decomposition: \(2 \mathrm{NaHCO}_{3}(s) \leftrightarrows \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). In the expression for \(K_{P}\), we only consider the concentrations of gaseous substances. In this decomposition, the gaseous products are \(CO_2\) and \(H_2O\). Therefore, the expression for the equilibrium constant, \(K_p\), for this decomposition is given by: \(K_{P}= [CO_{2}] [H_{2} O]\)
02

Expression for Kp of Decomposition (b)

Now consider the decomposition: \(2 \mathrm{CaSO}_{4}(s) \leftrightarrows 2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\). Similar to decomposition (a), we will only consider the gaseous products in our expression of \(K_p\). These gaseous products are \(SO_2\) and \(O_2\). Therefore, for this decomposition, the expression for the equilibrium constant, \(K_{P}\), is given by: \(K_{P}= [SO_{2}]^{2} [O_{2}]\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Equilibrium Expression
The equilibrium expression is a mathematical formula that quantifies the state of a chemical equilibrium. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant over time. The expression is written in terms of the equilibrium constant, denoted as K. In the context of gases, we refer to it as the constant for partial pressure, or \(K_P\).

For a general reaction, where \(aA + bB \rightleftharpoons cC + dD\), the equilibrium expression for \(K_P\) is written as \(K_{P} = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\), where the brackets denote the partial pressure of the gas. This equation signifies that equilibrium involves a delicate balance between the concentrations of reactants and products, with the exponent values representing the stoichiometry of the balanced equation. Solid and liquid phases are omitted in this expression as their concentrations do not change under equilibrium conditions.
Thermal Decomposition
Thermal decomposition refers to a process where a compound breaks down into two or more substances when heated. This type of reaction is endothermic; it absorbs energy from the surroundings, typically in the form of heat. In a chemical context, these reactions are interesting for the study of equilibrium, as they can reach a point where the reactants and products exist in a stable balance.

In the decomposition exercises provided, we observe compounds turning into simpler substances along with the evolution of gas. Such reactions are common in various industrial processes and can also be studied to understand the reactivity and stability of certain compounds under different temperature conditions. It's important to note that these reactions are reversible, leading to the establishment of a dynamic equilibrium at certain temperatures.
Partial Pressure
The concept of partial pressure is critical when discussing gases involved in a chemical reaction. It refers to the pressure one component of a gaseous mixture would exert if it were alone in the container. Dalton's Law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases.

When we calculate the equilibrium expression for reactions involving gases, it is the partial pressures that we incorporate into the equation. The equilibrium constant \(K_P\) takes into account these pressures. In the provided exercises, only the gases \(CO_2, H_2O, SO_2,\) and \(O_2\) are considered for the equilibrium expression, as solids and liquids do not contribute to the partial pressure in a system.
Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry when a reversible reaction proceeds in both the forward and reverse directions at equal rates, and the concentration of the reacting species becomes constant. This does not mean the reaction has stopped, but rather that the consumption and production of substances are happening at the same rate.

The equilibrium state is dynamic and can shift in response to changes in conditions such as temperature, pressure, and concentration according to Le Chatelier's Principle. Understanding chemical equilibrium helps chemists control reactions and optimize the yield of products in industrial settings.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A quantity of 0.20 mole of carbon dioxide was heated at a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Under this condition, the average molar mass of the gases was found to be \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is the \(K_{P}\) for the equilibrium if the total pressure was 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C} . \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are mixed initially at 0.350 atm and 0.762 atm, respectively, at \(350^{\circ} \mathrm{C}\). When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures, 1.112 atm?

Consider the following equilibrium process at \(700^{\circ} \mathrm{C}\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) $$ Analysis shows that there are 2.50 moles of \(\mathrm{H}_{2}, 1.35 \times\) \(10^{-5}\) mole of \(\mathrm{S}_{2}\), and 8.70 moles of \(\mathrm{H}_{2} \mathrm{~S}\) present in a 12.0-L flask at equilibrium. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

The equilibrium constant \(K_{\mathrm{c}}\) for the decomposition of phosgene, \(\mathrm{COCl}_{2}\), is \(4.63 \times 10^{-3}\) at \(527^{\circ} \mathrm{C}\) : $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.760 atm.

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00 -L flask. At \(648 \mathrm{~K}\), there is 0.0345 mole of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free