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Write the expressions for the equilibrium constants $K_{P}$ of the following thermal decomposition reactions:(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)$

(a) $K_{p}=P_{\mathrm{CO}_{2}} P_{\mathrm{H}_{2} \mathrm{O}}$(b) $K_{p}=P_{\mathrm{SO}_{2}}^{2} P_{\mathrm{O}_{2}}$

Chemistry 102

Chapter 14

Chemical Equilibrium

Rice University

University of Kentucky

University of Toronto

Lectures

10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

00:54

In chemistry, chemical equilibrium (also known as dynamic equilibrium) is a state of chemical stability in which the concentrations of the chemical substances do not change in the course of time due to their reaction with each other in a closed system. Chemical equilibrium is an example of dynamic equilibrium, a thermodynamic concept.

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in this problem, we are given to different decomposition hetero genius reactions. And we want to develop expressions for the equilibrium constant K p starting with the reaction in part A. We know that to write out KP, we only examine the species that are gasses. And so these two solids are not going to be present in our expression for KP. Because for KP, we express the gas concentrations in terms of partial pressures. And we cannot do that for solids so similar to Casey. We take the concentrations of the gas is appearing on the product side in the numerator and then we divide by the concentrations of the gas is on the reactant side and the denominator. We only have one solid species on the reactant side, so the denominator will be one. We're just left with the numerator. We see that the to gashes species appearing in this reaction, our co two and H 20 in the each heaviest documentary coefficient of one. So therefore, the expression for KP is equal to the partial pressure of co two gas times to the partial pressure of H 20 gas. And now for the reaction given in part B again see that we have to solid species and therefore no gas, A species on the reactant side. So the KP expression will have a denominator of one again. And on the products we have two guests species, and so we can write out their concentrations in terms of their partial pressures. So KP for this reaction is equal to the equilibrium. Partial pressure of eso to has a striking metric coefficient of two and the chemical reaction. So we square it, you know, we multiply it by the equilibrium. Partial pressure of 02 gas.

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