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Write the equilibrium constant expressions for $K_{\mathrm{c}}$ and $K_{P},$ if applicable, for the following reactions:(a) $2 \mathrm{NO}_{2}(g)+7 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$(b) $2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)$(c) $\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$(d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons$$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q)$

(a) $K_{c}=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{NO}_{2}\right]^{2}\left[\mathrm{H}_{2}\right]^{7}} ; K_{p}=\frac{\left(P_{\mathrm{NH}_{3}}\right)^{2}}{\left(P_{\mathrm{NO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2}}\right)^{7}}$(b) $K_{c}=\frac{\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{O}_{2}\right]^{3}} ; K_{p}=\frac{\left(P_{\mathrm{SO}_{2}}\right)^{2}}{\left(P_{\mathrm{O}_{2}}\right)^{3}}$(c) $K_{c}=\frac{[\mathrm{CO}]^{2}}{\left[\mathrm{CO}_{2}\right]} ; K_{p}=\frac{P_{\mathrm{CO}}^{2}}{P_{\mathrm{CO}_{2}}}$(d) $K_{c}=\frac{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right]}$

Chemistry 102

Chapter 14

Chemical Equilibrium

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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.

05:47

Write the equilibrium cons…

07:49

Write equilibrium constant…

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01:25

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Write the equilibrium-cons…

05:11

Write equilibrium-constant…

for each part of this problem, we're going to write out the equilibrium expressions for the equilibrium constants, K, C and K P. We're going to do this whenever it is possible to do so in the way that we know whether it whether or not it is possible to form these expressions, is by examining the speeds of matter for each one of the chemical species involved in the reaction. The equilibrium constants are defined in terms of the equilibrium concentrations for all the species involved in the reaction. And we know that we can only have concentrations for species that appears gases or in a quickest solution. And we can only have KP expressions for reactions that involve gases, because the KP involves the equilibrium, partial pressures of gases, and we can only measure gases in terms of partial pressures. So that is how we know whether it's possible to form Casey and KP. So starting with part A, we see that we have three gases and one liquid assistance. We have at least one gas. We know that we conform expressions for both Casey and KP. We know that we do not include liquid species, and in this case liquid water in our expression for for Casey and KP in part A because we cannot write liquids in terms of concentrations. So starting with the expression for Casey, we know that this these equilibrium concentrations are in terms of malaria the most per leader. So starting with the product side, which goes on the numerator of our expression, the only gas is an H three. So that would be Moller concentration on equilibrium of NH three, and we square it due to a stroke the metric coefficients to and then we go to the reactant side, since the other product is a liquid and we cannot rate in terms of concentrations for the first reactant, we have the equilibrium concentration of no to and we swear it. We want to play by the equilibrium concentration of H two guests and we raise it to the power of seven. And so now we can write out the expression for KP in terms of the gashes species, so that would be the equilibrium. Partial pressure of NH three squared, since that is the only product that appears as a gas for the reactant should be the equilibrium. Partial pressure of No. Two squared times the equilibrium partial pressure of H two to the power of seven. So those are the equilibrium expressions for both Casey and KP For the reaction given import A Now in part B, we see that we have two gases and two solids. Since we have at least one gas, we know that we conform expressions for both Casey and KP. Again, we do not include the solid species in either one of our expressions. So starting with K c in the numerator that corresponds to the products So that would be the equilibrium Moeller concentration of eso to gas squared invited by the gashes reactant which is equilibrium Mueller, concentration of 02 cubed and so KP is defined as equilibrium partial pressure ratios between those two gases. So that would be the equilibrium. Partial pressure of s 02 squared, divided by the equilibrium partial pressure of Oh, too cute. And now for the reaction given in part C, we have two gases and one solid. We conform Casey and KP against since we have at least one gas and we do not include that solid. So for Casey, that would be equal to the equilibrium of Moeller concentration of CEO. Guess Squared, invited by the equilibrium, more concentration of CEO to gas and for KP that would be equal to the equilibrium. Partial pressure of CEO guests squared, divided by the equilibrium, partial pressure of CEO to gas and finally, for part d, all of the species involved in the reaction or qui ists. We know that we can form an expression for K C based on a quiz species, because it is possible to measure the concentrations of Equus species in terms of moles per leader. However, since none of this species air gases, it is not possible to write an expression for KP because we have to measure the concentrations at equilibrium for KP in terms of partial pressures which can only be related to gashes, species and not acquiesce species. So we can Onley form the expression for Casey and that comes out to the equilibrium. Mueller concentrate Concentration of C six each, five C o minus times the equilibrium Mueller concentration of H plus, divided by equilibrium. More concentration of C six, h five c 00 h

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