4 - Take the original K to the 5/2 power to find K of given...

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17.14 Check that correct coefficients from the balanced equation are included as exponents in the mass action expression. a) 2 NO2Cl(g) ' 2 NO2(g) + Cl2(g) a) The given reaction 1/2 S2(g) + H2(g) ' H2S(g) is the reverse reaction of the original reaction multiplied by a factor of 1/2. The equilibrium constant for the reverse reaction is the inverse of the original constant. When a reaction is multiplied by a factor, K, the new equation is equal to the K of the original equilibrium raised to a power equal to the factor. For the reaction given in part a), take (1/K) b) The given reaction 5 H2S(g) ' 5 H2(g) + 5/2 S2(g) is the original reaction multiplied by 5/2.
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Unformatted text preview: Take the original K to the 5/2 power to find K of given reaction. 17.18 The concentration of solids and pure liquids do not change, so their concentration terms are not written in the reaction quotient expression. a) 2 Na2O2(s) + 2 CO2(g) ' 2 Na2CO3(s) + O2(g) = [H2O(g)] Only the gaseous water is used. The “(g)” is for emphasis. c) NH4Cl(s) ' NH3(g) + HCl(g) Qc = [NH3][HCl] 17.19 a) H2O(l) + SO3(g) ' H2SO4(aq) b) 2 KNO3(s) ' 2 KNO2(s) + O2(g) Qc = [O2] c) S8(s) + 24 F2(g) ' 8 SF6(g) Qc...
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