The heat of neutralization of solid metal hydroxide with an aqueous solution of an acidmight be determined directly in a one-step reactionМОН(s) + H* M'+ H-0(1)or indirectly using a two-step reactionМОН(s) —> М* + онОН + НH2O(1)Based on the net ionic equation of the reaction, the theoretical enthalpy of reaction,AHîheo!liheo kJ/mol], can be found from the enthalpies of formation asΔΗλ ΔΗ; ΗΟ() + ΔΗ Μ' (αq)] -ΔΜΟH (s)]|Based on the measured change in temperature, AT[K], and the heat capacity of the cup,Ccup[J/K], the experimental enthalpy of reaction, AHap[kJ/mol], canfound asAHerp(S„ Viot + Ceup) AT/nL.R.,where Vtot [mL] is the total volume of the solution, nL.R.[mmol] is the amount of thelimiting reagent, and specific heat of water is s, = 4.18 JML¯!K¯lThe percent error can be found according to the formulaAH-AHtheo x 100%%E =expAHiheoUsing the values for standard enthalpies of formation from the table 8-1 find the enthalpiesfor the reactions in the following questions.Question 1Consider two-step neutralization reaction of solid metal hydroxide MOH with anacid. The formation enthalpy of MOH is -438.15 kJ/mol and for Mt it is-249.07 kJ/mol. What is the enthalpy of dissociation?kJ/molWhat is the enthalpy of neutralization of an aqueous solution of solid metal hydroxideMOH with an acid?kJ/mol

The enthalpy of formation of MOH is -438.15 kJ/mol. The formation enthalpy of M+ is -249.07 kJ/mol.The enthalpy for formation and the enthalpy of dissociation is equal and opposite to each other. The dissociation enthalpy of MOH is calculated as, ∆dHMOH=-∆fHMOH Where, ∆fHMOH is the formation enthalpy of MOH. Substitute the formation enthalpy of MOH in the above formula. ∆dHMOH=--438.15 kJ/mol=438.15 kJ/molThe enthalpy of neutralization is calculated by the formula, ∆H=∆fHH2O+∆fHM+-∆fHMOH Where, ∆fHM+ is the formation enthalpy of an acid. ∆fHH2O is the formation enthalpy of water (-285.8 kJ/mol). Substitute the formation enthalpy of M+ and MOH in the above formula. ∆H=-285.8 kJ/mol+-249.07 kJ/mol--438.15 kJ/mol=-285.8 kJ/mol+189.08 kJ/mol=-96.72 kJ/molAnswer: The dissociation enthalpy of MOH and the enthalpy of neutralization is 438.15 kJ/mol and -96.72 kJ/mol, respectively.

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Consider two-step neutralization reaction of solid metal hydroxide MOH with an acid. The formation enthalpy of MOH is -438.15 kJ/mol and for M+ it is -249.07 kJ /mol. What is the enthalpy of dissociation? kJ/mol What is the enthalpy of neutralization of an aqueous solution of solid metal hydroxide MOH with an acid? kJ/mol

Consider two-step neutralization reaction of solid metal hydroxide MOH with an acid. The formation enthalpy of MOH is -438.15 kJ/mol and for M+ it is -249.07 kJ /mol. What is the enthalpy of dissociation? kJ/mol What is the enthalpy of neutralization of an aqueous solution of solid metal hydroxide MOH with an acid? kJ/mol

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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