The standard enthalpies of formation of ions in aqueous solutions are obtained b

Question

The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to H ions; that is. delta Hdegree f [H+ (aq)] = 0. For the following reaction HC1(g) rightarrow H+(aq) + Cl- (aq) delta H degree = -74.9 kJ/mol calculate delta H f for the Cl- ions. kJ/mol Given that A H i for OH ions is -229.6 kJ/mol. calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HC1) is titrated by 1 mole of a strong base (such as KOH) at 25 degree C. kJ/mol

Explanation / Answer

a)

The given reaction is

HCl (g) ---> H+ + Cl-

we know that

dHrxn = dHfo products - dHfo reactants

so

dHrxn = dHfo H+ + dHfo Cl- - dHfo HCl

Given

dHfo H+ = 0

dHrxn = -74.9

using standard values

dHfo HCl = -92.3

so

we get

-74.9 = 0 + dHfo Cl- + 92.3

dHfo Cl- = -167.2

dHfo for the Cl- ions is -167.2 kJ/mol

2)

now the nuetralization reaction is given by

H+ (aq) + OH- (aq)   --> H20 (l)

so

dHrxn = dHfo H20 - dHfo H+ - dHfo OH-

Given

dHfo OH- = -229.6

using standard values

dHfo H20 = -285.8

so

dHrxn = -285.8 -0 - ( -229.6)

dHrxn = -285.8 + 229.6

dHrxn = -56.2

so

the enthalpy of neutralization is -56.2 kJ/mol

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