3. You are given the following standard enthalpies of formation at 25°C. HF (aq)

Question

3. You are given the following standard enthalpies of formation at 25°C. HF (aq) OH (aq) F (aq) -320.1 kJ/mol -229.6 kJ/mol -329.1 kJ/mol -285.8 kJ/mol a. Calculate the standard enthalpy of neutralization of HF (aq) HF (aq) + OH' (aq) F (aq) + H2O (1) b. Using the value of -56.2 kJ/mol as the standard enthalpy change for the reaction H' (aq) + OH. (aq) H2O (1) calculate the standard enthalpy change for the reaction HF (aq) H+ (aq) + F. (aq) - 4. You are given the following information N2 (g) + 20 2 NO2(g) 2 (g) -4.8 kJ N:04 (g) N204(g) - determine the enthalpy change (AH) for the reaction: N2 (g) + 2 O2 (g) 2 NO2(g) -

Explanation / Answer

3 a)

standard enthalpy of neutralization of HF = standard enthalpies of formation of Products -standard enthalpies of formation of Reactants.

HF + OH- >>>>> F- + H2O
= (-329.1 + -285.5) - (-320.1 + -229.6) KJ/mol
= (-614.2) - (-549.7) KJ/mol
= -614.2 + 549.7 KJ/mol
= 64.5 KJ/mol

b) standard enthalpy change for the reaction (H+ + OH- yields H2O) = -56.2 KJ/mol

standard enthalpies of Products - standard enthalpies of Reactants = -56.2 KJ/mol
standard enthalpies of formation of Products = standard enthalpies of H2O = -285.8 KJ/mol
standard enthalpies of formation of Reactants = standard enthalpies (H+ + OH-)
standard enthalpies of formation of OH-....-229.6 kj/mol
Therefore, -56.2 KJ/mol = -285.8 KJ/mol - (H+ + -229.6 KJ/mol)
-56.2 KJ/mol = -56.2 KJ/mol - H+
H+ = -56.2 KJ/mol + 56.2 KJ/mol = 0 KJ/mol
standard enthalpies of formation of H+ = 0 KJ/mol

standard enthalpy change for the reaction (HF yields H+ + F-)
= (H+ + F-) - HF
= (0 + -329.1) - (-320.1) KJ/mol
= -329.1 + 320.1 KJ/mol
= -9.1 KJ/mol which is the standard enthalpy change for the reaction (HF yields H+ + F-)

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