at 25c the standard enthalpy of formation of HF(aq) is -320.1 kj/mol. of OH(aq) it is...

at 25c the standard enthalpy of formation of HF(aq) is -320.1 kj/mol. of OH(aq) it is -229.6Kj/mol. of F (aq) it is -329.2 Kj/mol.of H2O(l) it is -285.9Kj/mol.

a. calculate the standard enthalpy of neutralization of HF.
HF (aq) + OH- (aq) -->F-(aq) +H2O (l)

b.using the value of -56.2kj as the standard enthalpy change for the reaction
H+(aq)+ OH(aq)->H2O(l)

c. calculate the standard enthalpy of change for the reaction
HF->H+F

Solutions

Expert Solution

a. calculate the standard enthalpy of neutralization of HF.
HF (aq) + OH- (aq) -->F-(aq) +H2O (l)

b.using the value of -56.2kj as the standard enthalpy change for the reaction
H+(aq)+ OH(aq)->H2O(l)

c. calculate the standard enthalpy of change for the reaction
HF->H+F

first calculate the the standard enthalpy of neutralization of HF.
HF (aq) + OH- (aq) -->F-(aq) +H2O (l)

standard enthalpy change for the above reaction

= dH product - dH recants

= [-329.2 Kj/mol. + (-285.9Kj/mol)] – [(-320.1 kj/mol) + (-229.6Kj/mol)]

= [-329.2 Kj/mol. -285.9Kj/mol + 320.1 kj/mol + 229.6Kj/mol]

= -65.4 KJ/ mole

F-(aq) +H2O (l) --- > HF (aq) + OH- (aq) dH = + 65.4 KJ/ mole ---1

H+(aq)+ OH(aq)->H2O(l) dH = -56.2kj ----2

Now add this two equation we will get the following equation:

HF->H+F

F-(aq) +H2O (l) --- > HF (aq) + OH- (aq) dH = + 65.4 KJ/ mole ---1

H+(aq)+ OH(aq)->H2O(l) dH = -56.2kj ----2

HF->H+F dH = + 9.2 KJ/ mole

queen_honey_blossom answered 6 months ago

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