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The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to H io...Asked by Willy
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 H(f) [H+(aq)]=0
A.for this reaction: calculate delta H(f) for the Cl- ions.
HCl(g) ==H2O==>H+(aq)+Cl-(aq)
delta H= -74.9 kJ/mol
(answer has to be in kJ/mol)
B.given that delta H(f) for OH- ions is -229.6 kJ/mol, calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 25^C.
(answer has to be in kJ/mol)
A.for this reaction: calculate delta H(f) for the Cl- ions.
HCl(g) ==H2O==>H+(aq)+Cl-(aq)
delta H= -74.9 kJ/mol
(answer has to be in kJ/mol)
B.given that delta H(f) for OH- ions is -229.6 kJ/mol, calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 25^C.
(answer has to be in kJ/mol)
Answers
Answered by
Jaya
A: you can either look at the properties table or use -74.9kJ/mol= x +
92.3 kJ/mol and solve for x which is -167.16...... juse the equation deltaH= Sum deltaH of products - sum deltaH of reactants
idk how to do B yet
92.3 kJ/mol and solve for x which is -167.16...... juse the equation deltaH= Sum deltaH of products - sum deltaH of reactants
idk how to do B yet
Answered by
Jesse
For part B:
Heat of neutralization is -56.2
Heat of neutralization is -56.2
Answered by
Xerxes(NEUST)
First we need to get the balance equation for the given reaction
HCl + KOH -> KCl + H2O
Next is to get the heat change (q) of water from the reaction
So we have;
1mole × 36.45 g/mole HCl= 36.45 g
1mole × 56.1 g/mole KOH= 56.1 g
q=msdeltaT
q=(56.1g + 36.45g)× (4.184 J/g•°C)
× (25°C)
q= -9,681 J or -9.7 kJ/1 mole
q @constant pressure = deltaH(chnge)
From standard enthalpy of formation (appendices of your book)
HCl = -92.3 kJ/mol
K+ = -251.2 kJ/mol
-OH = -229.6 kJ/mol
KCl = -435.9 kJ/mol
H2O = -9.7 kJ/mol
H(soln') = H (prod) - H (reactant)
= (-435.9 + -9.7) - (-92.3 + -251.2 + -229.6)
= 127.5 kJ/mol
deltaH°= deltaH (rxn) = -deltaH (soln)
-74.9 = deltaH (rxn) = -127.5
deltaH (rxn) = -127.5 + 74.9
= -52.6 kJ/mol
deltaH (rxn) OR deltaH of (neutralization) is -52.6 kJ/mol
Maybe there is another way to solve it. But I hope this one will help ! Thank you.
HCl + KOH -> KCl + H2O
Next is to get the heat change (q) of water from the reaction
So we have;
1mole × 36.45 g/mole HCl= 36.45 g
1mole × 56.1 g/mole KOH= 56.1 g
q=msdeltaT
q=(56.1g + 36.45g)× (4.184 J/g•°C)
× (25°C)
q= -9,681 J or -9.7 kJ/1 mole
q @constant pressure = deltaH(chnge)
From standard enthalpy of formation (appendices of your book)
HCl = -92.3 kJ/mol
K+ = -251.2 kJ/mol
-OH = -229.6 kJ/mol
KCl = -435.9 kJ/mol
H2O = -9.7 kJ/mol
H(soln') = H (prod) - H (reactant)
= (-435.9 + -9.7) - (-92.3 + -251.2 + -229.6)
= 127.5 kJ/mol
deltaH°= deltaH (rxn) = -deltaH (soln)
-74.9 = deltaH (rxn) = -127.5
deltaH (rxn) = -127.5 + 74.9
= -52.6 kJ/mol
deltaH (rxn) OR deltaH of (neutralization) is -52.6 kJ/mol
Maybe there is another way to solve it. But I hope this one will help ! Thank you.
Answered by
Degla Malaki
I agree to this answer as i found it a little bit confusing when the question was asked. you have to answer this question using all the basic knowledge of thermochemical equation especially enthalphy of formation and reaction.
i agree to the answer posted by Xerxes.
i agree to the answer posted by Xerxes.
Answered by
Brian
I need notes on enthalpy of formation
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