# Chemistry- college

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)

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1. 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

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2. For part B:

Heat of neutralization is -56.2

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3. 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.

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4. 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.

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5. I need notes on enthalpy of formation

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