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: 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
For part B:
Heat of neutralization is -56.2
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 need notes on enthalpy of formation
A. Well, it seems like the H+ ion is the superstar of this reaction, getting assigned the value of zero. Poor Cl- is left in the dust. But don't worry, we can still calculate its enthalpy of formation!
Since the enthalpy change for the reaction is given as -74.9 kJ/mol, we can use this information to figure out the enthalpy of formation for Cl- ions.
In the reaction: HCl(g) + H2O -> H+(aq) + Cl-(aq), we have 1 mole of HCl producing 1 mole of H+ ions and 1 mole of Cl- ions.
So, the enthalpy change for the reaction can be broken down as follows:
-74.9 kJ/mol = 0 kJ/mol (for H+(aq)) + delta H(f) (for Cl-(aq))
Since the enthalpy change for H+(aq) is zero by definition, we can rearrange the equation and solve for the enthalpy of formation of Cl- ions:
delta H(f) (for Cl-(aq)) = -74.9 kJ/mol - 0 kJ/mol = -74.9 kJ/mol
So, the enthalpy of formation for Cl- ions is -74.9 kJ/mol.
B. Ah, the age-old neutralization reaction where acids and bases join forces to create water. It's like a superhero team-up!
Given that the enthalpy of formation for OH- ions is -229.6 kJ/mol, we can now calculate the enthalpy of neutralization.
In this case, the neutralization reaction would look like this:
H+(aq) + OH-(aq) -> H2O(l)
From the reaction, we know that 1 mole of H+(aq) will react with 1 mole of OH-(aq) to produce 1 mole of H2O(l).
The enthalpy change for this reaction can be calculated using the enthalpies of formation of the reactants and products.
delta H = delta H(f) (for H2O(l)) - delta H(f) (for H+(aq)) - delta H(f) (for OH-(aq))
delta H = 0 kJ/mol (for H2O(l)) - 0 kJ/mol (for H+(aq)) - (-229.6 kJ/mol) (for OH-(aq))
delta H = 229.6 kJ/mol
So, the enthalpy of neutralization when 1 mole of a strong monoprotic acid (like HCl) is titrated by 1 mole of a strong base (like KOH) at 25°C is 229.6 kJ/mol.
Hope that puts a smile on your face!
To answer these questions and calculate the enthalpies of formation, we will make use of Hess's Law and the standard enthalpies of formation of water and reactants involved. Hess's Law states that if a reaction can be expressed as the sum of two or more reactions, then the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.
A. To calculate delta H(f) for the Cl- ions, we can express the given reaction as a combination of two reactions:
1. HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq) (given)
2. HCl(g) + H2O(l) → H3O+(aq) + OH-(aq)
The standard enthalpies of formation of water, H3O+, and OH- ions are known. The enthalpy change of reaction 2 can be calculated using the known enthalpies of formation:
delta H(rxn2) = delta H(f) [H3O+(aq)] + delta H(f) [OH-(aq)] - delta H(f) [HCl(g)] - delta H(f) [H2O(l)]
delta H(rxn2) = (0 kJ/mol) + (-229.6 kJ/mol) - delta H(f) [HCl(g)] - (0 kJ/mol)
delta H(rxn2) = -229.6 kJ/mol - delta H(f) [HCl(g)]
We are given the enthalpy change for reaction 1, which is -74.9 kJ/mol. Now we can relate the enthalpies of reaction 1 and 2 using Hess's Law:
delta H(rxn1) = delta H(rxn2) + delta H(rxn2)
-74.9 kJ/mol = (-229.6 kJ/mol - delta H(f) [HCl(g)]) + (-229.6 kJ/mol - delta H(f) [HCl(g)])
Simplifying the equation:
-74.9 kJ/mol = -459.2 kJ/mol - 2 * delta H(f) [HCl(g)]
Now, we can solve for delta H(f) [Cl-(aq)]:
2 * delta H(f) [Cl-(aq)] = -459.2 kJ/mol - (-74.9 kJ/mol)
2 * delta H(f) [Cl-(aq)] = -384.3 kJ/mol
delta H(f) [Cl-(aq)] = -192.2 kJ/mol
Therefore, the delta H(f) for the Cl- ions is -192.2 kJ/mol.
B. To calculate the enthalpy of neutralization, we need to consider the enthalpy change for the neutralization reaction between HCl and KOH. The balanced equation for the reaction is:
HCl(aq) + KOH(aq) → H2O(l) + KCl(aq)
The enthalpy change for this reaction can be obtained by applying Hess's Law:
delta H(overall) = delta H(f) [H2O(l)] + delta H(f) [KCl(aq)] - delta H(f) [HCl(aq)] - delta H(f) [KOH(aq)]
delta H(overall) = 0 kJ/mol + 0 kJ/mol - delta H(f) [HCl(aq)] - 0 kJ/mol
delta H(overall) = - delta H(f) [HCl(aq)]
We know from part A that the delta H(f) [Cl-(aq)] is -192.2 kJ/mol, and since HCl(aq) dissociates completely in water to give Cl- ions, delta H(f) [HCl(aq)] = delta H(f) [Cl-(aq)].
Therefore, the enthalpy of neutralization is -192.2 kJ/mol.
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.