Not sure if these are right
1) What is the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol if 5.00 moles of HCl(aq) (acid) neutralized 5.00 moles of NaOH(aq) (base) and released 13093 J of heat?
13093/5= 2618.6 then convert to kj
2) Calculate the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information:
The temperature change equals 5.06°C,
50.0 mL of 1.00 M concentration of Acid
50.0 mL of 1.00 M concentration of Base
Heat capacity of the calorimeter is 6.50 J/°C.
The specific heat of water is 4.180 J/g°C
5*5*5.06*6.50= 822.25/4.180= 196.7
3) How much heat is gained (in Joules) by the water (qwater) when a chemical reaction takes place in 100.0 mL aqueous solution and has a temperature increase of 13.26 °C? (Remember cwater = 4.180 J/°Cg)
100.0*13.26= 1326/4.180= 317.22
Technically all three are wrong; however, #1 is ok but you didn't finish the problem. You answer of 2618.6 J is correct but that needs to be changed to kJ. That makes it 2.6186 kJ/mol and that needs to be rounded to 3 significant figures. I would round that to 2.62 kJ/mol.
#2. I think I remember doing this for you in the last day or so. I think you either didn't substitute correctly OR you made an algebra error.
#3. mass x specific heat H2O dT. You omitted the specific heat H2O from your work.
If you will show your work on #2 someone will point out the error(s).
I did show my work for #2
5*5*5.06*6.50= 822.25/4.180= 196.7
Yes, and I asked where did the 5*5 come from? I showed you how to do it in the response above that.
1) To find the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol, you need to divide the amount of heat released by the number of moles of the limiting reactant.
In this case, HCl and NaOH react in a 1:1 mole ratio, so both have the same number of moles (5.00 mol).
The heat released is given as 13093 J. To convert this to kJ, divide it by 1000:
13093 J / 1000 = 13.093 kJ
Now, divide the heat released by the number of moles:
13.093 kJ / 5.00 mol = 2.6186 kJ/mol
So, the Molar Enthalpy of Neutralization (ΔHn) is approximately 2.6186 kJ/mol.
2) The Molar Enthalpy of Neutralization (ΔHn) in kJ/mol can be calculated using the formula:
ΔHn = (qreaction - qcalorimeter) / n
where qreaction is the heat released by the reaction, qcalorimeter is the heat absorbed by the calorimeter, and n is the number of moles.
First, calculate qreaction by multiplying the volume of the solution by the specific heat of water and the temperature change:
qreaction = (50.0 mL + 50.0 mL) * (1.00 mol/L) * (4.18 J/g°C) * (5.06°C)
= 10.12 g * 4.18 J/g°C * 5.06°C
= 213.7968 J
Next, calculate qcalorimeter by multiplying the heat capacity of the calorimeter by the temperature change:
qcalorimeter = 6.50 J/°C * 5.06°C
= 32.89 J
Now, calculate the moles of the limiting reactant (assuming equal volumes of acid and base, and 1:1 mole ratio of reaction):
n = 1.00 mol/L * (0.050 L + 0.050 L)
= 0.100 mol
Finally, calculate ΔHn:
ΔHn = (213.7968 J - 32.89 J) / 0.100 mol
= 1809.068 J / 0.100 mol
= 18090.68 J/mol
Converting to kJ/mol:
18090.68 J/mol / 1000 = 18.09068 kJ/mol
Therefore, the Molar Enthalpy of Neutralization (ΔHn) is approximately 18.09068 kJ/mol.
3) To find the heat gained by the water (qwater), you can use the formula:
qwater = m * c * ΔT
where m is the mass of water, c is the specific heat of water, and ΔT is the temperature change.
Given that the volume of the aqueous solution is 100.0 mL, or 100.0 g (assuming density of water is 1 g/mL), and the specific heat of water is 4.180 J/g°C:
qwater = 100.0 g * 4.180 J/g°C * 13.26°C
= 5291.68 J
Therefore, the heat gained by the water (qwater) is 5291.68 J.