Hi Dr. Bob, I first want to tank you for all your help. Secondly, I don't understand why this was marked wrong.

When 27.20mL of .5 M H2SO4 is added to 27.20 of 1M KOH in a coffee cup calorimeter at 23.5 degrees C, the temp rises to 30.17 degrees C. Calculate delta H of this reaction.
(Assume that the total volume is the sum of the individual volumes and that density and specific heat capacity of the solution are the same as pure water d of water =1.00g/mL: c of water =4.184J/g*c))

my work: correct or no??

• 2KOH + H2SO4→ 2H2O + K2SO4
n(KOH) = 0.0272 * 1 = 0.0272mol
n(H2SO4) =0.0272 * 0.5 = 0.0136 mol
total volume = 27.2 + 27.2 = 54.4mL
mass of solution = 54.4mL * 1g/mL =54.4g
Q = mcΔT
= 54.4g * 4.18 J/g°C * (30.17 - 23.5)°C
= 1516.74J

The heat is lost from system to surrounding, so ΔH is-ve
For every 0.0136 mol of H2SO4 react, it releases 1516.74J of heat
For 1 mole of H2SO4,
ΔH= 1516.74/0.0136
= -111.525 kJ / mol
= -112 kJ/ mol.......thank you, sir•chemistry/check my work I have to submit today - DrBob222, Saturday, October 17, 2015 at 1:06am
The heat is lost from system to surrounding, so ΔH is-ve
For every 0.0136 mol of H2SO4 react, it releases 1516.74J of heat
For 1 mole of H2SO4,
ΔH= 1516.74/0.0136
= -111.525 kJ / mol
= -112 kJ/ mol.......thank you, sIr

This value is for 2 mols H2O so you need to to do -112 kJ/mol x 1/2 = about 56 kJ/mol.

•Is that a -56 ?•chemistry-Dr Bob - DrBob222, Sunday, October 18, 2015 at 11:43am
Yes, of course. I dropped the - sign on the last step.

it said hint:

What is the balanced equation for this acid base neutralization reaction? First find q by multiplying the total mass of the solution by the change in temperature and the specific heat capacity of water. Once you have found q , divide q ( in kj) by the moles of H2SO4 to find the heat of reaction in units of kj/mol . In the question , the volume of the solution are given in units of ml. How would you find the solution masses in grams from the density of water? How would you find the moles of H2SO4 from the volume and concentration of H2SO4?

thank you again, Dr. Bob

I think I gave you a bum steer. I gave you the kJ/mol H2O and the question is for kJ/rxn. Your answer of -112 kJ was right from the beginning.

thank you!!!!!

To calculate the delta H of the reaction, we can follow the steps outlined in the problem.

Step 1: Write the balanced equation
The balanced equation for the acid-base neutralization reaction is: 2KOH + H2SO4 → 2H2O + K2SO4

Step 2: Calculate the moles of H2SO4
Given that the volume of H2SO4 is 27.20 mL and the concentration is 0.5 M, we can calculate the moles of H2SO4 using the formula: moles = volume (in L) x concentration.
Converting the volume to Liters: 27.20 mL = 0.0272 L
Moles of H2SO4 = 0.0272 L x 0.5 M = 0.0136 mol

Step 3: Calculate the total volume
Since the total volume is the sum of the individual volumes of H2SO4 and KOH, we can calculate it as follows: total volume = 27.20 mL + 27.20 mL = 54.40 mL

Step 4: Calculate the mass of the solution
Given that the density of water is 1.00 g/mL, the mass of the solution can be calculated using the formula: mass = volume x density.
Mass of the solution = 54.40 mL x 1.00 g/mL = 54.40 g

Step 5: Calculate the heat transferred (q)
Using the formula: q = mass x specific heat x change in temperature, we can calculate the heat transferred.
Given that the specific heat capacity of water is 4.184 J/g°C, and the change in temperature is (30.17°C - 23.5°C) = 6.67°C, the calculation becomes:
q = 54.40 g x 4.184 J/g°C x 6.67°C = 1518 J

Step 6: Calculate delta H
Since heat (q) is lost from the system to the surroundings, delta H is negative.
To calculate delta H in kJ/mol, we divide the heat (q) by the moles of H2SO4.
Delta H = 1518 J / 0.0136 mol = -111.47 kJ/mol (rounded to two decimal places)

However, in your calculation, you divided the heat (q) by the moles of H2SO4 to obtain a positive value. In this case, since the reaction is exothermic (heat is lost), the value should be negative. The correct calculation should be:
Delta H = -111.47 kJ/mol

In the subsequent message, Dr. Bob corrected his mistake and stated that the correct value is -56 kJ/mol. This correction was made by multiplying the calculated value by 1/2, as the reaction only produces 1 mole of H2O for every 2 moles of H2SO4. Therefore, the corrected delta H value is -56 kJ/mol.