Help with Heat of Neutralization Homework

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In summary, the conversation is about a heat of neutralization lab where 25mL of HCl and 25mL of NaOH were reacted in a calorimeter, resulting in a temperature increase of 7°C. The question asked for the q_water value, which was calculated to be 1465.1J. The next question asked for the delta H_neutralization, which was found to be -1465.1J. The net ionic equation for the reaction was written as H+ + OH- --> H2O. Finally, the question asked for the delta H_neut per mole of water formed, which required knowing the concentrations of the solutions. The correct answer was found to be -58 kJ
  • #1
sp3sp2sp
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Homework Statement


Im confused about this and need help
We did heat of nuetralization lab where we reacted 25mL of HCL and 25mL of NaOH using calorimeter.
The temp difference was an increase of exactly 7degC
One question asks me to show q_water (J) and to assume 50g water for the calculation
so i plugged in 50g * 4.186J/gdegC * 7degC = 1465.1J

Next it asks me what delta H_nuetralization (J) is and I wrote:
delta H_neut = q_rxn = -q_water = -1465.1J

Next it asks me for net ionic. I wrote
H^(aq) + + OH^-(aq) --> H2O(l)

Then it asks me for delta H_neut (kJ/mol), per mole of water formed and this is where I am confused, because internet search shows value should be about -58kJ/mol

50g water = 2.78mol
so if 50g water produces -1465.1J, then 1 mol water would produce 1465.1J / 2.775mol =527.0J
= 0.527kJmol - this is way off so I must be doing something wrong somewhere.
thanks for any help

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  • #2
You didn't produce 50 g of water.

Were you given concentrations of NaOH and HCl?
 
  • #3
Yes the whole experiement was nuetralization reaction. I had 25.0mL of NaOH and 25.0mL of HCl. I understand that the acid and base nuetralize each other and this is exothermic so heat given off but I am still confused about the whole concept for the enthalpy. thanks for any more help
 
  • #4
ΔH = -58 kJ/mol means that 58 kJ of heat are released when 1 mole of H+ reacts with 1 mole of OH- to give 1 mole of water. This is what the question means when it asks for ΔH per mole of water formed - not the water initially present in the solutions. You need to work out the number of moles of water formed in the neutralization reaction, and divide 1465.1 J by that number. For that, you need the concentrations of the solutions, which Borek asked you for but you didn't give.
 

1. What is the heat of neutralization?

The heat of neutralization is the amount of heat released or absorbed when one mole of an acid reacts with one mole of a base to form a neutral solution.

2. Why is the heat of neutralization important?

The heat of neutralization is important because it tells us how much energy is released or absorbed during an acid-base reaction. This information helps us understand the thermodynamics of the reaction and can be useful in various industries, such as in the production of drugs and fertilizers.

3. How is the heat of neutralization calculated?

The heat of neutralization can be calculated by multiplying the moles of the acid or base by their respective heat capacity and temperature change, and then adding these values together. The heat capacity can be found in reference tables or can be experimentally determined.

4. What factors can affect the heat of neutralization?

The heat of neutralization can be affected by various factors such as the strength and concentration of the acid and base, the presence of any impurities or catalysts, and the temperature of the reactants and surroundings.

5. How can the heat of neutralization be used in real-life applications?

The heat of neutralization has practical applications in industries such as pharmaceuticals, agriculture, and energy production. It can be used to optimize reaction conditions, determine the purity of substances, and calculate the amount of energy released or absorbed in a reaction.

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