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In a calorimetry experiment to measure the heat of neutralization (delta H...

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haiiii | Student, College Freshman | (Level 2) eNoter

Posted April 17, 2012 at 11:37 PM via web

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In a calorimetry experiment to measure the heat of neutralization (delta H neutralization- (joules/mole)) 

25.0 mL of 2.0 M HCL reacts with 25.0 mL of 2.0 M NaOH. V total = 50 mL. The temperature of the solution goes from 18 degrees celsius to 30 degrees celsius. Specific heat = 4.18 Joules/ Degree gram; calculate delta H (joules/mole)

 

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troubleshooter | (Level 1) eNoter

Posted April 18, 2012 at 6:07 AM (Answer #1)

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I am not a teacher, but have taken college chemistry.

First, not required for the solution, but helpful to understand, is the reaction:

2 mole hydrochloric acid (HCl) reacts with 2 mole sodium hydroxide (NaOH) to yield 2 moles NaCl (salt) and 2 moles HOH (H2O) water.

 

Specific heat is specified for you in Joules/degree gram.

The equation is for grams, but the quantities are in mL.

The problem is inferring the HCl and NaOH are in aqueous solution.

For the purposes of this exercise I assume we are using a density of 1.00 for aqueous solutions.

So they will be the same; 1mL = 1 gram for water.

If the reagents were more concentrated, their density (weight per ml) would have to be factored in and calculated for the specific heat.

 

This yields 2 moles plus 2 moles yields 2 moles of product and creates X number of degrees of heat in the process in a sample of such and such weight.

 

Keep in mind you have to multiply the heat by the amount of grams!

(example: If experiment A heated a LARGE quantity of solution the same amount as another experiment B that heated a SMALL quantity of solution the same amount; the first experiment is producing MORE energy.  The amount of energy measured depends not only on the degrees it raised the solution, but also on what quantity it raised by that amount of degrees.)

Temperature change times mass that was changed represents the energy measured.

 

I will let you do your own math so you understand it better.

Figure take the degree change times the weight of the solution.

This is your degree gram part of the formula.

 

Now, from the formula;

Each degree gram represents 4.18 Joules.

So multiply your degree grams by 4.18 (Joules per each degree gram) to get your (total) Joules.

This is your TOTAL joules from the experiment, not the total per mole.

 

Now you have to figure out how many moles produced this total heat to figure how much heat energy came from each mole.

Also, I was not an A student.

I don't know if 2.0 mole A plus 2.0 mole B producing 2.0 mole C product, is counted as 2.0 mole or 4.0 mole.

I am pretty sure it is counted as 2.0 moles (how much heat energy per MOLE of reactant, not both added together)

 

Comments:

Note that the same amount of energy would be represented by an experiment that raised 100mL by ten degrees, or one that raised 1000mL by one degree.  The energy measured is how much weight is raised by how many degrees; and that rate of heating also differs with the specific heat properties of that material (in this case water solution).

Example: one experiment A that raises 100mL 10 degrees of one material (water, for instance) would not represent the same amount of energy as another experiment B that raises the same 100mL 10 degrees if it has a different specific heat than water (paraffin wax, for instance).

Wax has a much lower specific heat than water (for example), and the same amount of heating will raise the temperature of wax much higher than the equivalent amount of water; or alternately, less heat (Joules) is required to raise wax to the same temperature.

So wax heats much faster than water because of it's smaller specific heat.

 

This specific heat number is how you calculate what amount of heat represent how much energy (Joules) for that particular material.

That is why you were provided it so you can convert your temperature reading into a Joules quantity.

Then you were asked for how many Joules per mole (not total Joules), so you have to divide for your particular moles used.

 

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