Question

A coffee cup calorimeter was used to measure the heat evolved when 100 mL of a 0.5 M solution of NaOH (40.00 g/mole) was mixed with 100 mL of a 0.5 M HCl solution. The temperature of both solutions at the time of mixing was 21.2C. After the neutralization reaction took place, the final temperature was 24.5C. The density of the solution was 1.01 g/mL and the specific heat of the solution was 4.10 J/gC. The heat capacity of the calorimeter cup was 20 J/C.

a. Calculate enthalpy change ( H) for the neutralization per mole of NaOH.

b. Calculate enthalpy change ( H) for the neutralization per gram of NaOH.

Answer


A textbook (pardon the pun) calorimetry question.  I have answered a
similar question to this one before but I will still try my hand at it, first off let's figure out the total amount of energy generated from this reaction:

q(solution)= ms∆t
m = total mass of the final solution (100 ml + 100) x 1.01 = 202 g
s = specific heat 4.10 J/gC as given in the problem
∆t = 24.5-21.2C = 3.3C
q = (202g)(4.10 J/gC)(3.3C)
q = 2733.06 J = 2.73306 kJ

q(cup) = (20 J/C)(3.3C)
q = 66 J = 0.066 kJ

q(total) = 2733.06 J + 66 J = 2799.06 J

Here it is necessary to note that because the pressure is constant throughout the reaction, the values for the heat change of this reaction (∆q) is equal to the value for the enthalpy change for this reaction (∆H) with the modification that the value should be negative.  So, with the total value of ∆H at hand we need to determine how many mols of sodium hydroxide we have in this solution:

0.5 mol / L x 0.1 L = 0.05 mol

Easy enough?  With the mols of sodium hydroxide known (and knowing that they are equal to the mols of hydrochloric acid) we can just divide the joules by the number of mols to get the Joules / Mol.

-2799.06 J / 0.05 mol = -55981.2 J/mol = -55.9812 kJ/mol

So, the value above is your answer for your part A.  Part B just takes this another step further.  The mols of sodium hydroxide were determined previously above, this value is multiplied by the molecular weight of sodium hydroxide to get the grams of sodium hydroxide in this solution then the joules are divided by the grams of sodium hydroxide to get the joules per gram.  Conversely the value above could be divided by the molecular weight of sodium hydroxide as well to get the same value.

mol wt NaOH = 40.0 g / mol
40.0 g/mol x 0.05 mol = 2 grams
(-2799.06 J) / 2 grams = -1399.53 J/g

The value above being the answer to part B of your question.  The odd-ball part of this question came from the fact that the heat capacity of the calorimeter was given.  Usually this is neglected for 'bench scale' experiments such as the Styrofoam cup questions.  So the term has to be accounted for somewhere.  This comes into play if you write out all the components that are additive in making the q term.

q system = q solution + q calorimeter + q reaction

What we want in this case is the q of the reaction.  The q of the system is usually considered 0 because unless the problem says otherwise we assume that no heat is lost to the surroundings.  So, solve for reaction.

q reaction = - q solution - q calorimeter

Now you can see why the values suddenly become negative. Finally though, the neutralization of hydrochloric acid with sodium hydroxide solutions is well covered, hence there are plenty of literature values.  One such value I found(1) pegs the reaction at:

∆H = -56.2 kJ

That puts our value of -55.98 kJ well in agreement with this literature value.  Not bad for a calculation from a cup ;)

 

Reference
(1) Chang, Raymond. Chemistry. 6th ed. USA, McGraw-Hill. 1998

 

 

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