Enthalpy of Neutralization Reaction: A 25.0 mL sample of 0.200 M NaOH is mixed w

Question

Enthalpy of Neutralization Reaction:

A 25.0 mL sample of 0.200 M NaOH is mixed with a 25.0 mL sample of 0.200 M HNO3 in a coffee cup calorimeter.  NaOH and HNO3 will undergo Neutralization Reaction according to the following balanced equation:

NaOH(aq) +   HNO3(aq) --> NaCl (aq) +  H2O (l)  

Both solutions were initially at 35.00°C and Tmax of the resulting solution was recorded as 37.00°C (from the graph).

Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) density and the heat capacity of the resulting solution are the same as that of water. (density of water is 1.0 g/mL and heat capacity of water is 4.18 J/g °C)

Determine the H°rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HNO3. Show all your work.

(Hint: mass is total mass of acid+base and specific heat of solution is same as that of water, H°rxn = qsolution in kJ/ mols of NaOH)  

Explanation / Answer

Number of moles of NaOH , n = Molarity x volume in L

= 0.200 M x 25 mL x 10-3 L/mL

= 5x10-3 mol

Volume of solution , V = 25.0+25.0 = 50.0 mL

Mass of solution , m = volume x density

= 50.0 mL x 1.00 g/mL

= 50.0 g

Change in temperature = dt = final - initial = 37.00 - 35.00 = 2.00 oC

c = specific heat capacity = 4.18 J/goC

Amount of heat liberated , q = mcdt = 418 J

So this much amount of heat liberated by 5x10-3 mol of NaOH

For 1 mole of NaOH the heat liberated is = ( 418x1) / (5x10-3 ) J

= 83600 J

= 83.6 kJ

So H°rxn = -83.6 kJ/mol

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