Question

Andrea has been measuring the enthalpy change associated with reaction between HCl and NaOH in a coffee cup calorimeter. Andrea combined 51.67 mL of 1.00 M HCl and 38.49 mL of 1.00 M NaOH in a coffee cup calorimeter (mass of the coffee cups + a stir bar = 15.00 g). If the initial temperature of the acid/base solution was 22.52 oC, and the final observed temperature was 44.06 oC, what is the enthalpy change of the neutralization reaction, in Kilojoules (KJ) per mole of NaOH? Assume that the density (d = 1.00 g/mL) and specific heat capacity of the solution are identical to that of pure water (4.184 J/g-K). Provide your response to two digits after the decimal. You may wish to consult a periodic table. Important note: because this neutralization reaction is an exothermic process, use the negative sign (-) before type the number in your answer with no-space in between (i.e. -125.30). Do not use units (i.e. J) in your answer.

Answer 1

mass of the solution = 51.67+38.49= 90.16ml, density of the solution = 1 g/ml, mass of the solution = 90.16*1= 90.16 gm

Temperature rise = 44.06-22.52 =21.54 deg.c

enthalpy change of reaction =mass of the solution * specific heat* temperature difference= 90.16*4.184*21.54 joules=8125.5, since heat is relased, the reaction is exothermic, hence enthalpy change= -8125.5 Joules

moles of HCl taken = molarity of HCl* Volume of HCl in L= 1*51.67/1000 = 0.052 moles

moles of NaOH = molarity* Volume in L= 1*38.49/1000 = 0.0385 moles

The reaction between HCl and NaOH is HCl+ NaOH----->NaCl+ H2O

Theoretical molar ratio of HCl :NaOH= 1:1, actual ratio = 0.052 :0.0385

Limiting reactant is NaOH. All the NaOH is consumed.

hence enthalpy change per mole of NaOH= -8125.5/0.0385 J/mole =-211052.5 J/mole = -211.05 Kj/mole