Question

Andrea has been measuring the enthalpy change associated with reaction between HCl and NaOH in a coffee cup calorimeter. Andrea combined 45.57 mL of 1.00 M HCl and 39.1 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 16.07 oC, and the final observed temperature was 58.46 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. 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 solution(m) = V*density(d) = (45.57+39.1)*1 = 84.67 g

s = specific heat of water = 4.184 j/g.c

DT = 58.46 - 16.07 = 42.39

heat released(q) = m*s*dT

                 = 84.67*4.18*42.39

                 = 15 kj

no of mol of NaOH reacted = 39.1*1/1000 = 0.0391 mol

DHrxn = - q/n = -15/0.0391 = -383.63 kj/mol