Andrea has been measuring the enthalpy change associated with reaction between H
ID: 486004 • Letter: A
Question
Andrea has been measuring the enthalpy change associated with reaction between HCl and NaOH in a coffee cup calorimeter. Andrea combined 50.13 mL of 1.00 M HCl and 35.76 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 16.68 oC, and the final observed temperature was 61.67 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
Explanation / Answer
For HCl,
c= 1M = 1 mol L-1, M = 36.46 g mol-1, V = 0.05013 L
number of moles (n) = cV = 0.05013 mol
For NaOH,
c= 1M = 1 mol L-1, M = 39.99 g mol-1, V = 0.03576 L
number of moles (n) = cV = 0.03576 mol
Since, HCl (aq) + NaOH (aq) --> NaCl (aq) +H2O (l)
is a 1:1 reaction and NaOH is an limiting reagent (less number of moles of NaOH are involved in this reaction)
Therefore number of moles of NaCl = 0.03576 mol
mass of NaCl (m) = number of mol of NaCl X molar mass of NaCl
= 0.03576 mol X 58.44 g mol-1
= 2.0898 g
Total amount of heat abosrbed or released by the system (q) = mC(dT)
= 2.0898 g X 4.184 J g-1K-1 X(61.67-16.68)K
NB: temperature difference in K and C is the same, so there is no need to convert the temperatures to Kelvin scale
q= 393.38 J
Enthalpy change of neutralisation (dH) = |q|
since heat is liberated and dT is positive, this is an exothermic process
therefore dH = -q = -393.38 J
dH = -0.39 kJ