Chem 111 Lab: Enthalpy of a Reaction 3· If 15.0 mL of a 1.5M HCl solution at 21.
ID: 1037554 • Letter: C
Question
Chem 111 Lab: Enthalpy of a Reaction 3· If 15.0 mL of a 1.5M HCl solution at 21.5? is mixed with 25.0mL of a 1.5M NaOH solution at 21.5C that is in a calorimeter, and the final mixed solution temperature rises to 28.5°C, what is the AHn for this process? Assume that Cealorimetar 35.53C a. What is the balanced equation for the reaction? b. What is the source of the heat that is causing the increase in temperature? c. Calculate the amount of heat absorbed or lost for the HCI/NaOH solution. Assume aqueous conditions. (Cwater-4.18 Jlg K, d 1.0 g/mL). Calculate the amount of heat absorbed or lost by the calorimeter. The calorimeter's initial temperature is the same as the solution that is initially inside it. d. c. Determine the amount of heat absorbed or lost during this reaction. Which of the reactants is the limiting reagent? Determine the moles of the limiting reagent. f. Determine the amount of heat given off per mole of limiting reagent( ?11m). Make sure to include an appropriate sign indicating whether it is an endothermic (+) or exothermic (-) process. Provide your answer in units of kJ/mol g.Explanation / Answer
Ans. #a. Balanced equation: HCl(aq) + NaOH(aq) -------> H2O(l) + NaCl(aq)
Net ionic reaction: H+(aq) + OH-(aq) ---------> H2O(l)
#b. The reaction between H+ and OH-, i.e. acid-base neutralization is the source of energy release during the reaction; which in turn, increases the temperature of the reaction mixture.
#c. Total mass of reaction mixture (using density of 1.00 g/ mL)
= 15.0 g + 25.0 g = 40.0 g
Amount of heat gained/lost by the samples is given by-
q = m s dT - equation 1
Where,
q = heat change
m = Total mass of sample (acid + base)
s1 = specific heat of reaction mixture
dT = Final temperature – Initial temperature
Putting the values in equation 1-
q = 40.0 g x (4.184 J g-10C-1) x (28.5 – 21.5)0C = 1171.52 J = +1171.52 J
Since the sign of q is (+), the HCl/NaOH solution absorbs heat during its increase in temperature.
#d. Heat change for calorimeter, q2 = C calorimeter x dT
= 35.5 J 0C-1 x (28.5 – 21.5)0C = 248.5 J
Since the sign of q2 is (+), the calorimeter absorbs heat during its increase in temperature.
#e. Total heat absorbed by (Calorimeter + HCl/NaOH solution) must be equal to the total amount of heat released by the reaction.
So,
Heat released by the reaction = - (q +q2) = - (1171.52 J + 248.5 J) = -1420.02 J
Note: The (-ve) sign indicates release of heat.
#f. Moles of HCl = Molarity x Vol of soln. in liters = 1.5 M x 0.015 L = 0.0225 mol
Moles of NaOH = Molarity x Vol of soln. in liters = 1.5 M x 0.025 L = 0.0375 mol
# Net ionic reaction: H+(aq) + OH-(aq) ---------> H2O(l)
Following stoichiometry, 1 mol H+ (from HCl) is neutralized by 1 mol OH- (from NaOH). Since moles of HCl is less than that of NaOH in 1:1 stoichiometry reaction, HCl is the limiting reactant.
Moles of limiting reactant (HCl) = 0.0225 mol
#g. Heat given off per mol HCl, dH0 = (-q) / moles of HCl taken
= (-1420.02 J) / 0.0225 mol
= -63112.0 J/ mol
= -63.112 kJ/ mol
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