Hope this example will help you :
HCl(aq)is a strong monoprotic acid: HCl → H+(aq)+ Cl-(aq)
NaOH(aq)is a strong monobasic base: NaOH → Na+(aq)+ OH-(aq)
50.0 mL of 1.0 mol L-1NaOH(aq)is placed in the styrofoam cup.
The temperature of the NaOH(aq)is recorded.
1.0 mol L-1HCl(aq)at the same temperature is added 10.0 mL at a time.
The reaction mixture is stirred between each addition.
The maximum temperature the solution reaches is then recorded.
The results of the experiment are shown below:
Initially, the temperature of the reaction mixture in the calorimeter (styrofoam cup) increases as HCl(aq)is added.
Energy (heat) is being produced by the reaction.
The reaction is exothermic.
Maximum temperature reached is 24.6oC when 50.0 mL of HCl(aq)had been added.
When 50.0 mL of the acid has been added, all the base has been neutralised.
HCl(aq)+ NaOH(aq)→ NaCl(aq)+ H2O(l)
moles HCl(aq)added = moles of NaOH(aq)present in the calorimeter
Adding more acid doesn't increase the temperature in the calorimeter any further#.
We can calculate the molar heat of neutralisation for the reaction if we assume:
thedensityof each dilute aqueous solution is the same as water, 1 g mL-1at 25o
so, the mass of solution in grams = volume of solution in mLtheheat capacityof each solution is the same as for water, 4.18 JoC-1g-1
Extract the data needed to calculate the molar heat of neutralisation for this reaction:
V(NaOH)= volume of NaOH(aq)in the calorimeter = 50.0 mL
V(HCl)= volume of HCl(aq)added to achieve neutralisation = 50.0 mL
c(NaOH)= concentration of NaOH(aq)= 1.0 mol L-1
c(HCl)= concentration of HCl(aq)= 1.0 mol L-1
Ti= initial temperature of solutions before additions = 18.0oC
Tf= final temperature of solution at neutralisation = 24.6oC
d = density of solutions = 1 g mL-1(assumed)
Cg= specific heat capacity of solutions = 4.18 JoC-1g-1(assumed)
q = heat liberated during neutralisation reaction = ? JCheck the units for consistency and convert if necessary:
Convert volume of solutions (mL) to mass (g): density x volume = mass
since density = 1 g mL-1: 1 x volume (mL) = mass (g)
mass(NaOH)= 50.0 g
mass(HCl)= 50.0 gCalculate the heat produced during the neutralisation reaction:
heat produced = total mass x specific heat capacity x change in temperature
q = mtotalx Cgx ΔT
mtotal= mass(NaOH)+ mass(HCl)= 50.0 + 50.0 = 100.0 g
Cg= 4.18 JoC-1g-1
ΔT = Tf- Ti= 24.6 - 18.0 = 6.6oC
q = 100.0 x 4.18 x 6.6 = 2758.8 J
Calculate the moles of water produced:
OH-(aq)+ H+(aq)→ H2O(l)
1 mol OH-(aq)+ 1 mol H+(aq)→ 1 mol H2O
moles(H2O)= moles(OH-(aq))
moles(OH-(aq))= concentration (mol L-1) x volume (L) = 1.0 x 50.0/1000 = 0.050 mol
moles of water produced = 0.050 molCalculate the heat liberated per mole of water produced, ΔHneut:
ΔHneutwill be negative because the reaction is exothermic
ΔHneut= heat liberated per mole of water = -1 x q ÷ moles of water
ΔHneut= -1 x 2758.8 ÷ 0.050 = -55176.0 J mol-1
We can convert J to kJ by dividing by 1000:
ΔHneut= -55176.0 ÷ 1000 = 55.2 kJ mol-1
