How do you calculate the heat of neutralization? I have one sample problem. A st
ID: 573663 • Letter: H
Question
How do you calculate the heat of neutralization? I have one sample problem. A student conducts an investigation to determine the heat of neutralization when 60.0 mL of 0.500 M sulfuric acid is reacted with 50.0 mL of 1.00 M sodium hydroxide.
The following lab data was collected. • initial temperature of NaOH 23.4°C • initial temperature of H2SO4 23.4°C • stabilized temperature of mixture 29.3 °C
a. Assuming that the calorimeter has a specific heat capacity of 1.0 × 10 J/°C and that the density and specific heat capacity of the solution are the same as that of water, calculate the heat of neutralization for this reaction (q).
b. Calculate the molar enthalpy of solution (Hneut) for this reaction.
Explanation / Answer
Given that;
60.0 mL of 0.500 M sulfuric acid
50.0 mL of 1.00 M sodium hydroxide
initial temperature = 23.4°C
final temperature = 29.3 °C
specific heat capacity = 1.0 × 10 J/°C
number of moles = molaarity * volume in L
1.0 moles NaOH/liter x 50.0ml x 1 liter/1000 mL
= 0.050 moles NaOH
0.500 moles H2SO4/1liter x 60.0 mL x 1liter/1000 mL
= 0.030 moles H2SO4
The balance reaction of NaOH and H2SO4 is as follows:
2NaOH + H2SO4 == Na2SO4 + 2H2O
0.050 moles of NaOH x 1 mole H2SO4/2moles NaOH = 0.025 mole H2SO4
Since one has 0.030 moles H2SO4 , 0.050 moles NaOH is the limiting reactant.
Heat given off from neutralizaton:
Assume specific heat is that of water, namely 4.18 J/gC
Total volume of solution = 50 ml+60 ml=
110mL.
Density of wate ri s1.00 g/ ml then mass of 110 ml of this solution is 110 g.
And q= Mc*dT
M= mass of solution
C =specific heat
dT = temperature change
q= 110 g solution x 4.18 Joules/gC x (29.3-23.4)
= 2712.82 Joules
and ;
qCal = 1.0 × 10 J/°C x 5.9 (change in temperature) = 59 J
total heat = +59
=2771.82 Joules
Thus 2771.82 Joules of energy given off by the reaction.
If one wants the heat of neutralization per mole of water, then moles of
water from can be calculate das follows:
0.050 moles NaOH x 2 moles H20/2 moles NaOH = 0.050 moles H2O
Since heat was evolved the reaction is exothermic, so delta H is -2771.82
= - 2771.82 Joules/0.05 moles H2O = -55436.4 Joules/moles H2O
or -55436.4 Joules x 1 kilojoule/1000 Joules
= -55.44 kilojoules/mole H2O
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