The molar enthalpy of neutralization = 68.211 kJ/mol
The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released
Heat can be calculated using the formula:
A calorimeter is a device used to measure the specific heat of material
The heat is not wasted into the environment because the calorimeter is a closed system, so the reaction heat is the same as the heat absorbed by the solution and the calorimeter
[tex]\rm Q\:reaction=Q_{rxn}=-(q\:solution+q\:calorimeter)[/tex]
mass of solution = 59.21 + 50.03 = 109.24 g
c solution = 4,184 J / g ° C
ΔT = 26.8 ° C - 19.9 ° C = 6.9 ° C
Q = m. c ΔT
Q = 109.24. 4,184. 6.9
Q solution = 3153.7151 J
mass = 41.55 g
c the aluminum cup: 0.903 J / g ° C
ΔT = 26.8 ° C - 19.9 ° C = 6.9 ° C
Q calorimeter = m. c ΔT
Q calorimeter = 41.55. 0903. 6.9
Q calorimeter = 258.8856 J
So that:
reaction q = q rxn = - (q solution + q calorimeter)
q rxn = - (3153.7151+ 258.8856)
Q rxn = - 3412,6007 J
ΔH neutralization = change in enthalpy in the formation of 1 mole H₂O
Reactions that occur:
CH₃COOH (aq) + NaOH (aq) → CH₃COONa (aq) + H₂O (l)
The number of moles of H₂O formed is based on limiting reactants, so we look for moles each of the reactants
CH₃COOH = 59.21 g: 1 g / ml × 1.0 M = 59.21 mlmol = 0.05921 moles
NaOH = 50.03: 1 g / ml × 1.0 M = 50.03 mlmol = 0.05003 moles
NaOH mole is smaller, so it becomes a limiting reactant
So mole H₂O = mole NaOH = 0.05003 moles
then:
[tex]\rm \Delta H=\dfrac{-Q_{rxn}}{n}[/tex]
[tex]\rm \DeltaH=-[\dfrac{- 3412.6007}{0.05003}]\\\\\Delta H=\:68211.087\dfrac{J}{mole} =68.211\dfrac{kJ}{mol}[/tex]
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