Chemistry, 06.11.2019 06:31 maggie123433

The aluminum cup inside your calorimeter weighs 41.55 g. you add 59.21 g of 1.0 m acetic acid solution and 50.03 g of 1.0 m sodium hydroxide solution to the calorimeter. both solutions have an initial temperature of 19.9 oc, and the final temperature after addition is 26.8 oc. what is the molar enthalpy of neutralization, in units of kj/mol? assume that: the calorimeter is completely insulated the heat capacity of the empty calorimeter is the heat capacity of the aluminum cup: 0.903 j g-1 oc-1. the density of the two solutions is the same as that of water: 1.00 g/ml. the heat capacity of the two solutions is the same as that of water: 4.184 j g-1 oc-1.

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Answer from: paradisetiyae5261

The molar enthalpy of neutralization = 68.211 kJ/mol

Further explanation

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

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

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

$\rm Q\:reaction=Q_{rxn}=-(q\:solution+q\:calorimeter)$

Q solution

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

Q calorimeter = Q aluminum cup

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:

$\rm \Delta H=\dfrac{-Q_{rxn}}{n}$

$\rm \DeltaH=-[\dfrac{- 3412.6007}{0.05003}]\\\\\Delta H=\:68211.087\dfrac{J}{mole} =68.211\dfrac{kJ}{mol}$

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Answer from: Quest

The answer should be vaporization

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