Question

2/22 To plot against . just add each to the previous q to obtain Worksheet 7a (Gen Chem) Thermochemistry D) Calorimetry. 6. Determine the heat of reaction in kJ/mol H2O when $5.0 ml of 0.570 M HBr at 19.71°C is mixed with mL of 0.570 M KOH at 19.71°C in a coffee cup calorimeter. The temperature rises to 22.88°C. Hint: need to identify limiting reactant

Answer 1

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HBr is a strong acid and KOH is a strong base, so reaction between these two is acid-base neutralization reaction. Hence heat of reaction is also called as enthalpy of neutralization ΔH

ΔH = qrxn / n, where ΔH = enthalpy of reaction per mol, qrxn = total heat of reaction, n = moles reacted

qrxn = m x Cs x ΔT,

where m = mass of solution = 55 mL + 77 mL = 132 mL ~ 132 g (since water density 1g/mL)

Cs = specific heat capacity of water = 4.184 J/g.0C

ΔT = Tfinal - Tinitial = 22.88 - 19.71 = 3.17 0C

qrxn = 132 g x 4.184 J/g.0C x 3.17 0C = 1750.75296 J = 1.75 kJ

now we will calculate actual moles reacted

moles HBr = 55 mL x 10-3 L x 0.570 M = 0.03135 mol

moles KOH = 77 mL x 10-3 L x 0.570 M = 0.04389 mol

but 1 mol HBr reacts with 1 mol KOH only, so HBr is limiting reactant

ΔH = qrxn / n = 1.75 kJ / 0.03135 mol = 55.8 kJ/mol (this is almost near to standard value)

Answer: Heat of reaction = 55.8 kJ/mol

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