Solution:
The reaction between H2SO4 and KOH to form water is expressed as,
H2SO4 + 2KOH = 2H2O + K2SO4
Thus, number of moles of H2O is equal to number of moles of KOH.
The number of moles of H2SO4 = Molarity x Volume
= 0.500 mol /L. x 0.0100 L = 0.005 mol
The number of moles of KOH = Molarity x Volume
= 0.500 mol/L x 0.0100 L = 0.005 mol
Hence, the number of moles of H2O = 0.005 mol
Thus,
q = m c ΔT
q = 20 g x 4.184 J /°C g x 4 °C
= 334.72 J
(Since, mass of KOH and H2SO4 is equal to its volume because density of solution is considered as 1g/mL. Mass = 0.010 L + 0.010 L = 20 mL = 20 g )
Thus,
ΔH = - q /n = - 334.72 J /number of moles of water
ΔH = - 334.72 J /0.005 mol = - 66944 J/mol
ΔH = - 66.94 kJ/mol
Hence, molar heat of enthalpy (ΔH)for H2O = -66.94 kJ/mol
Total number of moles of product = moles of H2O and K2SO4
= 0.005 mol + 0.005 mol = 0.010 mol
Thus,
Molar enthalpy of products = - Q/number of moles
= - 1.386 kJ / 0.01 mol
= - 138.6 kJ/mol
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