Homework Answers
Here, use the specific heat of water = 4.184 J/g.C
Again -
dH = m C dT
=> dH = 152.5 grams (4.184 J/g.C) 6.9 C
=> dH = 4403 Joules released
Again, find moles H2SO4, using its molar mass:
4.5 g @ 98.08 g/mol = 0.04588 moles
Again, calculate the enthalpy change for theprocess
H2SO4(l)------>H2SO4(aq), in kJ/mol
4403 Joules released / 0.04588 moles =
dH = - 95968 Joules / mol (minus since its heat released)
dH = - 95.968 KJ / mol = -96 KJ/mol
So, Enthalpy change = - 96 KJ/mol
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