Volume of water 99.5 mL Mass of salt 5.56 g change in temperature Final temperat

Question

Volume of water

99.5 mL

Mass of salt

5.56 g

change in temperature

Final temperature = 20.4C

Initial temperature= 23.4C

Amount of Na2SO4. 10 H2O (s)

0.01725 moles

Solution Density

1.02 g mL-1

Volume of water

99.5 mL

Mass of salt

5.56 g

change in temperature

Final temperature = 20.4C

Initial temperature= 23.4C

t= -3.00C

Amount of Na2SO4. 10 H2O (s)

0.01725 moles

Solution Density

1.02 g mL-1

Calculate the heat of reaction, qrxn, in joules. Use the correct sign to indicate if heat is being absorbed or given off. Calculate standard solution enthalpy,

Explanation / Answer

q=msDT
mass = 5.56 + 99.5*1.02
Specific heat of water = 4.184 j/g.C
DT = 23.4-20.4 = -3
q = 107.05*4.184*(-3)
= ?1343.7 Joule
= ?1.3 kj
q = -DH
DH = + 1.3 Kj Per 0.01725 moles
standard solution enthalpy DHrxn = +1.3/0.01725
= +75.4 kj/mol

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