PROBLEM 2 A 74.8 g sample of copper at 143.2 oc is added to an insulated vessel

Question

PROBLEM 2 A 74.8 g sample of copper at 143.2 oc is added to an insulated vessel containing 165 mL of glycerol, C3HaO() (d 1.26 g/mL) at 24.8 oc. The final temperature is 31.1 .C. The specific heat of copper is o 385 Jg 1 c 1. What is the heat capacity of glycerol in PROBLEM 3 Use Hess's law to determine AH for the reaction C3H4(g)+2H20g)- C3H8(g). given that (a) H2(g) 1/202(g) H20 AH° 285.8 kJ (b) C3H4(g) +402(g)- 3CO2 (g)+2H20 AH 1937 kJ (c) C3H8(g)+502(g)- 3co2(g)+4H20(); A Ho 2219.1 kJ PROBLEM 4 Use standard enthalpies of formation to determine the standard enthalpy changes in the following reactions: (a) C3H8(g)+H2 (g) C2H6(g) CH4(g) (b) 2H2S(g) 302(g)-+2s02(g)+2H20(l) PROBLEM 5 A bomb calorimetry experiment is performed with xylose, C5H1005(s), as the combustible substance. The following are obtained: mass of xylose burned 1.183 g heat capacity of calorimeter: 4.728 kJpc initial calorimeter temperature: 23.29 oc final calorimeter temperature 27.19 oc (a) What is the heat of combustion of xylose, in kilojoules per mole? (b) Write the chemical equation for the complete combustion of xylose and represent the value of AH in this equation

Explanation / Answer

2) Use the principle of thermochemistry: heat lost by the hot substance = heat gained by the cold substance.

The copper block is at a higher initial temperature; hence it is the hot substance. The glycerol, is therefore the cold substance.

Let the heat capacity of glycerol be C J/mol.C.

Molar mass of copper = 63.55 g/mol.

Volume of glycerol taken = 165 mL; density of glycerol = 1.26 g/mL.

Therefore, mass of glycerol taken = (165 mL)*(1.26 g/mL) = 207.7 g.

Molar mass of glycerol = (3*12.01 + 8*1.008 + 1*15.9994) g/mol = 60.0934 g/mol.

Moles of glycerol in 207.7 g sample = (207.7 g)/(60.0934 g/mol) = 3.4563 mole.

Heat lost by copper block = (mass of copper)*(specific heat capacity of copper)*(change in temperature of copper) = (74.8 g)*(0.385 J g-1C-1)*(143.2 – 31.1)C = 3228.2558 J

Heat gained by glycerol = (moles of glycerol)*(specific heat capacity of glycerol)*(change in temperature of glycerol) = (3.4563 mole)*(C J mol-1 C-1)*(31.1 – 24.8)C = 21.77469*C J

Therefore,

3228.2558 = 21.77469*C

===> C = 3228.2558/21.77469 = 148.2572

The heat capacity of glycerol = 148.2572 J/mol.C (ans).

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