Question

NUMBER 1

A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 119 g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

Enter your answer in kilojoules per mole of compound to three significant figures.

NUMBER 2

The standard heat of formation, ΔH∘f, is defined as the enthalpy change for the formation of one mole of substance from its constituent elements in their standard states. Thus, elements in their standard states have ΔH∘f=0. Heat of formation values can be used to calculate the enthalpy change of any reaction.

Consider, for example, the reaction

2NO(g)+O2(g)⇌2NO2(g)

with heat of formation values given by the following table:

Substance	ΔH∘f (kJ/mol)
NO(g)	90.2
O2(g)	0
NO2(g)	33.2

Then the standard heat of reaction for the overall reaction is

ΔH∘rxn===ΔH∘f(products)2(33.2)−114 kJ−−ΔH∘f(reactants)[2(90.2)+0]

Part A

For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?

You do not need to look up any values to answer this question.

Check all that apply.

Na(s)+12F2(g)→NaF(s)

BaCO3(s)→BaO(s)+CO2(g)

2Na(s)+F2(g)→2NaF(s)

C(s,graphite)+O2(g)→CO2(g)

Na(s)+12F2(l)→NaF(s)

CO(g)+12O2(g)→CO2(g) Part B The combustion of heptane, C7H16, occurs via the reaction C7H16(g)+11O2(g)→7CO2(g)+8H2O(g) with heat of formation values given by the following table: Substance ΔH∘f (kJ/mol) C7H16 (g) -187.9 CO2(g) −393.5 H2O(g) −241.8 Calculate the enthalpy for the combustion of 1 mole of heptane. Express your answer to four significant figures and include the appropriate units.

Answer 1

q    = mcT

       = 119*4.18*(24.7-21)

       =1840.454J

      = 1840.454/2   = 920.227 J/mole    = 0.920227Kj/mole

    = -0.92Kj/mole

2NO(g)+O2(g)⇌2NO2(g)

ΔH∘rex   = ΔH∘f products - ΔH∘f reactants

              = 2*33.2 -(2*90.2+0)

              = -114 Kj

part-A

C(s,graphite)+O2(g)→CO2(g)

part-B

C7H16(g)+11O2(g)→7CO2(g)+8H2O(g)

ΔH∘rex   = ΔH∘f products - ΔH∘f reactants

              = 7*-393.5 + 8*-241.8 -(-187.9 + 11*0)

               = -4501 Kj