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Part A For which of the following reactions is Hrxn equal to Hf of the product(s

ID: 906902 • Letter: P

Question

Part A For which of the following reactions is Hrxn equal to Hf of the product(s)? You do not need to look up any values to answer this question. Check all that apply. Check all that apply. SO(g)+12O2(g)SO2(g) Na(s)+12Cl2(l)NaCl(s) 2Na(s)+Cl2(g)2NaCl(s) S(s)+O2(g)SO2(g) Na(s)+12Cl2(g)NaCl(s) SO3(g)12O2(g)+SO2(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 Hf (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. Hrxn =

Explanation / Answer

Part A

Note that Hf = enthalpy of formation; therefore, we must state the definition first:

- there must be only 1 mol of the desired product "formed"

- It must be formed by only elements

- It must be the most abundant/stable phase/allotrope found in nature

- It must be at STP

then... knowing this:

SO(g)+1/2O2(g)SO2(g) --> cant be since, SO is a compound, not an element

Na(s)+1/2Cl2(l)NaCl(s) --> this can't be, since Cl2 is not found as liquid in nature at STP

2Na(s)+Cl2(g)2NaCl(s) --> not possible, this gives 2 mol of NaCl, can't be

S(s)+O2(g)SO2(g) --> this can't be, since S is found as S8(s) in nature, the most stable allotrope

Na(s)+1/2Cl2(g)NaCl(s) ---> this is correct; since it produces 1 mol of NaCl from elements

SO3(g)12O2(g)+SO2(g) --> false, this is made up of compounds and not elements

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 Hf (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. Hrxn =

First, defint eht combustion enthalpy change

dH = Hproducts - Hreactants = 7CO2 + 8H2O - (C7H16 +11O2)

substitute values

dH = 7*-393.5 + 8*-241.8 - (-187.9 + 11*0) kJ/mol

dH = -4501 kJ/mol