Need help with these questions. Especially (a). I need a explanation plz. Here i

Question

Need help with these questions. Especially (a). I need a explanation plz.
Here is the table
9.73 Acetylene gas (ethyne; HC CH) burns in an oxy 73 Acetylene gas (ethyne; HC CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor. The heat of reaction for the combustion of acetylene is 1259 kJ/mol. (a) Cal- culate the C C bond energy, and compare your value with that in Table 9.2, p. 371. (b) When 500.0 g of acetylene burns, how many kilojoules of heat are given off? (c) How many grams of CO2 form? (d) How many liters of O at 298 K and 18.0 atm are consumed?

Explanation / Answer

heat of reaction can be calculated from bond energies. its is given by

Heat of reaction = Bond enegies of bodns broken - Bond energies of bonds formed

The reaction we have is combustion of acetylene

C2H2 + 2.5 O2 ---> 2CO2 + H2O

dH of reaction = BE ( C triple bond C) + 2 BE ( C-H) + 2.5 BE ( O=O) - 4 BE ( C=O) - 2 BE ( O-H)

- 1259 = BE ( C triple bond C)+ 2( 413) + 2.5 ( 498) - 4( 799) - 2( 467)      

BE ( C triple bond C) = 800 KJ/mol

Table value is 839 KJ/mol

b) C2H2 moles = mass / Moalr mass

           = 500 / 26.04 = 19.2

Heat released = dH of combustion x moles

              = 1259 KJ/mol x 19.2 mol

          = 24174 KJ/mol    of heat released

c) 1C2H2 gives 2CO2 , hence CO2 moles formed = 2 x 19.2 = 38.4

CO2 mass formed = moles x molar mass of CO2 = 38.4 x 44g/mol = 1689.6 g

d) O2 moles consumed = 2.5 x C2H2 moles = 2.5 x 19.2 = 48

we use PV = nRT   equation    where R is gas constant = 0.08206 liter atm/molK

18 atm x V = 48 mol x 0.08206 liter atm/molK x 298 K

V = 65.2 L

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