Question

water vapormate the heat for the combustion of 2.6 g of acetylen gas C,H to form carbon moles qcetne-0o.1mo d4 4744-y7og 2964 /mo se standard heat of formations to find the heat for the combustion of 2.6 g of acetylene gas Cally to form por. Explain why your answers for problem 4 and 5 are not exactly the same carbon dioxide and water va

Answer 1

4.

Balanced combuiton reaction is

2C₂H₂ + 5O₂$\rightarrow$ 4CO₂ + 2H₂O

Bond enthalpy of

C $\equiv$ C = 963 kJ/mol

O = O = 498 KJ/mol

C - H = 414 kJ/mol

O - H = 465 kJ/mol

C = O = 532 kJ/mol

$\Delta$H⁰ = $\sum$$\Delta$H⁰ (Reactant bonds) - $\sum$$\Delta$H⁰ (product bonds)

$\Delta$H⁰ = [2 $\Delta$H⁰(C $\equiv$C) + 4$\Delta$H⁰(C - H) + 5$\Delta$H⁰(O = O) ] - [8$\Delta$H⁰(C = O) + 4$\Delta$H⁰(O - H) ]

Substitute the bond enthalpy in above equation

[2(963 kJ/mol) + 4( 414 kJ/mol) + 5(498)] - [8(532 kJ/mol) + 4( 465 kJ/mol) ]

(6072 kJ/mol) - (6116) = -44 kJ/mol

$\Delta$H⁰ for your reaction is = -44 kJ/mol

-44 kJ/mol is for combution of 2 mole of C₂H₄

molar mass of C₂H₄ = 28.05 gm/mole then 2 mole C₂H₄ = 2 X 28.05 = 56.1 gm

for combution of 56.1 gm C₂H₄$\Delta$H⁰ for your reaction is = -44 kJ/mol then for 2.6 gm C₂H₄$\Delta$H⁰ for your reaction is = -44 X 2.6 / 56.1 = -2.04 KJ

for coombution of 2.6 gm acetylene gas $\Delta$H⁰ = -2.04 KJ

5.

Hof(C2H2(g)) = 226.73 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Hof(CO2(g)) = -393.509 KJ/mol

Hof(H2O(g)) = -241.818 KJ/mol

we have the Balanced chemical equation as:

2 C2H2(g) + 5 O2(g) ---> 4 CO2(g) + 2 H2O(g)

deltaHo rxn = 4*Hof(CO2(g)) + 2*Hof(H2O(g)) - 2*Hof( C2H2(g)) - 5*Hof(O2(g))

deltaHo rxn = 4*(-393.509) + 2*(-241.818) - 2*(226.73) - 5*(0.0)

deltaHo rxn = -2511.132 KJ/mol

Molar mass of C2H2 = 2*MM(C) + 2*MM(H)

= 2*12.01 + 2*1.008

= 26.036 g/mol

mass of C2H2 = 2.6 g

we have below equation to be used:

number of mol of C2H2,

n = mass of C2H2/molar mass of C2H2

=(2.6 g)/(26.036 g/mol)

= 9.986*10^-2 mol

delta Ho = deltaHorxn * number of mol

= -2511.132 KJ/mol * 9.986*10^-2 mol

= -250.8 KJ