Question

Ethylene on combustion gives co2 and water. Its enthalpy of combustion is -1410. Calculate the enthalpy of formation of ethylene

Answer 1

Answer : Enthalpy of formation for ethylene is 139.4 kJ/mol

Explanation :

The balanced chemical equation for combustion of ethylene can be written as,

$C_{2}H_{4}(g)+ 3O_{2}(g)\rightarrow 2CO_{2} (g)+2H_{2}O(g)$

The enthalpy for above reaction is -1410.

Enthalpy of a reaction is calculated using following formula.

$\bigtriangleup H_{rxn}= \sum \bigtriangleup H_{f}Products - \sum \bigtriangleup H_{f}Reactants$

$\bigtriangleup H_{rxn}= [2\times \bigtriangleup H_{f}CO_{2}+2\times \bigtriangleup H_{f}H_{2}O] - [\bigtriangleup H_{f}C_{2}H_{4}+3\times \bigtriangleup H_{f}O_{2}]$

The standard enthalpies of formation for H2O, CO2 and O2 can be taken from reference table.

The values are given below.

ΔHf CO₂ (g) = -393.5 kJ/mol

ΔHf H₂O (g) = -241.8 kJ/mol

ΔHf O₂ (g) = 0

ΔH rxn = -1410 kJ

Let us plug in the above values in the formula.

$-1410 kJ= [2mol\times (-393.5kJ/mol)+2mol\times (-241.8 kJ/mol)] - [\bigtriangleup H_{f}C_{2}H_{4}+3mol\times (0)]$

$-1410 kJ= [(-787kJ)+(-483.6kJ)] - [\bigtriangleup H_{f}C_{2}H_{4}+0]$

$-1410 kJ= [-1270.6kJ] - [\bigtriangleup H_{f}C_{2}H_{4}]$

$\bigtriangleup H_{f}C_{2}H_{4}=-1270.6 kJ + 1410 kJ$

$\bigtriangleup H_{f}C_{2}H_{4}=139.4 kJ$

Enthalpy of formation for ethylene is 139.4 kJ/mol