Question

Ethyne (C2 H2 (g), mc032-1.jpgHf = 226.77 kJ/mol) undergoes complete combustion in the presence of oxygen to produce carbon dioxide (CO2 (g), mc032-2.jpgHf = ‚393.5 kJ/mol ) and water (H2 O(g), mc032-3.jpgHf = ‚241.82 kJ/mol) according to the equation below.

Answer 1

2 C2H2 + 5 02  >  4 CO2 + 2 H2O

Products  - Reactants ( all units are kJ/mo1):

(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1

-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1

Answer 2

Answer: The $\Delta H$ for the given reaction is -2511.18 kJ/mol

Explanation:

Enthalpy of the reaction is equal to the total sum of the standard enthalpies of the  formation of products minus the total sum of the standard enthalpies of formation of reactants. It is represented by $\Delta H_{rxn}$

Mathematically,

$\Delta H_{rxn}=\Delta H_{products}-\Delta H_{reactants}$

For the given reaction:

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)$

The standard enthalpy of the elements present in their standard state is always zero.

$\Delta H_{rxn}$ for the reaction is calculated by:

$\Delta H_{rxn}=[4(\Delta H_{CO_2})+2(\Delta H_{H_2O})]-[2(\Delta H_{C_2H_2})+5(\Delta H_{O_2})]$

We are given:  

$\Delta H_{C_2H_2}=226.77kJ/mol\\\Delta H_{O_2}=0kJ/mol\\\Delta H_{CO_2}=-393.5kJ/mol\\\Delta H_{H_2O}=-241.82kJ/mol$

Putting values in above equation, we get:  

$\Delta H_{rxn}=[4(-393.5)+2(-241.82)]-[2(226.77)]kJ/mol\\\\\Delta H_{rxn}=-2511.18kJ/mol$

Hence, $\Delta H$ for the given reaction is -2511.18 kJ/mol