Acetylene burns in air according to the following equation: c2h2(g) + 5 2 o2(g) → 2 co2(g) + h2o(g) δh o rxn = −1255.8 kj given δh o f of co2(g) = −393.5 kj/mol and δh o f of h2o(g) = −241.8 kj/mol, find δh o f of c2h2(g).

Considering the combustion of acetylene, the standard enthalpy of formation of acetylene is 227.0 kJ/mol.

Let's consider the following thermochemical equation.

C₂H₂(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g)    ΔH°rxn = −1255.8 kJ

Given the standard enthalpies of formation (ΔH°f), and consider that the enthalpy of formation of simple substances in the standard state is zero, we can calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.

where,

We will use this equation to calculate the standard enthalpy of formation of acetylene.

Considering the combustion of acetylene, the standard enthalpy of formation of acetylene is 227.0 kJ/mol.

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The standard enthalpy of formation, , of C₂H₂ is 227 kJ/mol

From the question,

We are to determine the   of C₂H₂

From the given chemical equation,

C₂H₂(g) +  ⁵/₂O₂(g) → 2CO₂(g) + H₂O(g)               = −1255.8 kJ

The standard enthalpy of a reaction, , is given by the formula

From the question

of CO₂ = -393.5 kJ/mol

of H₂O = -241.8 kJ/mol

∴ = (2× of CO₂ + of H₂O) - ( of C₂H₂ + ⁵/₂× of O₂)

Putting the parameters into the equation, we get

-1255.8 = (2× -393.5 + -241.8) - ( of C₂H₂ + ⁵/₂× 0)

(NOTE: of O₂ = 0 kJ/mol)

-1255.8 = (-787 -241.8) -  ( of C₂H₂ + 0)

-1255.8 = -1028.8 -  ( of C₂H₂)

∴ of C₂H₂ = -1028.8 + 1255.8

of C₂H₂ = 227 kJ/mol

Hence, the standard enthalpy of formation, , of C₂H₂ is 227 kJ/mol

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c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current