What is the standard enthalpy of reaction when 6.45g of acetylene are consumed?

Acetylene can be formed by the following reaction:
#CaC_2(s)+5O_2(g)rarrC_2H_2(g)+Ca(OH)_s(s)#

Calculate #DeltaH_(rxn)# when 6.45g of #C_2H_2# are consumed in the equation below:
#2C_2H_2(g)+5O_2rarr4CO_2(g)+2H_2O(g)#

1 Answer
Ernest Z.
Nov 25, 2017

#Δ_text(rxn)H = "-311 kJ"#

Explanation:

You can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products.

The formula is

#color(blue)(bar(ul(|color(white)(a/a)Δ_text(rxn)H° = Δ_text(f)H_text(products)^@ - Δ_text(f)H_text(reactants)^@color(white)(a/a)|)))" "#

Step 1. Calculate #Δ_text(r)H^@"# for 1 mol of reaction

#color(white)(mmmmmmmmm)"2C"_2"H"_2"(g)" + "5O"_2"(g)" → "4CO"_2"(g)" + 2"H"_2"O(g")#
#Δ_text(f)H^@"/kJ·mol"^"-1": color(white)(ml) "226.75" color(white)(mmmll)0color(white)(mmmll)"-393.5" color(white)(mmll) "-241.8"#

#Δ_text(r)H^° = ["4(-393.5 - 2×241.8) -(2×226.75 - 5×0)"]color(white)(l)"kJ" = "[-2057.2 -" 453.50]color(white)(l)"kJ" = "-2511.1 kJ"#

Step 2. Calculate #Δ_text(r)H# for 6.45 g of #"C"_2"H"_2#

#Δ_text(r)H = 6.45 color(red)(cancel(color(black)("g C"_2"H"_2))) × (1 color(red)(cancel(color(black)("mol C"_2"H"_2))))/(26.04 color(red)(cancel(color(black)("g C"_2"H"_2)))) × ("-2511.1 kJ")/(2 color(red)(cancel(color(black)("mol C"_2"H"_2)))) = "-311 kJ"#

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