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

## Acetylene can be formed by the following reaction: $C a {C}_{2} \left(s\right) + 5 {O}_{2} \left(g\right) \rightarrow {C}_{2} {H}_{2} \left(g\right) + C a {\left(O H\right)}_{s} \left(s\right)$ Calculate $\Delta {H}_{r x n}$ when 6.45g of ${C}_{2} {H}_{2}$ are consumed in the equation below: $2 {C}_{2} {H}_{2} \left(g\right) + 5 {O}_{2} \rightarrow 4 C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$

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 ${\text{C"_2"H}}_{2}$

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"