# How would you use a standard enthalpies of formation table to determine the change in enthalpy for these reactions?

## $2 C O \left(g\right) + {O}_{2} \left(g\right) \to 2 C {O}_{2} \left(g\right)$ $C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$ $2 {H}_{2} S \left(g\right) + 3 {O}_{2} \left(g\right) \to 2 {H}_{2} O \left(l\right) + 2 S {O}_{2} \left(g\right)$

Jun 28, 2017

$- 566 k J$
$- 999 k J$
$- 1125 k J$

#### Explanation:

Our equation of interest to apply for this problem is:
DeltaH_(rxn)^° = sum(n*p roducts)-sum(m* react ants)

Where n and m are moles of substances. We'd subsequently look the values up, do the math, and voila, so in my text, here's what I get. The reason some reactants/products have 0 formation energy is because when the molecule is the most stable form, no energy is needed to form.

$2 C O \left(g\right) + {O}_{2} \left(g\right) \to 2 C {O}_{2} \left(g\right)$

DeltaH_(rxn)^° = (2*-393.5)-(0+2(-110.5)) = -566kJ

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

DeltaH_(rxn)^° = (-393.5+2*-285.8)-(-74.1+2*0) = -999kJ

$2 {H}_{2} S \left(g\right) + 3 {O}_{2} \left(g\right) \to 2 {H}_{2} O \left(l\right) + 2 S {O}_{2} \left(g\right)$

DeltaH_(rxn)^° = (2*-285.8+2*-296.9)-(2*-20.17+3*0) = -1125kJ