Question

Consider the following data:

Three entries. 1: Delta H f 1 of C O 2 equals negative 393.5 kilojoules per mole. 2: Delta H f 1 H 2 O equals negative 241.82 kilojoules per mole. 3: Delta H f 1 C 2 H 2 equals negative 84.68 kilojoules per mole.

Using these data, calculate es001-1.jpgHrxn for this chemical reaction. Round to the nearest whole number.

2C2H6(g) + 7O2(g) es001-2.jpg 4CO2(g) + 6H2O(g)

es001-3.jpg=
kJ

Answer 1

Use ΔHrxn = ΣnΔHf(products) − ΣnΔHf(reactants).

Products:
4 CO2: 4(−393.5) = −1574.0 kJ
6 H2O: 6(−241.82) = −1450.92 kJ
Sum(products) = −3024.92 kJ

Reactants:
2 C2H6: 2(−84.68) = −169.36 kJ
7 O2: 0 kJ
Sum(reactants) = −169.36 kJ

ΔHrxn = −3024.92 − (−169.36) = −2855.56 kJ ≈ −2856 kJ

Answer: −2856 kJ.