1. Calculate the enthalpy change for the reaction: (answer
= -137kJmol-1)
C2H4(g)(g) + H2(g)
---> C2H6(g)
given the following standard enthalpies of combustion:
DHoc
(C2H4(g)) = - 1411Jmol-1
DHoc(H2(g))
= - 286kJmol-1
DHoc (C2H6(g))
= - 1560kJmol-1
2. Calculate the standard enthalpy change of formation of methane from the
following enthalpies of combustion. (answer = -76kJmol-1)
DHoc (CH4(g))
= - 890Jmol-1
DHoc(H2(g))
= - 286kJmol-1
DHoc (C(s))
= - 394kJmol-1
3. Calculate the standard enthalpy of formation of ethanol, C2H5OH,
from the following standard enthalpies of combustion:
DHoc (C2H5OH(l))
= - 1367Jmol-1
DHoc(H2(g))
= - 286kJmol-1
DHoc (C(s))
= - 394kJmol-1
4. Methanol can be made in a two-step process.
Step 1: CH4(g) +H2O(g) == CO(g) + 3H2(g)
Step 2: CO(g) + 2H2(g) == CH3OH(g)
Use the following enthalpies of combustion to calculate the enthalpies of
reaction for each of the steps:
DHoc
(CH4(g)) = - 890mol-1
DHoc(H2(g))
= - 245kJmol-1
DHoc (CO(g))
= - 283kJmol-1
DHoc
(CH3OH(g)) = - 671Jmol-1