Data: C(graphite) + O2(g) forms CO2 (ga) Delta H = -393.5kJ
H2(g) + 1/2O2(g) forms H2O(l)Delta H = 285.8kJ.
CH3OH(l) + 3/2O2(g)forms CO2(g) + 2H2O(l)Delta H = -726.4
Using data above, calculate the enthalpy change for the raction below
C(graphite) + 2H2 +1/2 O2(g) forms CH3OH(l)
Label the equations you have above as 1, 2, 3, and 4.
Use 1 as is.
Multiply equation 2 and add to 1 (multiply delta H by 2 also).
Reverse equation 3 (and reverse delta H) and add in.
This should give you equation 4. Then add delta Hs as noted above to give total delta H for the reaction.
To calculate the enthalpy change for the given reaction:
C(graphite) + 2H2 + 1/2O2(g) → CH3OH(l)
You need to use the Hess's Law. Hess's Law states that the enthalpy change for a reaction is the same, regardless of the route taken.
Here's how you can calculate it using the given data:
1. Start by balancing the chemical equation using the given coefficients:
C(graphite) + 2H2 + 1/2O2(g) → CH3OH(l)
2. Determine the enthalpy change of each reaction given:
i. C(graphite) + O2(g) → CO2(g) ΔH = -393.5 kJ
ii. H2(g) + 1/2O2(g) → H2O(l) ΔH = -285.8 kJ
iii. CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l) ΔH = -726.4 kJ
3. Reverse reaction i because we need C(graphite) on the left side:
CO2(g) → C(graphite) + O2(g) ΔH = +393.5 kJ
4. Multiply reaction ii by 2 because we need 2H2 on the left side:
2[H2(g) + 1/2O2(g) → H2O(l)] ΔH = 2(-285.8 kJ) = -571.6 kJ
5. Multiply reaction iii by 2 because we need CH3OH(l) on the left side:
2[CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l)] ΔH = 2(-726.4 kJ) = -1452.8 kJ
6. Combine the reactions to cancel out the common compounds:
[C(graphite) + O2(g) → CO2(g)] + [2[H2(g) + 1/2O2(g) → H2O(l)]] + [2[CH3OH(l)+3/2O2(g) → CO2(g) + 2H2O(l)]] = ΔH
7. Add up the enthalpy changes:
+393.5 kJ + (-571.6 kJ) + (-1452.8 kJ) = -1630.9 kJ
Therefore, the enthalpy change for the reaction
C(graphite) + 2H2 +1/2 O2(g) → CH3OH(l) is -1630.9 kJ.