Question

Hess's Law Given the following data: 2C(s) + 2H2(g) + O2(g) → CH3OCHO(l) ΔH°=-366.0 kJ CH3OH(l) + O2(g) → HCOOH(l) + H2O(l) ΔH°=-473.0 kJ C(s) + 2H2(g) + 1/2O2(g) → CH3OH(l) ΔH°=-238.0 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH°=-286.0 kJ calculate ΔH° for the reaction: HCOOH(l) + CH3OH(l) → CH3OCHO(l) + H2O(l)

Answer 1

Given the following data:

2C(s) + 2H2(g) + O2(g) ---- > CH3OCHO(l) ΔH°=-366.0 kJ    -----1

CH3OH(l) + O2(g) ---- > HCOOH(l) + H2O(l) ΔH°=-473.0 kJ   ------2

C(s) + 2H2(g) + 1/2O2(g) ---- > CH3OH(l) ΔH°=-238.0 kJ ------3

H2(g) + 1/2O2(g) ---- > H2O(l) ΔH°=-286.0 kJ -------4

calculate ΔH° for the reaction:

HCOOH(l) + CH3OH(l) ----> CH3OCHO(l) + H2O(l)

As follows:

First reverse equation number 3 and 2 and multiply equation 3 and 4 by 2 , than add all :

HCOOH(l) + H2O(l) ---- > CH3OH(l) + O2(g) ΔH°=+ 473.0 kJ

2CH3OH(l) ---- >2 C(s) + 4H2(g) + O2(g) ΔH°=+476.0 kJ

2C(s) + 2H2(g) + O2(g) ---- > CH3OCHO(l) ΔH°=-366.0 kJ    

2H2(g) + O2(g) ---- > 2 H2O(l) ΔH°=-572.0 kJ

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HCOOH(l)+ CH3OH(l) ----> CH3OCHO(l) + H2O(l) ΔH°=11 kJ