3 Find the H for the reaction below given the following reac

3. Find the H for the reaction below, given the following reactions and subsequent H values: 2co,(g) + H20(g) C2H2(g) + 5/202(g) AH formation of CO2 =-393.5 ki dH formation of H2O(g) =-142.2 kj formation of C2H6(g) =-91.5 kj 4. Calculate the ton of C (s) + H2O (s) CO (g)+ H2 (s) based on the following reactions: AH formation of CO2 (0)-393.5 k AH formation of H2O--242.0 k 5. Calculate the of Fe,04 (s) + CO (g) 3 FeO (s) CO2 (g) + based on the following reactio &H; formation of CO (g) =-110.5 kj dH formation of FeO (s) =-39 k AH formation of Fe,046)--118 J AH formation of CO2 (g)=-393.5 k]

Solution

Ans. #3. Following Hess law of constant heat summation,

            dH_rxn = (sum of dH_f of products) – (sum of dH_f of reactants)

            Or, dH_rxn = (dH_f C₂H₂ + 2.5 x dH_f O₂) – (2 x dH_f CO₂ + dH_f H₂O)

            Or, dH_rxn = [(-91.5 kJ) + 0] – [2 x (-393.5 kJ) + (-142.2 kJ)]

            Or, dH_rxn = (-91.5 kJ) – (-929.2 kJ) = (-91.5 kJ) + 929.2 kJ

            Hence, dH_rxn = 837.7 kJ

Note: dH_f of O₂ (natural state of a gas) is taken to be zero.

            * dH_f C₂H₆ is mentioned instead of dH_f C₂H₂. It’s assumed to a typo error.

#4.      dH_rxn = (dH_f CO + dH_f H₂) – (dH_f C + dH_f H₂O)

Or, dH_rxn = [(-110.5 kJ) + 0 ] – [0 + (-242.0 kJ)]

Or, dH_rxn = (-110.5 kJ) + 242.0 kJ

Hence, dH_rxn = 131.5 kJ

Note: dH_f of H₂ (natural state of a gas) and C (natural state) are taken to be zero.

            * dH_f CO₂ is mentioned instead of dH_f CO. dHf CO₂ is of no use for this reaction. So, the value of dHf CO is taken as from question 5.

            * dH_f H­₂O(g) is a constant value. dH_f H₂O varies depending on the physical state of water- i.e. liquid and vapor. However, for one particular state, say vapor or liquid, its respective dH_f always remains constant. Question 3 and 4 mentions different values of dH_f H₂O for the vapor (gaseous) phase water – please recheck the values and phase of H₂O.

#5. dH_rxn = (3 x dH_f FeO + dH_f CO₂) – (dH_f FeSO₄ + dH_f CO)

Or, dH_rxn = [3x (-39.0 kJ) + (-393.5 kJ) ] – [ (-118.0 kJ) + (-110.5 kJ)]

Or, dH_rxn = (-510.5 kJ) + 228.5 kJ

Hence, dH_rxn = -282.0 kJ