Given the following standard heats of reactions (i) heat of formation of water =-68.3 kcal (ii) heat of combustion of acetylene =-310.6 kcal (iii) heat of combustion of ethylene =-337.2 kcal Calculate the heat of reaction for the hydrogenation of acetylene at constant volume (25° C ).

Question

Given the following standard heats of reactions (i) heat of formation of water =-68.3 kcal (ii) heat of combustion of acetylene =-310.6 kcal (iii) heat of combustion of ethylene =-337.2 kcal Calculate the heat of reaction for the hydrogenation of acetylene at constant volume (25° C ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

C 2 H 2+ H 2 longrightarrow C 2 H 4 Δ H°= Sigma Δ H text comb ° (reactants) - Sigma Δ H text comb ° (products) =-310.6-68.3-(-337.2) =-41.7 kcal

Q. Given the following standard heats of reactions (i) heat of formation of water =−68.3kcal (ii) heat of combustion of acetylene =−310.6kcal (iii) heat of combustion of ethylene =−337.2kcal Calculate the heat of reaction for the hydrogenation of acetylene at constant volume (25∘C).

C2​H2​+H2​⟶C2​H4​ ΔH∘=ΣΔHcomb ∘​ (reactants) −ΣΔHcomb ∘​ (products) =−310.6−68.3−(−337.2) =−41.7kcal

$C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}$
$Δ H^{\circ} = ΣΔ H_{comb}^{\circ}$ (reactants) $- ΣΔ H_{comb}^{\circ}$ (products)
$= - 310.6 - 68.3 - (- 337.2)$
$= - 41.7 k c a l$