Asked Jun 9, 2019

What is the standard enthalpy change for the combustion of gaeous propylene, C3H6

C3H6(g) + 9/2 O2 (g) = 3CO2 (g) +3H2O (l)

Substance  delta (KJ/mol)
C3H(G) +20.4
CO2 (g) -393.5
H2O (l) -285.8



Expert Answer

Step 1

STEP 1: write the balanced chemical equation



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CH(g) +02(g) -+ 3C02(g) + 3H2Om 2

Step 2

enthalpy change for the reaction can be calculated by given formula:


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AH=nH (products)-nAH (reactants)

Step 3

The heat of formation of an element in its standard state(here oxygen is in standard state) is zero

. First finding total heat of formation of products


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ДН (0,) 3 о kJ - mol 1 ZnАНf" (products) %3D Зтol x дН/"(CО2) + Зто1 x ДНr (H20) — 3х-393.51 + 3х-285.83%3D—2038.02KJ ZnАнf" (reactants) %3D 1mol x ДH/"(СзН6)+ —x дҢ/"(02) 2 3 1тol x ДН/" (Сзн6) - 20.4kJ ДНС— — 2038.02— 20.4 % — 2058.4kJ


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