From the enthalpies of reaction H2(g) + F2(g)---->2HF(g) delta H= -537 kJ C(s) +

Question

From the enthalpies of reaction
H2(g) + F2(g)---->2HF(g) delta H= -537 kJ
C(s) + 2F2(g)---> CF4 (g) delta H= -680 kJ
2C(s) + 2H2(g)----> C2H4(g) delta H= +52.3 kJ

calculate delta H for the reaction of ethylene with F2
C2H4(g)+6F2(g)--->2CF4(g) + 4HF(g)

Explanation / Answer

From the enthalpies of reaction H2(g) + F2(g)---->2HF(g) delta H= -537 kJ = delta h1 C(s) + 2F2(g)---> CF4 (g) delta H= -680 kJ = delta h2 2C(s) + 2H2(g)----> C2H4(g) delta H= +52.3 kJ= delta h3 Here we have to find delta h for the equation C2H4(g)+6F2(g)--->2CF4(g) + 4HF(g) In the above equation formation of 4 HF and 2 CF4 and deformation of C2H4 has happened. So, the delta h will be = (2*-537) + (2*-680) + (52.3) = -1074 - 1360 +52.3 = - 2381.7 KJ

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