Question

From the enthalpies of reaction H2(g)+F2(g)→2HF(g)ΔH=−537kJ C(s)+2F2(g)→CF4(g)ΔH=−680kJ 2C(s)+2H2(g)→C2H4(g)ΔH=+52.3kJ calculate ΔH for the reaction of ethylene with F2: C2H4(g)+6F2(g)→2CF4(g)+4HF(g)

Answer 1

Answer:

$\Delta H$ for the given reaction is -2486.3 kJ

Explanation:

The given equation can be written as a combination of the following equation:

$2H_{2}(g)+2F_{2}(g)\rightarrow 4HF(g)$  ; $\Delta H_{1}= (2\times -537kJ)=-1074 kJ$

$2C(s)+4F_{2}(g)\rightarrow 2CF_{4}(g)$  ;  $\Delta H_{2}=(2\times -680kJ)=-1360kJ$

$C_{2}H_{4}(g)\rightarrow 2C(s)+2H_{2}(g)$  ;  $\Delta H_{3}=-52.3kJ$

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$C_{2}H_{4}(g)+6F_{2}(g)\rightarrow 2CF_{4}(g)+4HF(g)$

$\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=(-1074-1360-52.3)kJ=-2486.3kJ$