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Q.

Given the following equations and ΔH values, determine the enthalpy of reaction at 298 K for the reaction:
C2H4(g)+6F2(g)2CF4(g)+4HF(g)
H2(g)+F2(g)2HF(g);ΔH1=537kJC(s)+2F2(g)CF4(g);ΔH2=680kJ2C(s)+2H2(g)C2H4(g);ΔH3=52kJ

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a

-1165

b

-2382

c

+1165

d

+2382

answer is B.

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Detailed Solution

Given,
H2(g)+F2(g)2HF(g);ΔH1=537kJC(s)+2F2(g)CF4(g);ΔH2=680kJ2C(s)+2H2(g)C2H4(g);ΔH3=52kJ
Change in enthalpy for the given reaction,
ΔH=537×2680×2+52=2382

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Given the following equations and ΔH∘ values, determine the enthalpy of reaction at 298 K for the reaction:C2H4(g)+6F2(g)⟶2CF4(g)+4HF(g)H2(g)+F2(g)⟶2HF(g);ΔH1∘=−537kJC(s)+2F2(g)⟶CF4(g);ΔH2∘=−680kJ2C(s)+2H2(g)⟶C2H4(g);ΔH3∘=52kJ