From given following equations and `DeltaH^(@)` values, determine the enthalpy of reaction at 298 K for the reaction :
`C_(2)H_(4)(g)+6F_(2)(g)rarr2CF_(4)(g)+4HF(g)`
`H_(2)(g)+F_(2)(g)rarr2HF(g), " "DeltaH_(1)^(@)=-537 kJ`
`C(s)+2F_(2)(g)rarrCF_(4)(g)," "DeltaH_(2)^(@)=-680 kJ`
`2C(s)+2H_(2)(g)rarrC_(2)H_(4)(g)," "DeltaH_(3)^(@)=52 kJ`

Options

A. `-1165 kJ`
B. `+1165 kJ`
C. `-2486 kJ`
D. `+2486 kJ`

Answer

C

Explanation

`{:(H_(2)(g) + F_(2)(g) rarr 2HF(g),DeltaH_(1)^(0) = -537 kJ),(C(s) + 2F_(2)(g) ro CF_(4)(g),DeltaH_(2)^(0) = -680 kJ),(2C(s) + 2H_(2)(g) rarr C_(2)H_(4)(g), DeltaH_(3)^(0) = 52 kJ):}/(C_(2)H_(4)(g) + 6F_(2)(g) rarr2CF_(4)(g) +4HF(g)DeltaH = {-DeltaH_(3)^(0) +2DeltaH_(2)^(0) + 2DeltaH_(1)^(0)})`
`DeltaH^(@) = -52 + 2 xx -680 + 2 xx -537 = -2486 kJ`.