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Difference between Internal Energy and Enthalpy

The enthalpy ...

Question

The enthalpy change for the reaction of 50.00 ml of ethylene with 50.00 ml of $H_{2}$ at 1.5 atm pressure is $Δ H_{-}$ = -0.31 kJ. The value of $Δ E$ will be

-0.3024 kJ

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0.3024 kJ

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-2.567 kJ

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-0.0076 kJ

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Solution

The correct option is A
-0.3024 kJ

$C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g) Δ n g = 1 - 2 = - 1; Δ H = - 0.31 K J m o l^{- 1} p = 1.5 a t m, Δ V = - 50 m L = - 0.050 L Δ H = Δ E + p Δ V - 0.31 = Δ E - 0.0076; Δ E = - 0.3024 K J$

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