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 $\Delta H = -0.31 \ kJ$. The value of $\Delta E$ will be (in $kJ$)

  • A
    $-0.3024$
  • B
    $0.3024$
  • C
    $2.567$
  • D
    $-0.0076$

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