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 $\Delta H_{-}$ = -0.31 kJ. The value of $\Delta E$ will be

• A. - 0.3024 kJ
• B. 0.3024 kJ
• C. 2.567 kJ
• D. -0.0076 kJ

Answer 1

The correct option is A

- 0.3024 kJ

$C_2{H}_4\left(g\right)+H_2\left(g\right)\to C_2{H}_6\left(g\right)\Delta ng=1-2=-1;\Delta H=-0.31KJmo{l}^{-1}p=1.5atm,\Delta V=-50mL=-0.050L\Delta H=\Delta E+p\Delta V-0.31=\Delta E-0.0076;\Delta E=-0.3024KJ$