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. 0.235 kJ
• B. 0.3024 kJ
• C. 2.567 kJ
• D. 0.0076 kJ

Answer 1

Answer: (A)

Use the following formula to find the value of $\Delta E$.
$\Delta H-\Delta E+P \Delta V$
where $\Delta H=-0.31 \,kJ\,mol ^{-1}$
$P=1.5\, atm$
$\Delta V=-50\, mL =-0.050\, L$
$\Delta E=?$
$\Delta H=\Delta E+P \Delta V$
or $-0.31=\Delta E+(1.5 \times-0.05)$
or $-0.31=\Delta E-0.075$
or $-0.31+0.075=\Delta E$
$\Delta E=0.235$