Question

$\Delta H$ for the reaction, $C$ (graphite) $+2 H_{2(g)} \rightarrow C H_{4(g)}$ at $298\, K$ and $1\, atm$ is $-17900\, cal$. The $\Delta E$ for the above conversion would be

• A. − 17900 cal
• B. 17900 cal
• C. 17308 cal
• D. − 17308 cal

Answer 1

Answer: (D)

$C$ (graphite) $+2 H _{2}(g) \rightarrow CH _{4}(g) ; \Delta H=-17900\, cal$
$\Delta E=? \Delta H=\Delta E+\Delta n_{g} R T$
where, $\Delta n_{g}=$ number of moles of gaseous products - number of moles of gaseous reactants
$=1-2=-1$
$\therefore \Delta H=\Delta E-1 \times R \times T$
$-17900=\Delta E-1 \times 298 \times 2$
$\Delta E=-17900 +596$
$\therefore \Delta E=-17304\, cal$