Question

$C _{6} H _{12}(l)+9 O _{2}( g ) \rightarrow 6 H _{2} O (l)+6 CO _{2}( g ) ; \Delta H _{298}=-936.9\, kcal$. Thus

• A. $-936.9=\Delta E -\left(2 \times 10^{-3} \times 298 \times 3\right) kcal$
• B. $+936.9=\Delta E +\left(2 \times 10^{-3} \times 298 \times 3\right) kcal$
• C. $-936.9=\Delta E -\left(2 \times 10^{-3} \times 298 \times 2\right) kcal$
• D. $-936.9=\Delta E +\left(2 \times 10^{-3} \times 298 \times 2\right) kcal$

Answer 1

Answer: (A)

$\Delta n_{g}=6-9=-3$
$\therefore \Delta H=\Delta U+\Delta n_{g} R T$
or, $\Delta U=\Delta H-\Delta n_{g} R T$
$=-936.9\, kCal =(-3 R \times 298 K )$
or, $-936.9 \,kcal =\Delta U+\left(-3 \times 2 \,cal \,k^{-1} m o l^{-1} \times 298 \,K \right) $
$=\Delta U-\left(3 \times 2 \times 298 \times 10^{-3}\right) kcal$
$\Delta U =\Delta E =$ change in internal energy