Question

What will be the heat change at constant volume for the reaction whose heat change at constant pressure is $ -560 \,kcal $ at $ 27\,{}^{\circ}C $ ? The reaction is
$ C_{8}H_{16}+12O_{2} \to8CO_{2}+8H_{2}O $
(Given $ R = 2\, cal\,mol^{-1}\,K^{-1} $ )

• A. $ -558200 $ calories
• B. $ 442800 $ calories
• C. $ -561800 $ calories
• D. $ 368240 $ calories

Answer 1

Answer: (A), (B), (C), (D)

$\Delta H =-560\, kcal$, $\Delta \, E=?$
$C_{8}H_{16\left(l\right)}+12O_{2\left(g\right)} \rightarrow 8CO_{2\left(g\right)}+8H_{2}O_{\left(l\right)}$
$\Delta H=\Delta E+\Delta n_{g} RT$
$\Delta E=\Delta H-\Delta n_{g} RT$
$=-560,000-\left(-4\times2\times300\right)$
$=-560,000+2400$
$=-5,57,600 \, cal$