Question

Heat of combustion $C\left(s\right),H_{2}\left(g\right)$ and $CH_4\left(g\right)$ are $-94 , -68$ and $-213\, kcal/mol$. Then $\Delta H^{\circ}$ for $C\left(s\right)+2H_{2}\left(g\right)\to\Delta CH_{4}\left(g\right)$is

• A. $-17\, kcal$
• B. $-111\, kcal$
• C. $-170\, kcal$
• D. $-85\, kcal$

Answer 1

Answer: (A)

For reaction,
$C\left(s\right)+2H_{2}\left(g\right)\to CH_{4}\left(g\right), \Delta H^{\circ}=?$
$\Delta H^{\circ}=-\left[\left(\Delta H^{\circ}_{c}ofCH_{4}\right)-\left(\Delta H^{\circ}_{c}of C+2\times\Delta H^{\circ}_{c}of H_{2}\right)\right]$
$C+O_{2}\to CO_{2};\Delta H=-94 kcal ...\left(i\right)$
$2H_2+O_{2}\to 2H_2O;\Delta H=-68\times2 kcal ...\left(ii\right)$
$CH_4+2O_{2}\to CO_2+2H_2O;\Delta H=-213 \,kcal ...\left(ii\right)$
$Eq \left(i\right)+\left(ii\right)-Eq\left(iii\right)$
$=-\left[\left(-213\right)-\left(-94+2\times-68\right)\right]kcal/mol$
$=-\left[-213+230\right]=-17\, kcal/mol$