Question

Calculate, the $\Delta G^o$ at $85{}^{o}C$ for the reaction:
$F{e}_2{O}_3\left(s\right)+3H_2\left(g\right)\to 2Fe\left(s\right)+3H_{2}O$
The data are: $\Delta H_{298}^o=-35kJ/mol$
$\Delta S_{358}^o=0.095kJmo{l}^{-1}{K}^{-1}$
$SubstanceF{e}_2{O}_3\left(s\right)Fe\left(s\right)H_{2}O\left(l\right)H_2\left(g\right)$
$C_p^o\left(J{K}^{-1}mol-1\right)103257529$

• A. $-4.11kJmo{l}^{-1}$
• B. $-63.91kJmo{l}^{-1}$
• C. $-63.91kJmo{l}^{-1}$
• D. $4.11kJmo{l}^{-1}$

Answer 1

The correct option is B $-63.91kJmo{l}^{-1}$

Given,
$F{e}_2{O}_3\left(s\right)+3H_2\left(g\right)\to 2Fe\left(s\right)+3H_{2}O$
We know, $\Delta C_p^o=(C_p{)}_P-(C_p{)}_R$
$\Rightarrow \Delta C_p=2\times 25+3\times 75-103-3\times 29=85J/molK$

Again, $\frac{\Delta H_{T_2}-\Delta H_{T_1}}{T_2-T_1}=\Delta C_p$
$\Rightarrow \frac{\Delta H_{358}-(-35)}{358-298}=85\times {10}^{-3}$
$\Rightarrow \Delta H_{358}^o=-29.9kJmo{l}^{-1}$

We know, $\Delta G^o=\Delta H^o-T\Delta S^o$
$=-29.9-358\times 0.095$$=-63.91kJmo{l}^{-1}$