Question

Given the following data :
Determine at what temperature the following reaction is spontaneous ?
$FeO\left(s\right)+C\left(graphite\right)⟶Fe\left(s\right)+CO\left(g\right)$

• A. $298K$
• B. $668K$
• C. $966K$
• D. $\Delta G^o$ is +ve, hence the reaction will never be spontaneous

Answer 1

Answer: (C)

$966K$

${\Delta }_{r}H$ = $H_{Reactants}-H_{Products}$
${\Delta }_{r}H$ = -110.5 - (-266.3) = 155.8 kJ/mol
Similarly, ${\Delta }_{r}S$ = $S_{Reactants}-S_{Products}$
${\Delta }_{r}S$ = (27.28 + 197.6) - (57.49 + 5.74) = 161.65 J/mol K
Now at ${\Delta }_{r}G$= 0, ${\Delta }_{r}H$ = $T{\Delta }_{r}S$
Therefore, $T$ = ${\Delta }_{r}H$/${\Delta }_{r}S$ = 155.8 x 1000/ 161.65 = 964 K.
Thus, above 964 K the given reaction is spontaneous which implies at 966K the reaction is spontaneous (Correct Option: C)