Data given for the following reaction is as follows :
FeO(s) + C(graphite) $\to$ Fe(s) + CO(g)
| Substance | $\Delta H^\circ$ (kJ mol$^{ - 1}$) |
$\Delta S^\circ$ (J mol$^{ - 1}$ K$^{ - 1}$) |
|---|---|---|
| $Fe{O_{(s)}}$ | $- 266.3$ | 57.49 |
| ${C_{(graphite)}}$ | 0 | 5.74 |
| $F{e_{(s)}}$ | 0 | 27.28 |
| $C{O_{(g)}}$ | $- 110.5$ | 197.6 |
The minimum temperature in K at which the reaction becomes spontaneous is ___________. (Integer answer)
Answer (integer)
964
Solution
${T_{\min }} = \left( {{{{\Delta ^0}H} \over {{\Delta ^0}S}}} \right)$<br><br>${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] -$<br><br>$= \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$<br><br>$= [0 - 110.5] - [ - 266.3 + 0]$<br><br>= 155.8 kJ/mol<br><br>${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] -$<br><br>$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$<br><br>$= [27.28 + 197.6] - [57.49 + 5.74]$<br><br>= 161.65 J/mol-K<br><br>${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$ K<br><br>$\simeq 964$ k (nearest integer)
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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