JEE MAIN - Chemistry (2021 - 27th August Evening Shift - No. 22)
Data given for the following reaction is as follows :
FeO(s) + C(graphite) $$\to$$ Fe(s) + CO(g)
The minimum temperature in K at which the reaction becomes spontaneous is ___________. (Integer answer)
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
964
Explanation
$${T_{\min }} = \left( {{{{\Delta ^0}H} \over {{\Delta ^0}S}}} \right)$$
$${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] - $$
$$ = \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$$
$$ = [0 - 110.5] - [ - 266.3 + 0]$$
= 155.8 kJ/mol
$${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] - $$
$$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$$
$$ = [27.28 + 197.6] - [57.49 + 5.74]$$
= 161.65 J/mol-K
$${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$$ K
$$ \simeq 964$$ k (nearest integer)
$${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] - $$
$$ = \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$$
$$ = [0 - 110.5] - [ - 266.3 + 0]$$
= 155.8 kJ/mol
$${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] - $$
$$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$$
$$ = [27.28 + 197.6] - [57.49 + 5.74]$$
= 161.65 J/mol-K
$${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$$ K
$$ \simeq 964$$ k (nearest integer)
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