Which is better reducing agent at 983 K, C or CO ?

Explanation

When carbon reacts with dioxygen tow reactions are possible:
`C(s) + O_(2)(g) to CO_(2) (g)`
`2C(s) + O_(2)(g) to 2CO(g)`
The CO thus produced can also react with `O_(2)` to form `CO_(2)` .
`2 CO(g) + O_(2) (g) to 2CO_(2)(g)` In the first reaction, the volume of `CO_(2)` produced is the same as that of `O_(2)` pordoued is the same as that of `O_(2)` used, therefore, `DeltaS` is very small and `Delta_(f)G^(@)` does not changes with tempreature. Thus, the graph of `Delta_(f)G^(@)` against T is almost horizontal .(Fig . 6.19)

The second reaction preduces two volumes of CO for every one volume of ` O _ 2 ` used. Thus, ` Delta S ` is positive and hence ` Delta _f G ^(@) ` becomes increasingly negative as the temperature increases. Consequenty, C, CO line on the Ellingham diagram slopes downwards . The third reaction produces two volumes of ` CO _ 2 `for every three volumes of reactants used. Thus, ` Delta S ` is - ve and hence ` Delta _ f G ^(@) ` becomes increasingly + ve as the temperature increases. Consequentrly, CO, `CO_ 2 ` line on the Ellingham diagram slopes upwards. The three lines cross at 983 K . Below this temperature, formation of ` CO _ 2 ` from C as well as from CO is energetically more favourable. In other words, below 983 K , both C and Co can act as reducing agents but since CO can be more easily oxidised to ` CO _ 2 ` than C to ` CO _2 `, therefore , below 983 K, CO is a more effective reducing agent than C. However, above 983 K, CO is more stable and hence its oxidation to ` CO _ 2` is less rapid than that of C to ` CO _ 2 `. Therefore, above 983 K, C is a better reducing agent than CO.