Question

$\Delta G ^ \circ \,Vs\, T$ plot in the Ellingham's diagram slopes downwards for the reaction

• A. $Mg + \frac {1}{2}O_2 \rightarrow MgO$
• B. $2Ag + \frac {1}{2}O_2 \rightarrow Ag_2O$
• C. $C + \frac {1}{2} O_2 \rightarrow CO$
• D. $CO +\frac{1}{2} O_2 \rightarrow CO_2$

Answer 1

The Correct Option is C

In Ellingham diagram, higher is the slope higher is the entropy change so $\Delta S \left( C _{( s )}+1 / 2 O _{2( g )} \longrightarrow CO _{( g )}\right)>\Delta S \left( C _{( S )}+ O _{2( g )} \longrightarrow CO _{2( g )}\right.$
as CO have higher slope.

Also, Carbon monoxide is a more effective reducing agent than carbon below $983 K$ but above this temperature the reverse is true.