question_answer

In the Ellingham diagram plotted between \[\Delta \,G{}^\circ \] versus T for a reaction, negative slope will be for reaction

A)  \[{{C}_{(s)}}+{{O}_{2(g)}}\to C{{O}_{2(g)}}\]

B)  \[2M{{g}_{(s)}}+{{O}_{2(g)}}\to 2MgO\]

C)  \[2{{C}_{(s)}}+{{O}_{2(g)}}\to 2C{{O}_{(g)}}\]

D)  \[2C{{O}_{(g)}}+{{O}_{2(g)}}\to 2C{{O}_{2}}_{(g)}\]