JEE MAIN - Chemistry (2016 - 9th April Morning Slot - No. 3)
The plot shows the variation of −$$ln$$ Kp versus temperature for the two reactions.
M(s) + $${1 \over 2}$$ O2(g) $$ \to $$ MO(s) and
C(s) + $${1 \over 2}$$ O2(g) $$ \to $$ CO(s)
_9th_April_Morning_Slot_en_3_1.png)
Identify the correct statement :
M(s) + $${1 \over 2}$$ O2(g) $$ \to $$ MO(s) and
C(s) + $${1 \over 2}$$ O2(g) $$ \to $$ CO(s)
Identify the correct statement :
At T > 1200 K, carbon will reduce MO(s) to M(s).
At T < 1200 K, the reaction
MO(s) + C(s) $$ \to $$ M(s) + CO(g) is spontaneous.
MO(s) + C(s) $$ \to $$ M(s) + CO(g) is spontaneous.
At T < 1200 K, oxidation of carbon is unfavourable
Oxidation of carbon is favourable at all temperatures.
Explanation
Carbon is a good reducing agent for oxides. The reason why carbon reduces metal oxide spontaneously at T < 1200 K is that $$ - {{\ln {K_p}} \over T}$$ line for CO has a negative slope.
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