The plot shows the variation of - ln K _ P Versus temperature for the two reactions. M(s) + (1)/(2) O _ 2 (g) → MO ( s ) and C(s) + (1)/(2) O _ 2 (g) → CO ( s ) Identify the correct statement:

Correct answer: C. At T < 1200K , the reaction MO(s) + C(s) → M(s) + CO ( g ) is spontaneous = Δ G ^ 0 + RTlnQ At Equilibrium = 0 and Q = K ∆ G ^ o = - RTln Kp ∴- ln K _ p = ∆ G ^ o RT ∆ G ^ o ∝ Temperature. At high temperature ∆ G ^ o value is very high so non-feasible or non-spontaneous process takes place.

JEE Main 20169 AprilOnline

Exam: JEE Main
Year: 2016
Date: 9 April
Mode: Online
Subject: Chemistry
Chapter: General Principles and Processes of Isolation of Metals

JEE Main 2016 · 9 April · Online · Chemistry · General Principles and Processes of Isolation of Metals

The plot shows the variation of - ln K _ P Versus temperature for...

The plot shows the variation of

- ln {K}_{P}

Versus temperature for the two reactions.

M\left(s\right) + \dfrac{1}{2}{O}_{2}\left(g\right) \to MO ( s )

and

C\left(s\right) + \dfrac{1}{2}{O}_{2}\left(g\right) \to CO ( s )

The plot shows the variation of - ln K _ P Versus temperature for the two reactions. M(s) + (1)/(2) O _ 2 (g) → MO ( s

Identify the correct statement:

Options

AAt $T < 1200K$ , oxidation of carbon is unfavourable
BOxidation of carbon is favourable at all temperatures
CAt $T < 1200K$ , the reaction $MO\left(s\right) + C\left(s\right) \to M\left(s\right) + CO ( g )$ is spontaneous
DAt $T > 1200K$ , carbon will reduce ${MO}_{\left(s\right)}to{M}_{\left(s\right)}$ .

Correct answer

C. At $T < 1200K$ , the reaction $MO\left(s\right) + C\left(s\right) \to M\left(s\right) + CO ( g )$ is spontaneous

Step-by-step solution

\DeltaG = \Delta{G}^{0} + RTlnQ

At Equilibrium

\DeltaG = 0

\text{ and } Q = K

∆ {G}^{o} = - RTln Kp

∴- ln {K}_{p} = \dfrac{∆ {G}^{o}}{RT}

∆ {G}^{o} ∝

Temperature. At high temperature

∆ {G}^{o}

value is very high so non-feasible or non-spontaneous process takes place.

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