The plot shows the variation of − ln Kp versus temperature for the two reactions. M(s) + 1 \over 2 O_2 (g) \to MO(s) and C(s) + 1 \over 2 O_2 (g) \to CO(s) Identify the correct statement :

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

At T > 1200 K, carbon will reduce MO(s) to M(s).

At T < 1200 K, oxidation of carbon is unfavourable

Thermodynamics PYQ — JEE Main 2016 · 9 Apr · Shift 1 · Q2 with Solution

View written solutionFree

Correct answer: B

Use the relation between $K_p$ and spontaneity
For any reaction,

\Delta G^\circ = -RT \ln K_p

Given that the graph is of $-\ln K_p$ vs $T$ ,

-\ln K_p = \frac{\Delta G^\circ}{RT}

So:

If $-\ln K_p < 0$ , then $\ln K_p > 0 \Rightarrow K_p > 1$ , hence $\Delta G^\circ < 0$ and the reaction is spontaneous.
If $-\ln K_p > 0$ , then $K_p < 1$ , hence $\Delta G^\circ > 0$ and the reaction is not spontaneous.

Interpret the two given reactions

The two oxidation reactions are:

\text{(i)}\quad M(s)+\frac12 O_2(g) \to MO(s)

\text{(ii)}\quad C(s)+\frac12 O_2(g) \to CO(g)

For reduction of the oxide by carbon:

MO(s)+C(s) \to M(s)+CO(g)

This reaction is obtained by:

reversing reaction (i):

MO(s) \to M(s)+\frac12 O_2(g)

adding reaction (ii):

C(s)+\frac12 O_2(g) \to CO(g)

Hence,

\Delta G^\circ_{\text{reduction}} = \Delta G^\circ_{C\to CO} - \Delta G^\circ_{M\to MO}

So carbon will reduce $MO$ when oxidation of carbon is more favourable than oxidation of metal, i.e. when the carbon line lies below the metal line on an Ellingham-type plot.

Read the graph information
The two lines intersect at about $1200\,\text{K}$ .

For $T<1200\,\text{K}$ , the line for $C + \frac12 O_2 \to CO$ lies below that for $M + \frac12 O_2 \to MO$ .
Therefore,

\Delta G^\circ_{C\to CO} < \Delta G^\circ_{M\to MO}

which gives

\Delta G^\circ_{MO + C \to M + CO} < 0

So the reduction reaction is spontaneous for $T<1200\,\text{K}$ .

For $T>1200\,\text{K}$ , the opposite is true, so carbon cannot reduce $MO$ spontaneously.

Check each option

A: At $T>1200\,\text{K}$ , carbon will reduce $MO(s)$ to $M(s)$ .
This is false, because above the intersection temperature the reduction is not spontaneous.

B: At $T<1200\,\text{K}$ , the reaction

MO(s)+C(s) \to M(s)+CO(g)

is spontaneous.
This is true.

C: At $T<1200\,\text{K}$ , oxidation of carbon is unfavourable.
False. In fact, carbon oxidation is favourable there relative to metal oxidation.

D: Oxidation of carbon is favourable at all temperatures.
This is not the correct conclusion from the graph; favourability depends on the sign of $\Delta G^\circ$ , and the statement as given is not supported as the correct option here.

Final answer The correct statement is:

\boxed{\text{B}}

More from Thermodynamics

View written solutionFree

Correct answer: B

Use the relation between $K_p$ and spontaneity
For any reaction,

\Delta G^\circ = -RT \ln K_p

Given that the graph is of $-\ln K_p$ vs $T$ ,

-\ln K_p = \frac{\Delta G^\circ}{RT}

So:

If $-\ln K_p < 0$ , then $\ln K_p > 0 \Rightarrow K_p > 1$ , hence $\Delta G^\circ < 0$ and the reaction is spontaneous.
If $-\ln K_p > 0$ , then $K_p < 1$ , hence $\Delta G^\circ > 0$ and the reaction is not spontaneous.

Interpret the two given reactions

The two oxidation reactions are:

\text{(i)}\quad M(s)+\frac12 O_2(g) \to MO(s)

\text{(ii)}\quad C(s)+\frac12 O_2(g) \to CO(g)

For reduction of the oxide by carbon:

MO(s)+C(s) \to M(s)+CO(g)

This reaction is obtained by:

reversing reaction (i):

MO(s) \to M(s)+\frac12 O_2(g)

adding reaction (ii):

C(s)+\frac12 O_2(g) \to CO(g)

Hence,

\Delta G^\circ_{\text{reduction}} = \Delta G^\circ_{C\to CO} - \Delta G^\circ_{M\to MO}

So carbon will reduce $MO$ when oxidation of carbon is more favourable than oxidation of metal, i.e. when the carbon line lies below the metal line on an Ellingham-type plot.

Read the graph information
The two lines intersect at about $1200\,\text{K}$ .

For $T<1200\,\text{K}$ , the line for $C + \frac12 O_2 \to CO$ lies below that for $M + \frac12 O_2 \to MO$ .
Therefore,

\Delta G^\circ_{C\to CO} < \Delta G^\circ_{M\to MO}

which gives

\Delta G^\circ_{MO + C \to M + CO} < 0

So the reduction reaction is spontaneous for $T<1200\,\text{K}$ .

For $T>1200\,\text{K}$ , the opposite is true, so carbon cannot reduce $MO$ spontaneously.

Check each option

A: At $T>1200\,\text{K}$ , carbon will reduce $MO(s)$ to $M(s)$ .
This is false, because above the intersection temperature the reduction is not spontaneous.

B: At $T<1200\,\text{K}$ , the reaction

MO(s)+C(s) \to M(s)+CO(g)

is spontaneous.
This is true.

C: At $T<1200\,\text{K}$ , oxidation of carbon is unfavourable.
False. In fact, carbon oxidation is favourable there relative to metal oxidation.

Final answer The correct statement is:

\boxed{\text{B}}

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