Identify the correct statement(s)

Low pressure is favourable for evaporation of

H_{2} O (l) .

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On adding graphite to an equilibria involving

C (s), O 2 (g)

and

C O_{2} (g)

C O_{2} (g)

will be formed.

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For the reaction:

3 F e (s) + 4 H_{2} O (g) ⇌ F e_{3} O_{4} (s) + 4 H_{2} (g)

K_{P}

and

K_{C}

will be same.

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Normal boiling point of water will be more than standard boiling point of water.

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Solution

The correct options are
A Low pressure is favourable for evaporation of $H_{2} O (l) .$
C For the reaction: $3 F e (s) + 4 H_{2} O (g) ⇌ F e_{3} O_{4} (s) + 4 H_{2} (g)$ $K_{P}$ and $K_{C}$ will be same.
D Normal boiling point of water will be more than standard boiling point of water.
Low pressure will favour the vapourisation.
No effect of adding C (graphite) as it is in solid state.
In option C, $K_{P} = K_{C} \times [R T]^{Δ n_{g}}$
$K_{P} = K_{C}$ as $△ n_{g} = 0$
$A (l) ⇌ A (g)$
On increasing pressure, Boiling point increases. So normal boiling point (boiling point at 1 atm) > Standard boiling point (Boiling point at 1 bar).

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