For the reaction 2O3 g - 3O2 g- H is negative- S and G are respectively-

2O_3 (g) → 3O_2 (g) nbsp; ; H = ve (given); now, S = + ve nbsp; as the moles of gas are increasing. We also know, and G= H T S from the given value ...

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Standard XIII

Chemistry

Spontaneity

For the react...

Question

For the reaction $2 O_{3} (g) \to 3 O_{2} (g), △ H$ is negative, $△ S$ and $△ G$ are respectively:

+ve, +ve

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+ve, -ve

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-ve, +ve

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-ve, - ve

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Solution

The correct option is B +ve, -ve
$2 O_{3} (g) \to 3 O_{2} (g); △ H = - v e (g i v e n); n o w, △ S = + v e$
as the moles of gas are increasing.

We also know,
and $△ G$ = $△ H$ - $T △ S$
from the given values
$△ G$ = $- v e$ - $(+ v e)$
= $- v e$
So, $△ G$ will be negative for this.

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Similar questions

Q. For the reaction

2 O_{3} (g) \to 3 O_{2} (g), △ H

is negative,

△ S

and

△ G

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Q. For the reaction

2 O_{3} (g) \to 3 O_{2} (g), △ H

is negative,

△ S

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2 O_{3} (g) \to 3 O_{2} (g)

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(fast)
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O_{3} (g) + [O] k^{'} - -------- \to 2 O_{2} (g)

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Spontaneity

Standard XIII Chemistry

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For the reaction 2O3(g)→3O2(g),△H is negative, △S and △G are respectively:

The correct option is B +ve, -ve2O3(g)→3O2(g);△H=−ve(given);now,△S=+ve as the moles of gas are increasing. We also know, and △G= △H - T△S from the given values △G =−ve - (+ve) =−ve So, △G will be negative for this.

For the reaction $2 O_{3} (g) \to 3 O_{2} (g), △ H$ is negative, $△ S$ and $△ G$ are respectively:

The correct option is B +ve, -ve
$2 O_{3} (g) \to 3 O_{2} (g); △ H = - v e (g i v e n); n o w, △ S = + v e$
as the moles of gas are increasing.

We also know,
and $△ G$ = $△ H$ - $T △ S$
from the given values
$△ G$ = $- v e$ - $(+ v e)$
= $- v e$
So, $△ G$ will be negative for this.