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Entropy Change in Isothermal Process

The standard ...

Question

The standard entropies of $C O_{2} (g), C (s) a n d O_{2} (g)$ are $213.50, 5.74$ and $205 J / K$ respectively. What will be the standard entropy of formation of $C O_{2}$ :

A

- 2.76 J / K

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B

5.76 J / K

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C

2.76 J / K

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D

- 5.76 J / K

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Solution

The correct option is C $2.76 J / K$
The reaction taking place is
$C (s) + O_{2} (g) \to C O_{2} (g)$

We know,
$Δ S = \sum S_{p r o d u c t s} - \sum S_{r e a c t a n t s}$

here in this case of formation of $C O_{2}$ ,
putting the values,
$Δ S = [213.50] - [5.74 + 205]$
$= 2.76 J / K$

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Similar questions

Q. The standard entropies of

C O_{2} (g), C (s) a n d O_{2} (g)

213.50, 5.74

205 J / K

respectively. The standard entropy of formation of

C O_{2}

Q. The standard entropies of

C O_{2} (g), C (s) a n d O_{2} (g)

213.50, 5.74

205 J / K

respectively. The standard entropy of formation of

C O_{2}

Q. Show that the reaction,

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g)

300 K

is spontaneous and exothermic, when the standard entropy change is

- 0.094 k J {m o l}^{- 1} K^{- 1}

. The standard Gibbs free energies of formation for

{C O}_{2}

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Q. Show that the reaction

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g)

300 K

is spontaneous and exothermic, when the standard entropy change is

- 0.094 k J m o l^{- 1} K^{- 1}

. The values of standard Gibbs free energy of formation of

C O_{2}

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- 137.2 k J m o l^{- 1}

, respectively.

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H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) + 68.3 k c a l

{C H}_{4} + 2 O_{2} (g) \to {C O}_{2} (g) + 2 H_{2} O (l) + 210.8 k c a l

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