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Heat of Reaction

The value of ...

Question

The value of ${Δ H}_{t r a n s i t i o n}$ of $C (g r a p h i t e) ⟶ C (d i a m o n d)$ is $1.9 k J / m o l$ at $25^{o} C$ and entropy of graphite is higher than entropy of diamond. This implies that:

C (d i a m o n d)

is more thermodynamically stable than

C (g r a p h i t e)

25^{o} C

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C (g r a p h i t e)

is more thermodynamically stable than

C (d i a m o n d)

25^{o} C

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diamond will provide more heat on complete combustion at

25^{o} C

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{Δ G}_{t r a n s i t i o n}

C (d i a m o n d) ⟶ C (g r a p h i t e)

- v e

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Solution

The correct options are
B $C (g r a p h i t e)$ is more thermodynamically stable than $C (d i a m o n d)$ at $25^{o} C$
C diamond will provide more heat on complete combustion at $25^{o} C$
D ${Δ G}_{t r a n s i t i o n}$ of $C (d i a m o n d) ⟶ C (g r a p h i t e)$ is $- v e$
${Δ H}_{t r a n s i t i o n}$ of $C (g r a p h i t e) ⟶ C (d i a m o n d)$ is $1.9 k J / m o l$ at $25^{o} C$ so its an endothermic process.
Internal energy of diamond is more than graphite so it has higher combustion energy.
Also, as internal energy of graphite is less than diamond so $C (g r a p h i t e)$ is more thermodynamically stable than $C (d i a m o n d)$ at $25^{o} C$ .
${Δ G}_{t r a n s i t i o n}$ of $C (d i a m o n d) ⟶ C (g r a p h i t e)$ is $- v e$ as for this process $Δ H$ is negative and $Δ S$ is positive so $Δ G = Δ H - T Δ S$ will always negative.

Suggest Corrections

Similar questions

Δ H_{t r a n s i t i o n}

C (g r a p h i t e) ⟶ C (d i a m o n d)

1.9 k J / m o l

25^{\circ} C

Assertion- $C (g r a p h i t e) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 393.5 k . J$

$C (d i a m o n d) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 395.4 k . J$

Reason(R): graphite is more stable than diamond, so heat of combustion is less.

C_{d i a m o n d} \to C_{g r a p h i t e}; Δ H = - v e .

C (d i a m o n d) ⇌ C (g r a p h i t e)

Δ_{r} = - 1.9 k J / m o l e

3.5

2.3 g / m L

C_{(g r a p h i t e)}

C_{(d i a m o n d)}

T = 298 K

Δ_{f} G^{\circ} C_{(g r a p h i t e)} = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} C_{(d i a m o n d)} = 2.9 k J m o l^{- 1}

1 b a r

C_{(g r a p h i t e)}

C_{(d i a m o n d)}

2 \times 10^{- 6} m^{3} m o l^{- 1}

C_{(g r a p h i t e)}

C_{(d i a m o n d)}

T = 298 K

C_{(g r a p h i t e)}

C_{(d i a m o n d)}

1 J = 1 k g m^{2} s^{- 2}; 1 P a = 1 k g m^{- 1} s^{- 2}; 1 b a r = 10^{5} P a]

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