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Work Done in Isothermal Reversible Process

Which of the ...

Question

Which of the following sets represent the normal physical states of the elements concerned at $25^{\circ} C$ and one bar pressure, with $Δ H^{\circ}$ (formation) $= 0$

C (d i a m o n d), S_{8} (s), N a (s)

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C (d i a m o n d)), C (g r a p h i t e), B r_{2} (I)

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C (g r a p h i t e), B r_{2} (g), P_{4} (w h i t e)

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C (g r a p h i t e), B r_{2} (I), I_{2} (s)

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Solution

The correct option is D $C (g r a p h i t e), B r_{2} (I), I_{2} (s)$
$Δ_{f} H^{\circ} = 0$
$C_{g r a p h i t e}$ has $Δ_{f} H^{\circ} = 0$ not diamond because it is more stable than diamond
$B r_{2}$ ; standard physical state in liquid
$I_{2} \to s o l i d$

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Similar questions

Q. The standard state Gibbs free energies of formation of

C (g r a p h i t e)

and

C (d i a m o n d)

at T = 298 K are

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}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by

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

. If

C (g r a p h i t e)

is converted to

C (d i a m o n d)

isothermally at T = 298 K, the pressure at which

C (g r a p h i t e)

is in equilibrium with

C (d i a m o n d)

is:
[Useful information:

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

]

Q. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are

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}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by

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

. If C(graphite) is converted to C(diamond) isothermally at T = 298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is
[Useful information:

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

Calculate $Δ H$ for $C_{(g r a p h i t e)} \to C_{d i a m o n d}$ from the following data:

i) $C (g r a p h i t e) + O_{2} \to C O_{2} (g), Δ H = - 94 K c a l$

ii) $C (d i a m o n d) + O_{2} \to C O_{2} (g), Δ H = - 68 K c a l$

Q. Determine the heat of transformation of

C_{(d i a m o n d)} \to C_{(g r a p h i t e)}

from the following data

C_{(d i a m o n d)} + O_{2} (g) \to C {C O}_{2} (g); (Δ H = - 94.5 k c a l)

C_{(g r a p h i t e)} + O_{2} (g) \to C {C O}_{2} (g); (Δ H = - 94.0 k c a l)

Q. The standard state Gibbs free energies of formation of

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

and

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

T = 298 K

are

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}

The standard state means that the pressure should be

1 b a r

, and substance should be pure at a given temperature. The conversion of graphite

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

to diamond

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

reduces its volume by

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

. If

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

is converted to

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

isothermally at

T = 298 K

, the pressure at which

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

is in equilibrium with

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

, is?

[Useful information:

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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Which of the following sets represent the normal physical states of the elements concerned at 25∘C and one bar pressure, with ΔH∘ (formation)=0

The correct option is D C(graphite),Br2(I),I2(s)ΔfH∘=0Cgraphite has ΔfH∘=0 not diamond because it is more stable than diamondBr2 ; standard physical state in liquidI2→solid

Which of the following sets represent the normal physical states of the elements concerned at $25^{\circ} C$ and one bar pressure, with $Δ H^{\circ}$ (formation) $= 0$

The correct option is D $C (g r a p h i t e), B r_{2} (I), I_{2} (s)$
$Δ_{f} H^{\circ} = 0$
$C_{g r a p h i t e}$ has $Δ_{f} H^{\circ} = 0$ not diamond because it is more stable than diamond
$B r_{2}$ ; standard physical state in liquid
$I_{2} \to s o l i d$