Thermodynamics
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Question : 21 of 40
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The standard state Gibbs free energies of formation of C(graphite) and C(diagram) as T = 298 K are :
Δ f G ° [ C ( g r a p h i t e ) ] = 0 k J m o l − 1
Δ f G ° [ 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 by2 × 10 − 6 m 3 m o l − 1
If C(graphite) is converted to C(diamond) isothermally at T = 298 K, 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 ]
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
If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is: [Useful information:
[JEE Adv 2017 P2]
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