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barxatty [35]
3 years ago
7

For a given reaction, ΔH=78.7kJ/mol, and the reaction is spontaneous at temperatures above the crossover temperature, 461K. The

value of ΔS= _______ J/mol⋅K, assuming that ΔH and ΔS do not vary with temperature.
Chemistry
1 answer:
TiliK225 [7]3 years ago
7 0

Answer:

171 J/mol.K

Explanation:

The spontaneity of a reaction depends on the Gibbs free energy (ΔG). If ΔG < 0, the reaction is spontaneous. The Gibbs free energy is related to the enthalpy (ΔH) and the entropy (ΔS) of the reaction through the following expression:

ΔG = ΔH - T.ΔS

where,

T is the absolute temperature

If ΔG < 0,

ΔH - T.ΔS < 0

ΔH < T.ΔS

ΔS > ΔH/T = (78.7 × 10³ J/mol)/461 K = 171 J/mol.K

ΔS must be at least 171 J/mol.K for the reaction to be spontaneous above 461 K.

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Answer:

The heat of the reaction is 105.308 kJ/mol.

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Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

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Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

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Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ

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Answer:

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Since, atomic number of an atom is the number of electrons and number of protons for neutral atoms.

So, the number of protons = number of electrons = 15

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So, number of neutrons + number of protons = 30

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