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Reil [10]
3 years ago
6

Calculate the free-energy change for the dissolution of 1 mole of nabr and 1 mole of nai at 298 k, given that the values of δh°s

oln are –1 and –7 kj/mol for nabr and nai. the corresponding values of δs°soln are +57 and +74 j/(mol·k).
Chemistry
1 answer:
Georgia [21]3 years ago
7 0
The free energy, which has a symbol of ΔG° at standard conditions has the formula:

ΔG° = n(ΔH° - TΔS°)

For NaBr:
ΔG° = (1 mol)[(-1 kJ/mol)(1000 J/1 kJ) - (298 K)(57 J/mol·K)]
<em>ΔG° = -17,986 J</em>

For NaI:
ΔG° = (1 mol)[(-7 kJ/mol)(1000 J/1 kJ) - (298 K)(74 J/mol·K)]
<em>ΔG° = -29,052 J</em>
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Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

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    No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{12 g}{60 g/mol}

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

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Hence, molarity = \frac{\text{no. of moles}}{volume}

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As reaction equation for the given reaction is as follows.

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HBr

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