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cupoosta [38]
3 years ago
5

Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2

07.0 91.3 33.2 -285.8 S°(J/mol∙K) 146.0 210.8 240.1 70.0 Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -207.0 91.3 33.2 -285.8 S°(J/mol∙K) 146.0 210.8 240.1 70.0 -151 kJ -85.5 kJ +50.8 kJ +222 kJ +101.6 kJ
Chemistry
2 answers:
siniylev [52]3 years ago
8 0

Answer:

\Delta G\°_{rxn}=50.8\frac{kJ}{mol}

Explanation:

Hello,

In this case, given the reaction:

2 HNO_3(aq) + NO(g) \rightarrow  3 NO_2(g) + H_2O(l)

We compute the ΔH°rxn by using the given enthalpies of formation:

\Delta H\°_{rxn}=3*\Delta _fH_{NO_2}+\Delta _fH_{H_2O}-2\Delta _fH_{HNO_3}-\Delta _fH_{NO}\\\\\Delta H\°_{rxn}=3*33.2+(-285.8)-2*(-207.0)-91.3=136.5kJ/mol

Similarly, we compute ΔS°rxn:

\Delta S\°_{rxn}=3*S_{NO_2}+S_{H_2O}-2S_{HNO_3}-S_{NO}\\\\\Delta S\°_{rxn}=3*240.1+70.0-2*146.0-210.8 =287.5J/mol*K

Finally we compute ΔG°rxn:

\Delta G\°_{rxn}=\Delta H\°_{rxn}-T\Delta S\°_{rxn}=136.5\frac{kJ}{mol}-298K*287.5\frac{J}{mol*K}*\frac{1kJ}{1000J}\\   \\\Delta G\°_{rxn}=50.8\frac{kJ}{mol}

Best regards.

Mkey [24]3 years ago
7 0

Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

<em>ΔG°rxn = +50.8 kJ/mol</em>

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<u>For propanol: </u>Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

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Mass of n-propanol = 1.00 g

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1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

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Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)</u>

<u></u>

<u>For ethanol:</u>

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1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

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<u></u>

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1 g of methanol on combustion gives 33.4 kJ of energy

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