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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Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
