Question

Визначте, які з наведених термохімічних рівнянь відповідають екзо- термічним процесам, ендотермічним? a) 4Р(тв.) + 502(г.) = 2P2O (тв.); ДН = -2984 кДж/моль;б) 2Al(тв.) + Fe,0,(тв.) = Al,О,(тв.) + 2Fe(тв.); ДН = -850 кДж/моль; B) 5(тв.) + O2(г.) = SO,(г.); ДН-297 кДж/моль; г) 2HgO(тв.) = 2Hg(p.) + O2(г.); ДН = +362 кДж д) H (г.) + Cl (г.) = 2НСІ(г.); ∆H = -185,6 кДж/моль;e) CH12O6(TB) + 60,(r.) = 6CO2(г.) + 6H2O(р.); ДН = -2800 кДж/моль. 25 Б.​

Answer 1

Ответ:
a) Екзотермічний процес, оскільки ∆Н від'ємний.

б) Екзотермічний процес, оскільки ∆Н від'ємний.

B) Екзотермічний процес, оскільки ∆Н від'ємний.

г) Ендотермічний процес, оскільки ∆Н додатний.

д) Екзотермічний процес, оскільки ∆Н від'ємний.

е) Ендотермічний процес, оскільки ∆Н додатний.

Answer 2

Exothermic and Endothermic Reactions

To determine which of the given thermochemical equations correspond to exothermic or endothermic processes, we need to understand the concepts of exothermic and endothermic reactions.

Exothermic reactions release energy in the form of heat to the surroundings. The enthalpy change (∆H) for an exothermic reaction is negative, indicating that the reaction releases heat.

Endothermic reactions absorb energy from the surroundings in the form of heat. The enthalpy change (∆H) for an endothermic reaction is positive, indicating that the reaction absorbs heat.

Now let's analyze each of the given thermochemical equations to determine if they correspond to exothermic or endothermic processes.

a) 4P(s) + 5O2(g) → 2P2O5(s); ∆H = -2984 kJ/mol This equation represents the formation of P2O5 from P and O2. The negative value of ∆H (-2984 kJ/mol) indicates that the reaction is exothermic, as it releases heat to the surroundings. [[1]]

b) 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s); ∆H = -850 kJ/mol This equation represents the reaction between Al and Fe2O3 to form Al2O3 and Fe. The negative value of ∆H (-850 kJ/mol) indicates that the reaction is exothermic, as it releases heat to the surroundings. [[2]]

c) 5S(s) + O2(g) → SO3(g); ∆H = -297 kJ/mol This equation represents the formation of SO3 from S and O2. The negative value of ∆H (-297 kJ/mol) indicates that the reaction is exothermic, as it releases heat to the surroundings. [[3]]

d) 2HgO(s) → 2Hg(l) + O2(g); ∆H = +362 kJ This equation represents the decomposition of HgO into Hg and O2. The positive value of ∆H (+362 kJ) indicates that the reaction is endothermic, as it absorbs heat from the surroundings. [[4]]

e) C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l); ∆H = -2800 kJ/mol This equation represents the combustion of glucose (C6H12O6) to form CO2 and H2O. The negative value of ∆H (-2800 kJ/mol) indicates that the reaction is exothermic, as it releases heat to the surroundings. [[5]]

In summary: - The equations a), b), and c) correspond to exothermic processes. - The equations d) and e) correspond to endothermic processes.

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