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harina [27]
3 years ago
9

Why was the term inert gases dropped? Why was the term inert gases dropped? The term "inert" was dropped because hydrogen was ad

dviced to be included to "inert" gases. The term "inert" was dropped because it no longer described all the group 8A elements. The term "inert" was dropped because some organic gases such as methane were named "inert", too, and confusion happened. The term "inert" was dropped because author's rights for the term ended and scientists obtained possibility to change it.
Chemistry
1 answer:
xxMikexx [17]3 years ago
8 0

Answer:

B) The term "inert" was dropped because it no longer described all the group 8A elements.

Explanation:

Inert elements in chemistry simply refers to elements that are chemically inactive and are not expected to form any compounds. this is the general belief for the group 8 elements as they all have complete duplet/octet configurations (and ideally, they ought to be very stable with no tendency to form compounds by participating in the loss and gain of electrons). However the discovery of compounds like xenon tetrafluoride (XeF4) proved this to be wrong.

Again, the reason the term - inert gses was droppedis beacause this term is not strictly accurate because several of them do take part in chemical reactions.

After dropping the term - Inert gases, they are now referred to as noble gases.

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\Delta _rH=-1124.14kJ/mol

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Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}

In such a way, by using the NIST database, we find that:

\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol

Thus, we plug in the enthalpies of formation to obtain:

\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol

Best regards!

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