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nlexa [21]
3 years ago
15

Standard Heat of Formation: The enthalpy change for the formation of 1 mol of a substance in its standard state from its constit

uent elements in their standard states. True/ false?
Chemistry
1 answer:
12345 [234]3 years ago
6 0

The given statement about standard heat of formation is true.

Explanation:

Yes, it is true that standard heat of formation is same as standard enthalpy of formation. As enthalpy is termed for change in the heat energy occurred in the process of completion of any chemical reaction, the standard enthalpy or heat of formation is termed as the heat energy required to form 1 mole of any substance in its standard state from its original, parent or constituent elements in their standard state. Basically, it defines the amount of energy required by the system or compound for its formation from different elements under different synthesis conditions but the nature of the parent samples and temperature and pressure should be kept standard.

So the given statement is true.

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Describe how the molecular structures of alkenes and alkynes differ from the structure of alkanes
evablogger [386]

Alkanes are saturated hydrocarbon that contains only single bonds, whereas Alkenes and Alkynes are unsaturated hydrocarbons which contain one or more double bond and triple bonds.

<u>Explanation:</u>

  • A saturated hydrocarbon with an only single bond is called alkanes. Ethane consisting of two carbon atoms that are bonded with a single bond and six hydrogen atoms sharing the other valence electron of carbon atoms. The molecular structure of alkane is CnH2n+2.
  • An unsaturated hydrocarbon with a two bond is called alkenes. Ethene consisting of two carbon atoms double-bonded to each other. The molecular structure of alkene is CnH2n.
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4 0
3 years ago
A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K
Ilya [14]

Answer:

-0.050 kJ/mol.K

Explanation:

  • A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
  • The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

8 0
3 years ago
Read 2 more answers
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