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Orlov [11]
3 years ago
15

Using the information in the table to the right, calculate the enthalpy of combustion of 1 mole of acetylene for the reaction

Chemistry
1 answer:
barxatty [35]3 years ago
8 0

The enthalpy of combustion of 1 mole of acetylene for the reaction using the information in the table (attached as figure) is - 1,256 kJ/mol

<h3>What is Enthalpy of combustion ?</h3>

The enthalpy of combustion of a substance is defined as the heat energy given out when one mole of a substance burns completely in oxygen.

Hence, The enthalpy of combustion of 1 mole of acetylene for the reaction using the information in the table is - 1,256 kJ/mol

Learn more about Enthalpy here ;

brainly.com/question/16720480

#SPJ1

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Furkat [3]
First of all, we convert the amount of item given in moles into grams. Each mol of steam is 18 grams of steam. Such that the heat released during the evolution of per gram of steam is,
                H = (2.01 J/goC)(160 - 100) + 2260 J/g + (4.179 J/gC)(100 - 0) + (333.5 J/g) + (2.09 J/gC)(45oC)
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What does chemical reaction describe
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3 0
3 years ago
The entropy of a system at 337 K increases by 221.7 J/mol•K. The free energy value is found to be –717.5 kJ/mol. Calculate the c
OlgaM077 [116]

<u>Answer:</u> The change in enthalpy for the given system is -642.8 kJ/mol

<u>Explanation:</u>

To calculate the change in enthalpy for given Gibbs free energy, we use the equation:

\Delta G=\Delta H-T\Delta S

where,

\Delta G = Gibbs free energy = -717.5 kJ/mol = -717500 J/mol    (Conversion factor: 1 kJ = 1000 J)

\Delta H = change in enthalpy = ?

T = temperature = 337 K

\Delta S = change in entropy = 221.7 J/mol.K

Putting values in above equation, we get:

-717500J/mol=\Delta H-(337K\times 221.7J/mol.K)\\\\\Delta H=-642787J/mol=-642.8kJ/mol

Hence, the change in enthalpy for the given system is -642.8 kJ/mol

4 0
4 years ago
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