Question
Asked Nov 21, 2019
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Using the table of standard formation enthalpies, calculate the reaction enthalpy of this reaction under standard conditions:
 
8H2S+4O2-->S8+8H2O
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Expert Answer

Step 1

Using the table of standard formation enthalpies, the reaction enthalpy of this reaction under standard conditions is to be determined:

8H2S+4O2→S8+8H2O          

Step 2

The enthalpy of reaction is the change in enthalpy (total heat content) of a reaction that occurs at constant pressure.

Mathematically, the enthalpy of a reaction (ΔHrxn) is the difference between the total enthalpy of formation of the products and the total enthalpy of formation of reactants.

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-ΣΗ, (products) - ΣΗ, (re actants) . ΔΗ. xn

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Step 3

Now, for the given reaction, standard formation enthalpies are-

Hf° (H2S) = -20.1 kJ/mol

Hf° (O2) = 0 kJ/mol

Hf° (S8) = 0 kJ/mol

Hf° (H2O) = -285.8

Therefore,

ΔHrxn = [Hf° (S8) + 8(Hf°...

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ΔΗ ΣΔΗ, (products) - Σ ΔΗ, (reactants) - [Η (5,) + (H * (H,Ο)]-[s(H, (H.S)) +4(Η, * (0,)] = [08(-285.8 kJ/mol)] - [8(-20.1 kJ/mol)+4(0)] Ε Χη =-2286.4 kJ/mol 160.8 kJ/mol = -2125.6 kJ/mol

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