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Asked Dec 9, 2019
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b Answered:E Enthalpy - Chemistry x
101 Chem101
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Question 21 of 29
Using the equations
Si(s) + 2 Cl2 (g) → SICI4 (s) AH° = -640.1 kJ/mol
2 Fe (s) + 3 Cl2 (g) → 2 FeCl3 (s) AH° = -800.0 kJ/mol
Determine the enthalpy for the reaction
3 SiCl: (s) + 4 Fe (s) → 4 FECI3 (s) + 3 Si (s).
kJ/mol
1
3
4
8
9.
x 100
+/-
8:49 PM
O Type here to search
12/8/2019
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b Answered:E Enthalpy - Chemistry x 101 Chem101 app.101edu.co Submit Question 21 of 29 Using the equations Si(s) + 2 Cl2 (g) → SICI4 (s) AH° = -640.1 kJ/mol 2 Fe (s) + 3 Cl2 (g) → 2 FeCl3 (s) AH° = -800.0 kJ/mol Determine the enthalpy for the reaction 3 SiCl: (s) + 4 Fe (s) → 4 FECI3 (s) + 3 Si (s). kJ/mol 1 3 4 8 9. x 100 +/- 8:49 PM O Type here to search 12/8/2019 LO

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Expert Answer

Step 1

Given information:

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Si(s)+ 2Cl, (g) –→ SiCl, (s).. .AH = -640.1kJ/mol 2Fe(s)+3Cl, (g)→ 2F€C1, (s).... .AH = -800.0 kJ/mol 3SİC1, (s) + 4Fe(s) → 4F¢C1; (s)+ 3Si(s) eq (1) eq (2) eq(3) %3D

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Step 2

In a chemical equation, if the equation is reversed then the sign for enthalpy change for that equation will also changes also if the equation is multiplied by some numerical value then the value for enthalpy change will also get multiplied.

For calculating enthalpy change for equation (3), first equation (1) is reversed and multiplied by 3 as follows:

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3SiCl, (s) → 3Si(s)+ 6Cl, (g ). . AH = (+640.1×3) kJ/mol eq (4)

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Step 3

Now multiplying equation (2) by 2 as the number of FeCl3 molecules in target equat...

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4Fe(s)+6C1, (8) → 4F¢CI, (s). . AH=(-800.0× 2) kJ/mol eq (5)

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