8 **Problem 8: Calculation of Enthalpy of Decomposition of Aluminum Chloride****Reaction:**\[ 2\text{AlCl}_3 (s) \rightarrow 2\text{Al} (s) + 3\text{Cl}_2 (g) \]Use the following four equations and their standard enthalpies of formation (\( \Delta H_f^\circ \)):a) \[ 2\text{Al} (s) + 6\text{HCl} (aq) \rightarrow 2\text{AlCl}_3 (aq) + 3\text{H}_2 (g) \] \(\Delta H_f^\circ = -1049 \, \text{kJ/mol}\)b) \[ \text{HCl} (aq) \rightarrow \text{HCl} (g) \] \(\Delta H_f^\circ = 74.8 \, \text{kJ/mol}\)c) \[ \text{H}_2 (g) + \text{Cl}_2 (g) \rightarrow 2\text{HCl} (g) \] \(\Delta H_f^\circ = -1845 \, \text{kJ/mol}\)d) \[ \text{AlCl}_3 (s) \rightarrow \text{AlCl}_3 (aq) \] \(\Delta H_f^\circ = -323 \, \text{kJ/mol}\)To solve, use Hess's Law to combine these reactions to match the desired overall decomposition reaction.

This problem is based on Hess's law.We have,(a) 2Al(s) + 6HCl → 2AlCl3(aq) + 3H2 ; ∆Hf1° = -1049 KJmol(b) HCl (aq) → HCl(g) ; ∆Hf2° = 74.8 KJmol(c) H2 (g) + Cl2(g) → 2HCl(g) ; ∆Hf3° = -1845 KJmol(d) AlCl3(S) → 2Al Cl3 (aq) ; ∆Hf4° = -323 KJmol(e) 2AlCl3(S) → 2Al(S) + 3Cl2 ; ∆Hd° Solving 2x(d) + 2(b) - 3(c) - (a), we get (e)So, ∆Hd° = 2x(-323)+2x(74.8)-3x(-18+5)-(104g)or, ∆Hd° = 6087.6 KJmol

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8. Calculate the enthalpy of decomposition of aluminum chloride in the following reaction: 2AICI; (s) 2A1 (s) + 3C12 (g) ---> Use the following four equations: a) 2A1 (s) + 6HCI (aq) 2AICI3 (aq) + 3H2 (g) | AH°;= -1049 kJ/mol b) HC1 (aq) → HC1 (g) AHf = 74.8 kJ/mol AH = -1845 kJ/mol c) H2 (g) + Cl2 (g) → 2HC1 (g) AH; = -323 kJ/mol d) AIC13 (s) → AICI3 (aq)

8. Calculate the enthalpy of decomposition of aluminum chloride in the following reaction: 2AICI; (s) 2A1 (s) + 3C12 (g) ---> Use the following four equations: a) 2A1 (s) + 6HCI (aq) 2AICI3 (aq) + 3H2 (g) | AH°;= -1049 kJ/mol b) HC1 (aq) → HC1 (g) AHf = 74.8 kJ/mol AH = -1845 kJ/mol c) H2 (g) + Cl2 (g) → 2HC1 (g) AH; = -323 kJ/mol d) AIC13 (s) → AICI3 (aq)

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Crystal Structures

Solids are present all around us. They have different properties and applications. Diamond and graphite are two well-known solids. Diamond is the hardest material in the world and is used in making jewelry. Graphite on the other hand is soft. The lead of your pencil is graphite. Why are they so different in appearance and properties despite the fact they are chemically the same? Both are allotropes of carbon. The answer lies in the structure, the way constituent particles are arranged in the solids.

Question

**Problem 8: Calculation of Enthalpy of Decomposition of Aluminum Chloride**

**Reaction:**
\[ 2\text{AlCl}_3 (s) \rightarrow 2\text{Al} (s) + 3\text{Cl}_2 (g) \]

Use the following four equations and their standard enthalpies of formation (\( \Delta H_f^\circ \)):

a) \[ 2\text{Al} (s) + 6\text{HCl} (aq) \rightarrow 2\text{AlCl}_3 (aq) + 3\text{H}_2 (g) \]
\(\Delta H_f^\circ = -1049 \, \text{kJ/mol}\)

b) \[ \text{HCl} (aq) \rightarrow \text{HCl} (g) \]
\(\Delta H_f^\circ = 74.8 \, \text{kJ/mol}\)

c) \[ \text{H}_2 (g) + \text{Cl}_2 (g) \rightarrow 2\text{HCl} (g) \]
\(\Delta H_f^\circ = -1845 \, \text{kJ/mol}\)

d) \[ \text{AlCl}_3 (s) \rightarrow \text{AlCl}_3 (aq) \]
\(\Delta H_f^\circ = -323 \, \text{kJ/mol}\)

To solve, use Hess's Law to combine these reactions to match the desired overall decomposition reaction.

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