13.) For the RAM : -> %3D AHp f AgBr= -100.4kJ △Hp g Brg aー+30,9J + 30.9kJ Frd fhe he at f Ranacion for AzIGS) g formatian for AgICS) -54.0=(-100,4) -(5• 30.9) - X 61.85=-X 1 - 61.858y=X
13.) For the RAM : -> %3D AHp f AgBr= -100.4kJ △Hp g Brg aー+30,9J + 30.9kJ Frd fhe he at f Ranacion for AzIGS) g formatian for AgICS) -54.0=(-100,4) -(5• 30.9) - X 61.85=-X 1 - 61.858y=X
Chemistry
10th Edition
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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See image below. Can you please show me how to solve this (i.e. why we don't add the 100.4, etc.)?
![**Title: Calculating the Heat of Formation for Silver Iodide (AgI)**
**Introduction:**
The following are calculations related to the reaction involving silver iodide (AgI) and its heat of formation. This reaction is analyzed based on the given thermodynamic data.
**Reaction Studied:**
\[
\text{AgI(s)} + \frac{1}{2} \text{Br}_2(\text{g}) \rightarrow \text{AgBr(s)} + \frac{1}{2} \text{I}_2(\text{s})
\]
**Data Provided:**
- \( \Delta H_f \) of AgBr = \( -100.4 \, \text{kJ} \)
- \( \Delta H_f \) of \(\frac{1}{2} \text{Br}_2 (\text{g}) \) = \( + 30.9 \, \text{kJ} \)
**Calculation Steps:**
1. Establish the reaction enthalpy equation:
\[
\Delta H_{\text{reaction}} = \Delta H_f (\text{AgBr}) + \Delta H_f \left(\frac{1}{2} \text{Br}_2(\text{g})\right)
\]
\[
-54.0 = (-100.4) + 30.9
\]
2. Calculate the heat of formation for AgI(s):
\[
\Delta H_f (\text{AgI}) = -54.0 = (-100.4) + 30.9 - x
\]
3. Simplify the equation to find \( x\):
\[
61.85 = -x
\]
\[
x = -61.85 \, \text{kJ/mol}
\]
**Conclusion:**
The heat of formation (\( \Delta H_f \)) for silver iodide (\(\text{AgI}\)) is calculated to be \(-61.85 \, \text{kJ/mol}\). This negative sign indicates an exothermic formation process.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F88f794be-122d-44d8-bdbd-e59efaff39a0%2F290d1fa9-5467-4077-a4e3-753d443be56c%2Fsbuthmd.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating the Heat of Formation for Silver Iodide (AgI)**
**Introduction:**
The following are calculations related to the reaction involving silver iodide (AgI) and its heat of formation. This reaction is analyzed based on the given thermodynamic data.
**Reaction Studied:**
\[
\text{AgI(s)} + \frac{1}{2} \text{Br}_2(\text{g}) \rightarrow \text{AgBr(s)} + \frac{1}{2} \text{I}_2(\text{s})
\]
**Data Provided:**
- \( \Delta H_f \) of AgBr = \( -100.4 \, \text{kJ} \)
- \( \Delta H_f \) of \(\frac{1}{2} \text{Br}_2 (\text{g}) \) = \( + 30.9 \, \text{kJ} \)
**Calculation Steps:**
1. Establish the reaction enthalpy equation:
\[
\Delta H_{\text{reaction}} = \Delta H_f (\text{AgBr}) + \Delta H_f \left(\frac{1}{2} \text{Br}_2(\text{g})\right)
\]
\[
-54.0 = (-100.4) + 30.9
\]
2. Calculate the heat of formation for AgI(s):
\[
\Delta H_f (\text{AgI}) = -54.0 = (-100.4) + 30.9 - x
\]
3. Simplify the equation to find \( x\):
\[
61.85 = -x
\]
\[
x = -61.85 \, \text{kJ/mol}
\]
**Conclusion:**
The heat of formation (\( \Delta H_f \)) for silver iodide (\(\text{AgI}\)) is calculated to be \(-61.85 \, \text{kJ/mol}\). This negative sign indicates an exothermic formation process.
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