Consider the following equations: 3A + 6B → 3D ΔΗ -404 kJ E + 2F → A AH = -114.7 kJ C → E+ 3D AH = 68.2 kJ Suppose the first equation is reversed and multiplied by , the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction? (If a box is not needed, leave it blank.) Net reaction: + + What is the overall heat of this reaction? Overall heat = kJ

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Chapter1: Chemical Foundations
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**Chemical Reaction and Enthalpy Change Analysis**

Consider the following equations:

1. \(3A + 6B \rightarrow 3D \quad \Delta H = -404 \, \text{kJ}\)

2. \(E + 2F \rightarrow A \quad \Delta H = -114.7 \, \text{kJ}\)

3. \(C \rightarrow E + 3D \quad \Delta H = 68.2 \, \text{kJ}\)

**Problem Statement:**

Suppose the first equation is reversed and multiplied by \(\frac{1}{6}\), the second and third equations are divided by 2, and the three adjusted equations are added. Determine the net reaction.

(If a box is not needed, leave it blank.)

**Net Reaction:**

\[ \boxed{} + \boxed{} + \boxed{} \]

\[ \rightarrow \boxed{} + \boxed{} + \boxed{} \]

**What's the overall heat of this reaction?**

Overall heat = \(\boxed{} \, \text{kJ}\)
Transcribed Image Text:**Chemical Reaction and Enthalpy Change Analysis** Consider the following equations: 1. \(3A + 6B \rightarrow 3D \quad \Delta H = -404 \, \text{kJ}\) 2. \(E + 2F \rightarrow A \quad \Delta H = -114.7 \, \text{kJ}\) 3. \(C \rightarrow E + 3D \quad \Delta H = 68.2 \, \text{kJ}\) **Problem Statement:** Suppose the first equation is reversed and multiplied by \(\frac{1}{6}\), the second and third equations are divided by 2, and the three adjusted equations are added. Determine the net reaction. (If a box is not needed, leave it blank.) **Net Reaction:** \[ \boxed{} + \boxed{} + \boxed{} \] \[ \rightarrow \boxed{} + \boxed{} + \boxed{} \] **What's the overall heat of this reaction?** Overall heat = \(\boxed{} \, \text{kJ}\)
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