Given the thermochemical equations X, +3 Y, – 2XY3 AH1 = -380 kJ X, +2Z, ► 2 XZ, AH2 = -110 kJ 2 Y, + Z, → 2 Y,Z AH3 = -280 kJ Calculate the change in enthalpy for the reaction. 4XY, + 7Z, - > 6 Y,Z + 4XZ, ΔΗ kJ

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Chapter1: Chemical Foundations
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Given the thermochemical equations:

\[ 
X_2 + 3Y_2 \rightarrow 2XY_3 \quad \Delta H_1 = -380 \, \text{kJ} 
\]

\[ 
X_2 + 2Z_2 \rightarrow 2XZ_2 \quad \Delta H_2 = -110 \, \text{kJ} 
\]

\[ 
2Y_2 + Z_2 \rightarrow 2Y_2Z \quad \Delta H_3 = -280 \, \text{kJ} 
\]

Calculate the change in enthalpy for the reaction:

\[ 
4XY_3 + 7Y_2 \rightarrow 6Y_2Z + 4XZ_2 
\]

\[ 
\Delta H = \underline{\hspace{2cm}} \, \text{kJ} 
\]
Transcribed Image Text:Given the thermochemical equations: \[ X_2 + 3Y_2 \rightarrow 2XY_3 \quad \Delta H_1 = -380 \, \text{kJ} \] \[ X_2 + 2Z_2 \rightarrow 2XZ_2 \quad \Delta H_2 = -110 \, \text{kJ} \] \[ 2Y_2 + Z_2 \rightarrow 2Y_2Z \quad \Delta H_3 = -280 \, \text{kJ} \] Calculate the change in enthalpy for the reaction: \[ 4XY_3 + 7Y_2 \rightarrow 6Y_2Z + 4XZ_2 \] \[ \Delta H = \underline{\hspace{2cm}} \, \text{kJ} \]
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