Question

The \( \Delta_f H^\theta \) of \( AO_3(s) \), \( BO_2(s) \), and \( ABO_3(s) \) is -635, \( x \), and -1210 kJ mol\(^{-1}\) respectively. The reaction:

\( ABO_3 (s) \rightarrow AO_3 (s) + BO_2 (g) \)

Has an enthalpy change of \( \Delta_r H^\theta = 175 \) kJ mol\(^{-1}\). What is the value of \( x \) (in kJ mol\(^{-1}\))?

• A. \( -750 \)
• B. \( +400 \)
• C. \( -400 \)
• D. \( +750 \)

Hint:

Hess's Law states that the enthalpy change of a reaction is the sum of the enthalpy changes of its components. Use the equation: \[ \Delta_r H^\theta = \sum \Delta_f H^\theta (\text{products}) - \sum \Delta_f H^\theta (\text{reactants}) \] to solve enthalpy-related problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Given Data 
We are given the standard enthalpies of formation (\( \Delta_f H^\theta \)): \[ \Delta_f H^\theta (AO_3) = -635 \text{ kJ mol}^{-1} \] \[ \Delta_f H^\theta (ABO_3) = -1210 \text{ kJ mol}^{-1} \] \[ \Delta_r H^\theta = 175 \text{ kJ mol}^{-1} \] We need to determine the enthalpy of formation of \( BO_2 \), denoted as \( x \). 

Step 2: Applying Hess's Law 
From the reaction: \[ ABO_3 (s) \rightarrow AO_3 (s) + BO_2 (g) \] Using the enthalpy change equation: \[ \Delta_r H^\theta = \sum \Delta_f H^\theta (\text{products}) - \sum \Delta_f H^\theta (\text{reactants}) \] \[ 175 = (-635 + x) - (-1210) \] 

Step 3: Solve for \( x \) 
\[ 175 = -635 + x + 1210 \] \[ x = 175 - 1210 + 635 \] \[ x = -400 \text{ kJ mol}^{-1} \] 

Step 4: Verify the Correct Answer 
Thus, the value of \( x \) is \( -400 \) kJ mol\(^{-1}\), which matches Option (3).