Question

Given: $\Delta_r H^{+}(\text{Ag}^+)=105.6\ \text{kJ mol}^{-1}$, $\Delta_r H^{+}(\text{Cl}^-)=-167.2\ \text{kJ mol}^{-1}$, and $\Delta_r H^{+}(\text{AgCl})=-127.1\ \text{kJ mol}^{-1}$, find the value of $\Delta_r H$.

• A. \( 6.0\,\text{kJ mol}^{-1} \)
• B. \( 65.5\,\text{kJ mol}^{-1} \)
• C. \( -65.5\,\text{kJ mol}^{-1} \)
• D. \( -6.0\,\text{kJ mol}^{-1} \)
• E. \( 100\,\text{kJ mol}^{-1} \)

Hint:

When calculating enthalpy changes for reactions, remember that the enthalpy of a compound is the sum of the enthalpies of its constituent ions or elements.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The standard enthalpy of reaction (\(\Delta_r H^\circ\)) can be calculated from the standard enthalpies of formation (\(\Delta_f H^\circ\)) of the reactants and products using Hess's Law.
Step 2: Key Formula or Approach:
The formula for the standard enthalpy of reaction is:
\[ \Delta_r H^\circ = \sum (\text{m} \cdot \Delta_f H^\circ_{\text{products}}) - \sum (\text{n} \cdot \Delta_f H^\circ_{\text{reactants}}) \] where 'm' and 'n' are the stoichiometric coefficients of the products and reactants, respectively.
Step 3: Detailed Explanation:
The given reaction is:
\[ \text{Ag}^+ (\text{aq}) + \text{Cl}^- (\text{aq}) \rightarrow \text{AgCl} (\text{s}) \] Here, the reactants are Ag\(^+\)(aq) and Cl\(^-\)(aq), and the product is AgCl(s). All stoichiometric coefficients are 1.
Applying the formula:
\[ \Delta_r H^\circ = [1 \times \Delta_f H^\circ(\text{AgCl, s})] - [1 \times \Delta_f H^\circ(\text{Ag}^+, \text{aq}) + 1 \times \Delta_f H^\circ(\text{Cl}^-, \text{aq})] \] Substitute the given values:
\[ \Delta_f H^\circ(\text{AgCl, s}) = -127.1 \text{ kJ mol}^{-1} \] \[ \Delta_f H^\circ(\text{Ag}^+, \text{aq}) = 105.6 \text{ kJ mol}^{-1} \] \[ \Delta_f H^\circ(\text{Cl}^-, \text{aq}) = -167.2 \text{ kJ mol}^{-1} \] Now, plug them into the equation:
\[ \Delta_r H^\circ = [-127.1] - [105.6 + (-167.2)] \] \[ \Delta_r H^\circ = -127.1 - [105.6 - 167.2] \] \[ \Delta_r H^\circ = -127.1 - [-61.6] \] \[ \Delta_r H^\circ = -127.1 + 61.6 \] \[ \Delta_r H^\circ = -65.5 \text{ kJ mol}^{-1} \] Step 4: Final Answer:
The calculated standard enthalpy of reaction is -65.5 kJ mol\(^{-1}\). This corresponds to option (C). Although the provided answer key states the question was cancelled, the calculation leads to a valid answer among the options. The cancellation might be due to a printing error or other issues in the original exam paper.