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: Understand the reaction involved.
The problem provides the enthalpy changes for the dissolution of ions and the formation of the salt. We are interested in calculating the overall enthalpy change for the reaction where silver chloride forms from silver ions and chloride ions: \[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl (s)} \]

Step 2: Use the provided enthalpy values.
We are given the following enthalpy values: \[ \Delta_r H^{+} (\text{Ag}^+) = 105.6\,\text{kJ mol}^{-1}, \quad \Delta_r H^{+} (\text{Cl}^-) = -167.2\,\text{kJ mol}^{-1}, \quad \Delta_r H^{+} (\text{AgCl}) = -127.1\,\text{kJ mol}^{-1} \]

Step 3: Write the equation for the overall enthalpy change.
To find the enthalpy change for the formation of \( \text{AgCl} \), we sum the individual enthalpy changes: \[ \Delta_r H = \Delta_r H^{+} (\text{Ag}^+) + \Delta_r H^{+} (\text{Cl}^-) + \Delta_r H^{+} (\text{AgCl}) \]

Step 4: Substitute the values.
Substitute the known values into the equation: \[ \Delta_r H = 105.6 + (-167.2) + (-127.1) \]

Step 5: Perform the calculation.
\[ \Delta_r H = 105.6 - 167.2 - 127.1 = 6.0\,\text{kJ mol}^{-1} \]

Step 6: Final conclusion.
The calculated enthalpy change for the formation of \( \text{AgCl} \) is \( 6.0\,\text{kJ mol}^{-1} \), so the correct option is: \[ \boxed{(1)\ 6.0\,\text{kJ mol}^{-1}} \]