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Calculate DeltaH(f)^(@) for chloride ion...

Calculate `DeltaH_(f)^(@)` for chloride ion from the following data `:`
`(1)/(2)H_(2(g))+(1)/(2)Cl_(2(g))rarr HCl_((g)),DeltaH_(f)^(@)=-92.4kJ`
`HCl_((g))+nH_(2)O rarr H_((aq.))^(+)+Cl_((aq.))^(-),DeltaH^(@)=-74.8kJ`
`DeltaH_(f)^(@) H_((aq.))^(+)=0.0kJ`

A

`-167.2` kJ

B

`-165.2` kJ

C

`-157.2` kJ

D

`-147.2` kJ

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the standard enthalpy of formation (ΔH_f^(@)) for the chloride ion (Cl^-), we can use the provided reactions and their enthalpy changes. ### Step-by-Step Solution: 1. **Identify the Reactions and Their Enthalpy Changes**: - Reaction 1: \[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g), \quad \Delta H = -92.4 \, \text{kJ} \] - Reaction 2: \[ HCl(g) + n H_2O \rightarrow H^+(aq) + Cl^-(aq), \quad \Delta H = -74.8 \, \text{kJ} \] - Given: \[ \Delta H_f^(@) H^+(aq) = 0.0 \, \text{kJ} \] 2. **Set Up the Equation for Reaction 1**: - The enthalpy change for Reaction 1 can be expressed as: \[ \Delta H_{reaction 1} = \Delta H_f^(@) HCl - \Delta H_f^(@) \left(\frac{1}{2} H_2\right) - \Delta H_f^(@) \left(\frac{1}{2} Cl_2\right) \] - Since both \(H_2\) and \(Cl_2\) are in their most stable forms, their standard enthalpy of formation is 0: \[ \Delta H_{reaction 1} = \Delta H_f^(@) HCl - 0 - 0 = \Delta H_f^(@) HCl \] - Therefore, we have: \[ \Delta H_f^(@) HCl = -92.4 \, \text{kJ} \] 3. **Set Up the Equation for Reaction 2**: - The enthalpy change for Reaction 2 can be expressed as: \[ \Delta H_{reaction 2} = \Delta H_f^(@) H^+ + \Delta H_f^(@) Cl^- - \Delta H_f^(@) HCl - n \Delta H_f^(@) H_2O \] - Given that \(\Delta H_f^(@) H^+ = 0.0 \, \text{kJ}\) and assuming that the enthalpy of water is negligible for this calculation (since it is in large excess), we can simplify this to: \[ \Delta H_{reaction 2} = 0 + \Delta H_f^(@) Cl^- - \Delta H_f^(@) HCl \] - Therefore, we have: \[ \Delta H_{reaction 2} = \Delta H_f^(@) Cl^- - (-92.4 \, \text{kJ}) \] 4. **Substituting the Known Values**: - We know that \(\Delta H_{reaction 2} = -74.8 \, \text{kJ}\), so we can substitute this into the equation: \[ -74.8 \, \text{kJ} = \Delta H_f^(@) Cl^- + 92.4 \, \text{kJ} \] 5. **Solving for \(\Delta H_f^(@) Cl^-\)**: - Rearranging the equation gives: \[ \Delta H_f^(@) Cl^- = -74.8 \, \text{kJ} - 92.4 \, \text{kJ} \] - Calculating this gives: \[ \Delta H_f^(@) Cl^- = -167.2 \, \text{kJ} \] ### Final Answer: \[ \Delta H_f^(@) Cl^- = -167.2 \, \text{kJ} \]
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