The `DeltaH_(f)^(0)(KF,s)` is `-563 kJ mol^(-1)`. The ionization enthalpy of `K(g)` is `419 kJ mol^(-1)`. and the enthalpy of sublimation of potassium is `88 kJ mol^(-1)`. The electron affinity of `F(g)` is `322 kJ mol^(-1)` and `F-F` bond enthalpy is `158 kJ mol^(_1)`. Calculate the lattice enthalpy of `KF(s)`.
The given data are as follows:
(i) `K(s)+1//2F_(2)(g)rarrKF(s)" "DeltaH_(f)^(0)= -563 kJ mol^(-1)`
(ii) `K(g)rarrK^(+)(g)+e^(-)" "Delta_("Ioniz")^(0)=419 kJ mol^(-1)`
(iii) `K(s)rarrK(g)" "DeltaH_("sub")^(0)=88kJ mol^(-1)`
(iv) `F(g)+e^(-)rarrF^(-)(g)" "DeltaH_(eg)^(0)= -322 kJ mol^(-1)`
(v) `F_(2)(g)rarr2F(s)" "DeltaH_("diss")^(0)= 158 kJ mol^(-1)`
(vi) `K^(+)(g)+F^(-)(g)rarrKF(s)" " DeltaH_(L)^(0)=?`

Explanation

To calculate the lattice enthalpy of KF(s), we will use Hess's law and the given thermodynamic data. We will manipulate the provided reactions to derive the lattice enthalpy.

### Step-by-Step Solution:

1. **Write down the given reactions and their enthalpy changes:**
- (i) \( K(s) + \frac{1}{2} F_2(g) \rightarrow KF(s) \)
\(\Delta H_f^{0} = -563 \, \text{kJ mol}^{-1}\)
- (ii) \( K(g) \rightarrow K^+(g) + e^- \)
\(\Delta H_{ionization}^{0} = 419 \, \text{kJ mol}^{-1}\)
- (iii) \( K(s) \rightarrow K(g) \)
\(\Delta H_{sub}^{0} = 88 \, \text{kJ mol}^{-1}\)
- (iv) \( F(g) + e^- \rightarrow F^-(g) \)
\(\Delta H_{eg}^{0} = -322 \, \text{kJ mol}^{-1}\)
- (v) \( F_2(g) \rightarrow 2F(g) \)
\(\Delta H_{diss}^{0} = 158 \, \text{kJ mol}^{-1}\)
- (vi) \( K^+(g) + F^-(g) \rightarrow KF(s) \)
\(\Delta H_L^{0} = ?\)

2. **Manipulate the reactions:**
- Start with reaction (i) and rearrange it to find the lattice enthalpy:
\[ \Delta H_L^{0} = \Delta H_f^{0} - \Delta H_{ionization}^{0} - \Delta H_{sub}^{0} - \Delta H_{eg}^{0} - \frac{1}{2} \Delta H_{diss}^{0} \]

3. **Substitute the values into the equation:**
- Substitute the known values into the equation:
\[ \Delta H_L^{0} = -563 - 419 - 88 + 322 - \frac{1}{2} \times 158 \]

4. **Calculate the values step-by-step:**
- Calculate \( -563 - 419 = -982 \)
- Calculate \( -982 - 88 = -1070 \)
- Calculate \( -1070 + 322 = -748 \)
- Calculate \( -\frac{1}{2} \times 158 = -79 \)
- Finally, calculate \( -748 - 79 = -827 \)

5. **Final result:**
- The lattice enthalpy of \( KF(s) \) is:
\[ \Delta H_L^{0} = -827 \, \text{kJ mol}^{-1} \]

### Summary:
The lattice enthalpy of KF(s) is \(-827 \, \text{kJ mol}^{-1}\).