Question

The lattice enthalpy of ki will be, if the enthalpy of (i) δfh (ki) = –78.0 kcal mol–1 (ii) ionisation energy of k to k+ is 4.0 ev (iii) dissociation energy of i2 is 28.0 kcal mol–1 (iv) sublimation energy of k is 20.0 kcal mol–1 (v) electron gain enthalpy for i to i– is –70.0 kcal mol–1 (vi) sublimation energy of i2 is 14.0 kcal mol–1 (1 ev = 23.0 kcal mol–1

Answer 1

Answer : The lattice enthalpy will be 141.0 kcal/mol.

Explanation :

Let us convert the given information in the equation form.

(1) Enthalpy of formation is the energy evolved when 1 mol of KI is formed from K and I₂ in their standard states. The equation for this reaction is given below.

$K(s) +\frac{1}{2} I_{2} (s) \rightarrow KI(s) .......\bigtriangleup H_{f} = -78.0 kcal/mol$ ..........Equation 1

(2) Ionisation energy is the amount of energy needed when 1 mol of gaseous K atoms are converted to gaseous K⁺ ions. The equation for this reaction is written below.

$K (g) \rightarrow K^{+} (g) + e^{-}...........I.E. = 4 ev$ .......Equation 2

Let us convert I.E from e.v. to kcal unit.

We have been given that , 1 ev = 23.0 kcal

IE for K is $4 ev \times\frac{23 kcal}{1ev} = 92 kcal/mol$

(3) The dissociation energy of I₂ is the energy absorbed when 1 mol I₂ dissociates to form I atoms. The equation for this is given below.

$I_{2} (g) \rightarrow 2I(g) ........ \bigtriangleup H_{dissociation} =  28.0 kcal/mol$ ...........Equation 3

(4) The sublimation energy is the energy needed to convert K from solid state to gaseous state. The equation for this is written as,

$K(s) \rightarrow K(g) ........... \bigtriangleup H_{sublimation} = 20.0 kcal/mol$ ............Equation 4

(5) Electron gain enthalpy is the amount of energy evolved when 1 mol atoms of I accept electron to form I⁻ ions

$I (g) + e^{-} \rightarrow I^{-} (g) ..........\bigtriangleup H_{EG} = -70.0 kcal/mol$ ..........Equation 5

(6) Sublimation of I₂ is the conversion of solid I₂ into gaseous I₂.

$I_{2} (s) \rightarrow I_{2} (g) ........ \bigtriangleup H_{sub} = 14.0 kcal/mol$ ..........Equation 6

We will reverse equations 2,3,4,5 6 and we will divide equation 3 and 6 by 2

We will keep equation 1 as it is .

Given below are the modified equations.

$K(s) +\frac{1}{2} I_{2} (s) \rightarrow KI(s) .......\bigtriangleup H_{f} = -78.0 kcal/mol$

$K^{+} (g) + e^{-} \rightarrow K (g) ...........I.E. = -92.2 kcal/mol$

$I(g) \rightarrow \frac{1}{2} I_{2} (g) ........ \bigtriangleup H =  -14.0 kcal/mol$

$K(g) \rightarrow K(s) ........... \bigtriangleup H = -20.0 kcal/mol$

$I^{-} (g) \rightarrow I(g) + e^{-}..........\bigtriangleup H = +70.0 kcal/mol$

$\frac{1}{2} I_{2} (g) \rightarrow \frac{1}{2} I_{2} (s) ........ \bigtriangleup H_{sub} = -7.0 kcal/mol$

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$K^{+} (g) + I^{-} (g) \rightarrow KI (s) .................\bigtriangleup H = 141 cal/mol$

We get above equation on adding all the other equations.

Lattice enthalpy is the amount of energy evolved when 1 mol of solid KI is formed from gaseous K+ and I- ions which is 141 kcal/mol from the above equation.

The lattice enthalpy of KI will be 141 kcal/mol

Answer 2

Answer : The lattice enthalpy will be 141.0 kcal/mol.

Explanation :

Let us convert the given information in the equation form.

(1) Enthalpy of formation is the energy evolved when 1 mol of KI is formed from K and I₂ in their standard states. The equation for this reaction is given below.

..........Equation 1

(2) Ionisation energy is the amount of energy needed when 1 mol of gaseous K atoms are converted to gaseous K⁺ ions. The equation for this reaction is written below.

.......Equation 2

Let us convert I.E from e.v. to kcal unit.

We have been given that , 1 ev = 23.0 kcal

IE for K is  

(3) The dissociation energy of I₂ is the energy absorbed when 1 mol I₂ dissociates to form I atoms. The equation for this is given below.

...........Equation 3

(4) The sublimation energy is the energy needed to convert K from solid state to gaseous state. The equation for this is written as,

............Equation 4

(5) Electron gain enthalpy is the amount of energy evolved when 1 mol atoms of I accept electron to form I⁻ ions

..........Equation 5

(6) Sublimation of I₂ is the conversion of solid I₂ into gaseous I₂.

..........Equation 6

We will reverse equations 2,3,4,5 6 and we will divide equation 3 and 6 by 2

We will keep equation 1 as it is .

Given below are the modified equations.

-----------------------------------------------------------------------------------------

We get above equation on adding all the other equations.

Lattice enthalpy is the amount of energy evolved when 1 mol of solid KI is formed from gaseous K+ and I- ions which is 141 kcal/mol from the above equation.

The lattice enthalpy of KI will be 141 kcal/mol