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Heat of Formation

The enthalpy ...

Question

The enthalpy changes for the following processes are listed below
$C l_{2} (g) = 2 C l (g),$ $242.3 k j m o l^{- 1}$
$I_{2} (g) = 2 I (g),$ $151.0 k j m o l^{- 1}$
$I C l (g) = I (g) + C l (g),$ $211.3 k j m o l^{- 1}$
$I_{2} (s) \to I_{2} (g),$ $62.76 k j m o l^{- 1}$
Given that the standard stales for iodine and chlorine are $I_{2} (s)$ and $C l_{2} (g)$ , Calculate the standard enthalpy formation for $I C l (g)$ .

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Solution

Enthalpy of formation of $I C l (g)$ = [ $H_{S}$ of $I_{2} (s)$ + $I E$ of $C l_{2} (g)$ + $H_{d i s s}$ of $I_{2} (g)$ ] $- Δ H_{I C l (g)}$

$H_{S} ⟶$ Heat of Sublimation, $I . E ⟶$ Ionisation Enthalpy, $H_{d i s s} ⟶$ Enthalpy of dissociation

Reaction:-
$\frac{1}{2} I_{2} (s) + \frac{1}{2} C l_{2} (g) ⟶ I C l (g)$
$Δ H_{f} = [\frac{1}{2} H_{S} + \frac{1}{2} I E + H_{d i s s} I_{2} (g)] - Δ H_{I C l}$
$= [\frac{1}{2} \times 62.76 + \frac{1}{2} \times 242.3 + \frac{1}{2} \times 151] - 211.3$
$= [31.38 + 121.15 + 75.5] - 211.3$
$= 16.73 K J m o l^{- 1}$

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Similar questions

Q. The enthalpy changes for the following precesses are listed below :

C l_{2} (g) \to 2 C l (g), 242.3 k J {m o l}^{- 1}

I_{2} (g) \to 2 I (g), 151.0 k J {m o l}^{- 1}

I C l (g) \to I (g) + C l (g), 211.3 k J {m o l}^{- 1}

I_{2} (s) \to I_{2} (g), 62.76 k J {m o l}^{- 1}

Given that the standard states for iodine and chlorine are

I_{2} (g)

C l_{2} (g),

the standard enthalpy for the formation of

C l (g)

Q. The enthalpy changes for the following processes are listed below:

{C l}_{2} (g) ⟶ 2 C l (g); Δ H^{o} = 242.3 k J {m o l}^{- 1}

I_{2} (g) ⟶ 2 I (g); Δ H^{o} = 151.0 k J {m o l}^{- 1}

{I C l} (g) ⟶ I (g) + C l (g); Δ H^{o} = 211.3 k J {m o l}^{- 1}

I_{2} (s) ⟶ I_{2} (g); Δ H^{o} = 62.76 k J {m o l}^{- 1}

Given that the standard states for iodine and chlorine are

I_{2} (s)

{C l}_{2} (g)

, the standard enthalpy of formation for

I C l (g)

Q. The enthalpy changes for the following processes are listed below:

C l_{2} (g) = 2 C l (g), 242.3 k J m o l^{- 1}

I_{2} (g) = 2 I (g), 151.0 k J m o l^{- 1}

I C l (g) = I (g) + C l (g), 211.3 k J m o l^{- 1}

I_{2} (s) = I_{2} (g), 62.76 k J m o l^{- 1}

Given that the standard states for iodine and chlorine are

I_{2} (s)

C l_{2} (g)

, the standard enthalpy of formation for

I C l (g)

Q. The enthalpy changes for the following processes are listed below:

C l_{2} (g) - 2 C l (g), 242.3 k J m o l^{- 1}

I_{2} (g) - 2 I (g), 151.0 k J m o l^{- 1}

I C I (g) - I (g) + C l_{(} g), 211.3 k L m o l^{- 1}

I_{2} (s) - I_{2} (g), 62.75 k J m o l^{- 1}

Given that the standard states for iodine and chlorine are

I_{2} (s)

C l_{2} (g)

, the standard enthalpy of formation for

I C I (g)

The enthalpy changes for the following processes are listed below:

C l_{2} (g) \to 2 C l (g);

242.3 k J

^{- 1}

I_{2} (g) \to 2 I (g);

151.0 k J m o l

^{- 1}

I C l (g) \to I (g) + C l (g);

211.3 k J

^{- 1}

I_{2} (s) \to I_{2} (g);

62.76 k J m o l

^{- 1}

Given that the standard states for iodine and chlorine are

I_{2}

C l_{2}

(g) , the standard enthalpy of formation of

I C l

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