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Enthalpy

The enthalpy ...

Question

The enthalpy changes for the following processes are listed below:
$C l_{2} (g) \to 2 C l (g);$ $242.3 k J$ mol $^{- 1}$
$I_{2} (g) \to 2 I (g);$ $151.0 k J m o l$ $^{- 1}$
$I C l (g) \to I (g) + C l (g);$ $211.3 k J$ mol $^{- 1}$
$I_{2} (s) \to I_{2} (g);$ $62.76 k J m o l$ $^{- 1}$
Given that the standard states for iodine and chlorine are $I_{2}$ (s) and $C l_{2}$ (g) , the standard enthalpy of formation of $I C l$ (g) is :

A

+ 16.8

k J

mol

^{- 1}

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B

+ 244.8

k J

mol

^{- 1}

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C

- 14.6

k J

mol

^{- 1}

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D

- 16.8

k J

mol

^{- 1}

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Solution

The correct option is A $+ 16.8$ $k J$ mol $^{- 1}$

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Similar questions

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 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)

C l_{2} (g)

, Calculate the standard enthalpy formation for

I C l (g)

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 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)

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