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Standard XII

Chemistry

Heat of Formation

The bond diss...

Question

The bond dissociation energies for ${C l}_{2}, I_{2}$ and $I C l$ are $242.3, 151$ and $211.3$ $k J / m o l$ respectively. The enthalpy of sublimation of iodine is $62.8 k J / m o l$ . What is the standard enthalpy of formation of $I C l (g)$ ?

- 211.3 k J / m o l

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419 C a l / m o l

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16.8 k J / m o l

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33.5 k J / m o l

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Solution

The correct option is C $16.8 k J / m o l$
${C l}_{2} (g) \to 2 C l (g); Δ H_{1} = 243.3 k J / m o l$

$I_{2} (g) \to 2 I (g); Δ H_{2} = 151 k J / m o l$

$I C l (g) \to I (g) + C l (g); Δ H_{3} = 211.3 k J / m o l$

$I_{2} (s) \to I_{2} (g); Δ H_{4} = 62.8 k J / m o l$

Required equation:
$\frac{1}{2} I_{2} (s) + \frac{1}{2} {C l}_{2} (g) \to I C l (g); Δ H = ?$

$Δ H = \frac{62.8 + 151 + 242.3}{2} - 211.3$

$= 16.75 k J / m o l$

Hence, option C is correct.

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Similar questions

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 :

Q. Enthalpy of formation of

C_{2} H_{6} = - 83 k J / m o l

Enthalpy of sublimation of graphite =

719 k J / m o l

Enthalpy of bond dissociation of

H_{2} = 435 k J / m o l

C - H bond enthalpy =

414 k J / m o l

Calculate the bond enthalpy of C - C

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)

and

C l_{2} (g)

, Calculate the standard enthalpy formation for

I C l (g)

Q. Bond dissociation enthalpies of

H_{2_{(} g)}

and

N_{2_{(} g)}

are 436.0

K J

m o l^{- 1}

and 941.8

K J

m o l^{- 1}

, respectively and enthalpy of formation of

N H_{3_{(} g)}

- 46 K J

m o l^{- 1}

What is the enthalpy of atomization of

N H_{3_{(} g)}

and the average bond enthalpy of

N - H

bond respectively (in

K J

m o l^{- 1}

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)

and

C l_{2} (g)

, the standard enthalpy of formation for

I C l (g)

is:

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The bond dissociation energies for Cl2,I2 and ICl are 242.3,151 and 211.3 kJ/mol respectively. The enthalpy of sublimation of iodine is 62.8 kJ/mol. What is the standard enthalpy of formation of ICl(g)?

The correct option is C 16.8 kJ/molCl2(g)→2Cl(g);ΔH1=243.3kJ/molI2(g)→2I(g);ΔH2=151kJ/molICl(g)→I(g)+Cl(g);ΔH3=211.3kJ/molI2(s)→I2(g);ΔH4=62.8kJ/molRequired equation:12I2(s)+12Cl2(g)→ICl(g);ΔH=?ΔH=62.8+151+242.32−211.3 =16.75kJ/molHence, option C is correct.

The bond dissociation energies for ${C l}_{2}, I_{2}$ and $I C l$ are $242.3, 151$ and $211.3$ $k J / m o l$ respectively. The enthalpy of sublimation of iodine is $62.8 k J / m o l$ . What is the standard enthalpy of formation of $I C l (g)$ ?