Hess's law problem. Please help.

Question

5. Using the following data and Hess’s law, calculate the standard heat of formation (ΔHof) for the compound, ICl (g), in KJ/mol:


Cl2 (g) → 2 Cl (g) ΔHorxn = 242.3 KJ

I2 (g) → 2 I (g) ΔHorxn = 151.0 KJ

ICl (g) → I (g) + Cl (g) ΔHorxn = 211.3 KJ

I2 (s) → I2 (g) ΔHorxn = 62.8 KJ

Answer 1

Hello John

 

The question states that you need to determine the heat of formation for the compound, ICl (g).  From the 4 thermochemical equations listed, the 3rd equation has ICl (g), however, it is in the reactant side and we need it on the product side.  Therefore, we need to flip equation #3, but remember to reverse the sign of ΔH°_f whenever you flip an equation.

 

Flip #3:  I (g)  +  Cl  (g)   -->   ICl (g)    ΔH°_f   = - 211.3 kJ/mol

 

Since this reaction has I (g) and Cl (g), looking over the remaining equations, you can see that we have 2Cl (g) and 2I (g) in equations #1 and #2, respectively.

 

Therefore, to match these coefficients to equation #3, we need to multiply equation #3 by a factor of 2.  As a result, we have to multiply ΔH°_f by 2 as well.

 

Double equation #3:  2I (g) + 2Cl (g) --> 2ICl (g)   ΔH°_f = - 211.3 kJ/mol x 2  =  - 422.6 kJ/mol

 

Equations #1, 2 and 4 will be written as given in order to cancel the common elements.

 

#3:  2I (g) + 2Cl (g) --> 2ICl (g)  ΔH°_f =  - 422.6 kJ/mol

#1:  Cl₂ (g)  -->  2Cl (g)  ΔH°_f = + 242.3 kJ/mol

#2:  I₂ (g)  --> 2I (g)  ΔH°_f = + 151.0 kJ/mol

#4:  I₂ (s)  -->  I₂ (g)  ΔH°_f = + 62.8 kJ/mol

 

As you add these 4 equations and cancel the common elements, the resultant thermochemical equation and the heat of formation is:

 

I₂ (s)  +  Cl₂ (g)  -->  2 ICl (g)  ΔH°_f = - 422.6 + 242.3 + 151.0 + 62.8  =  + 33.5 kJ/mol

Thank you J.R.S for point that out! I appreciate it. John, the Delta;H° of + 33.5 kJ/mol is for 2 moles of ICl. For 1 mole, it would be +33.5 / 2 = + 16.8 kJ/mol