Chemical Thermodynamics Question 327
Question: The enthalpy change states for the following processes are listed below:
$ Cl_2(g)\to 2Cl(g); $
$ 242.3kJmo{{l}^{-1}} $
$ I_2(g)\to 2I(g); $
$ 151.0kJmo{{l}^{-1}} $
$ ICl(g)\to I(g)+Cl(g); $
$ 211.3kJmo{{l}^{-1}} $
$ I_2(s)\to I_2(g); $
$ 62.76kJmo{{l}^{-1}} $
Given that the standard states for iodine chlorine are $ I_2(s) $ and $ Cl_2(g) $ , the standard enthalpy o formation for ICl (g) is:
Options:
A) +244.8 kJ $ mo{{l}^{-1}} $
B) -14.6 kJ $ mo{{l}^{-1}} $
C) -16.8kJ $ mo{{l}^{-1}} $
D) + 16.8kJ $ mo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I_2+Cl_2\to 2ICl $
$ \Delta H={e_{I_2S\to g}}+{e_{I-,I}}+{e_{Cl-,Cl}}-2\times {e_{1-,Cl}} $
$ =62.76+151.0+242.3-2\times 211.3 $
$ =33.46kJ $ for $ 2molICl $
$ \therefore \Delta H/mol=\frac{33.46}{2}=16.73kJ,mo{{l}^{-1}} $
