ThermochemistryMedium
Question
The enthalpy changes for the following processes are listed below:
Cl2(g) = 2Cl(g), 242.3 kJ mol-1
I2(g) = 2I(g), 151.0 kJ mol-1
ICl(g) = I(g) + Cl(g), 211.3 kJ mol-1
I2(s) = I2(g), 62.76 kJ mol-1
Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is
Cl2(g) = 2Cl(g), 242.3 kJ mol-1
I2(g) = 2I(g), 151.0 kJ mol-1
ICl(g) = I(g) + Cl(g), 211.3 kJ mol-1
I2(s) = I2(g), 62.76 kJ mol-1
Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is
Options
A.-14.6 kJ mol-1
B.-16.8 kJ mol-1
C.+ 16.8 kJ mol-1
D.+ 244.8 kJ mol-1
Solution
I2(S) + Cl2 → ICl(g)ᐃH =
-[μI - Cl]=
- (211.3)= 228.03 - 211.3
ᐃH = 16.73
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