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07/10/2020 11:10 am
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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:
(a) +16.8 kJ mol-1
(b) +244.8 kJ mol-1
(c) –14.6 kJ mol-1
(d) –16.8 kJ mol-1
1 Answer
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07/10/2020 11:22 am
(a) +16.8 kJ mol-1
Explanation:
I2(s) + Cl2(g) → 2ICl(g)
ΔrH = [ΔH(I2(s) → I2(g)) + ΔHI-I + ΔHCl-Cl] - [ΔHI-Cl]
= 151.0 + 242.3 + 62.76 - 2 × 211.3 = 33.46
DfH°(ICl) = 33.46/2
= +16.8 kJ mol-1