Question

The enthalpy changes for the following processes are listed below:

Cl2(g) $\rightarrow$ 2Cl(g); 242.3 kJ mol–1 ; I2(g) $\rightarrow$ 2I(g); 151.0 kJ mol–1

ICl(g) $\rightarrow$ I(g) + Cl(g); 211.3 kJ mol–1 ; I2(s) $\rightarrow$ I2(g); 62.76 kJ mol–1

Given that the standard state 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. +16.8 kJ mol–1
• C. +244.8 kJ mol–1
• D. -14.6 kJ mol–1

Answer 1

Answer: (B)

+16.8 kJ mol–1

Answer 2

$\frac{1}{2}$I2(s) + $\frac{1}{2}$CI2(g) $\rightarrow$ ICI(g) $\Delta$Hf, ICI(g) =

$\left\lbrack \frac{1}{2}\Delta H_{I_{2}}(s) \rightarrow I_{2}(g) + \frac{1}{2}\Delta H_{I - I} + \frac{1}{2}\Delta H_{Cl - Cl} \right\rbrack$

– $\left\lbrack \Delta H_{l - Cl} \right\rbrack$

= $\left\lbrack \frac{1}{2} \times 62.76 + \frac{1}{2}151.0 + \frac{1}{2} \times 242.3 \right\rbrack$ – $\lbrack 211.3\rbrack$ = 16.73 kJ/mol.