Question

The enthalpy change states for the following processes are listed below:

\[\ce{Cl2(g) = 2Cl (g)}\]     242.3 kJ mol^-1

\[\ce{I2(g) = 2I(g)}\]     151 kJ mol^-1

\[\ce{ICI (g) = I(g) + Cl(g)}\]   242.3 kJ mol^-1

\[\ce{I2(s) = I2(g)}\]   62.76 kJ mol^-1

Given that the standard states for iodine chlorine are I₂ (s) and Cl₂ (g), the standard enthalpy of formation for ICl (g) is:

Options

• 244. 8 kJ mol^-1

• - 14.6 kJ mol^-1

• - 16.8 kJ mol^-1

• 16.8 kJ mol^-1

Answer 1

16.8 kJ mol^-1 

Explanation:

\[\ce{Cl2(g) = 2Cl (g)}\]           242.3 kJ mol^-1

\[\ce{I2(g) = 2I(g)}\]                 151 kJ mol^-1

\[\ce{I2(s) = 2I(g)}\]                 62.76 kJ mol^-1

\[\underline{\ce{2I (g) +  2Cl(g) -> 2ICl(g)}}\]   2 × (- 242.3 kJ mol^-1)
\[\ce{I2(s) + Cl2(g) -> 2ICl(g)}\]

Δ H = 242.3 + 151 + 62.76 - (2 × 242.3)

Δ H = 33.46

Δ H = `(33.46)/2`

Δ H = 16.73