The standard enthalpy of formation for ICl(g) is calculated from: Cl_2(g) = 2Cl(g), H = 242.3 I_2(g) = 2I(g), H = 151.0 ICl(g) = I(g) + Cl(g), H = 211 - Chemistry MCQ AIEEE 2006 2026 | ExamTest.live

hardMCQAIEEE 20062026Chemistry

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The standard enthalpy of formation for ICl (g) is calculated from: Cl_{2} (g) = 2 Cl (g), Δ H = 242.3 I_{2} (g) = 2 I (g), Δ H = 151.0 ICl (g) = I (g) + Cl (g), Δ H = 211.3 I_{2} (s) = I_{2} (g), Δ H = 62.76 Standard states are I_{2} (s) and Cl_{2} (g) .

Solution & Step-by-step Explanation

Reaction: \frac{1}{2} I_{2} (s) + \frac{1}{2} Cl_{2} (g) \to ICl (g) Δ H_{f} = [\frac{1}{2} Δ H_{s u b} (I_{2}) + \frac{1}{2} Δ H_{d i ss} (I_{2}) + \frac{1}{2} Δ H_{d i ss} (Cl_{2})] - [Bond Energy ICl] $ Δ H_{f} = [\frac{1}{2} (62.76) + \frac{1}{2} (151.0) + \frac{1}{2} (242.3)] - 211.3 $ Δ H_{f} = [31.38 + 75.5 + 121.15] - 211.3 Δ H_{f} = 228.03 - 211.3 = + 16.73 kJ/mol

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The standard enthalpy of formation for ICl (g) is calculated from: Cl_{2} (g) = 2 Cl (g), Δ H = 242.3 I_{2} (g) = 2 I (g), Δ H = 151.0 ICl (g) = I (g) + Cl (g), Δ H = 211.3 I_{2} (s) = I_{2} (g), Δ H = 62.76 Standard states are I_{2} (s) and Cl_{2} (g) .

-14.6

-16.8

+16.8

+244.8

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