1/2 H2(g) + 1/2 I2(s) -> HI(g) ΔH = 26 kJ/molrxn 1/2 H2(g) + 1/2 I2(g) -> HI(g) ΔH = -5.0 kJ/molrxn — Thermodynamics Chemistry Question
Question
1/2 H2(g) + 1/2 I2(s) → HI(g) ΔH = 26 kJ/molrxn
1/2 H2(g) + 1/2 I2(g) → HI(g) ΔH = -5.0 kJ/molrxn
Based on the information above, what is the enthalpy change for the sublimation of iodine, represented below?
I2(s) → I2(g)
A.
15 kJ/molrxn
B.
21 kJ/molrxn
C.
31 kJ/molrxn
D.
42 kJ/molrxn
E.✓ Correct
62 kJ/molrxn
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