Question

The standard enthalpy change for the following reaction is 576 kJ at 298 K. 2 NaI(s) 2 Na(s) + I2(s) ΔH° = 576 kJ What is the standard enthalpy change for this reaction at 298 K? Na(s) + 1/2 I2(s) NaI(s)

Answer 1

ΔH° = -288kJ

Step-by-step explanation:

You can make algebraic opperations of the chemical reations to obtain the ΔH of a particular reaction.

From:

2NaI(s) → 2Na(s) + I2(s) ΔH° = 576 kJ

The ΔH of the inverse reaction:

2Na(s) + I2(s) → 2NaI(s) ΔH° = -576 kJ

Dividing the last reaction twice:

Na(s) + 1/2 I2(s) → NaI(s) ΔH° = -576 kJ / 2

ΔH° = -288kJ