Answer:
The answer to your question is ΔH = -44.36 kJ
Explanation:
Data
H NaOH (s) = -4.26 x 10² kJ/mol
H Na⁺(aq) = -2.4034 x 10² kJ/mol
H OH⁻(aq) = -2.3002 x 10² kJ/mol
Balanced chemical reaction
NaOH (s) ⇒ Na⁺(aq) + OH⁻(aq)
Process
1.- Write the equation to calculate the enthalpy
ΔH = ∑H products - ∑H reactants
2.- Substitution
ΔH = (-2.4034 x 10²) + (-2.3002 x 10²) - (-4.26 x 10²)
-Simplification
ΔH = -2.4034 x 10² - 2.3002 x 10² + 4.26 x 10²
-Result
ΔH = -44.36 kJ
