Respuesta :
Following information should be given in the problem.
delta H of formation Naâ‚‚Oâ‚‚ (s)= -510.9 kJ/mol ;
delta H of formation[NaOH (aq)] = -469.2 kJ/ mol ;
delta H of formation [Hâ‚‚O(l)] = -285.8 kJ/mol
Step 1 : Writing balanced equation
The balanced chemical equation for the reaction between sodium peroxide and water is given below.
[tex] 2 Na_{2} O_{2} (s) + 2 H_{2} O (l) -------> 4 NaOH (aq) + O_{2} (g) [/tex]
Step 2 : Finding moles of Oâ‚‚ gas
Moles of Oâ‚‚ = [tex] \frac{mass of O_{2}}{Molar mass of O_{2}} [/tex]
Moles of Oâ‚‚ = [tex] \frac{327.2 g}{32 g/mol} [/tex]
Moles of Oâ‚‚ = 10.225 mol
Step 3: Finding standard enthalpy of the reaction using the given enthalpy of formation values
The formula to calculate standard enthalpy of the reaction is as follows
ΔH rxn = ∑ΔHf products - ∑ΔHf reactants
ΔH rxn = ( 4×ΔHf NaOH + ΔHf O₂ ) - ( 2×ΔHf Na₂O₂ + 2×ΔHf H₂O )
ΔH rxn = ( 4×-469.2 kJ/mol + 0 ) - (2× -510.9 kJ/mol + 2×-285.8 kJ/mol )
ΔH rxn = ( -1876.8 kJ/mol) - ( -1593.4 kJ/mol)
ΔH rxn = -283.4 kJ/mol
Step 4 : Finding amount of energy released for the calculated moles of Oâ‚‚
From step 3 we know that when 1 mol of Oâ‚‚ is released, -283.4 kJ are released.
We have 10.225 mol Oâ‚‚
Let us set up the equation for calculating heat for 10.225 mol Oâ‚‚
10.225 mol O₂ × [tex] \frac{-283.4 kJ}{mol} = -2897.8 kJ [/tex]
-2897.8 kJ of heat is released during the reaction under standard conditions
