Question

Solid sodium peroxide (Na2O2) reacts with liquid water yielding aqueous sodium hydroxide and oxygen gas. How much heat is released if 327.2 g of oxygen gas is produced from the reaction of sodium peroxide and water under standard-state conditions?

Answer 1

Following information should be given in the problem.

delta H of formation Na₂O₂ (s)= -510.9 kJ/mol ;

delta H of formation[NaOH (aq)] = -469.2 kJ/ mol ;

delta H of formation [H₂O(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.

`2 Na_(2) O_(2) (s) + 2 H_(2) O (l) -------> 4 NaOH (aq) + O_(2) (g)`

Step 2 : Finding moles of O₂ gas

Moles of O₂ =
`(mass of O_(2))/(Molar mass of O_(2))`

Moles of O₂ =
`(327.2 g)/(32 g/mol)`

Moles of O₂ = 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 O₂

From step 3 we know that when 1 mol of O₂ is released, -283.4 kJ are released.

We have 10.225 mol O₂

Let us set up the equation for calculating heat for 10.225 mol O₂

10.225 mol O₂ ×
`(-283.4 kJ)/(mol) = -2897.8 kJ`

-2897.8 kJ of heat is released during the reaction under standard conditions