Answer :
2H2O2 ===> 2H2O + O2 ... TARGET EQUATION
We want to use the other equations that are given in order to make the target equation.
1) 2H2 + O2 ==> 2H2O ∆H = -572 kJ
2) H2 + O2 ==> H2O2 ∆H = -188 kJ
Reverse eq. 2 and multiply by 2 to get 2H2O2 on the left
2H2O2 ==> 2H2 + 2O2 ∆H = +376 kJ (change sign of ∆H and multiply by 2)
Copy eq.1
2H2 + O2 ==> 2H2O ∆H = -572
Add the two to get...
2H2O2 + 2H2 + O2 ==> 2H2 + 2O2 + 2H2O and cancel like terms on each side to get...
2H2O2 ==> 2H2O + O2 <==TARGET EQUATION
∆H = +376 kJ - 572 kJ = -196 kJ
We want to use the other equations that are given in order to make the target equation.
1) 2H2 + O2 ==> 2H2O ∆H = -572 kJ
2) H2 + O2 ==> H2O2 ∆H = -188 kJ
Reverse eq. 2 and multiply by 2 to get 2H2O2 on the left
2H2O2 ==> 2H2 + 2O2 ∆H = +376 kJ (change sign of ∆H and multiply by 2)
Copy eq.1
2H2 + O2 ==> 2H2O ∆H = -572
Add the two to get...
2H2O2 + 2H2 + O2 ==> 2H2 + 2O2 + 2H2O and cancel like terms on each side to get...
2H2O2 ==> 2H2O + O2 <==TARGET EQUATION
∆H = +376 kJ - 572 kJ = -196 kJ
The heat of reaction for the reaction 2H₂O₂ → 2H₂O +O₂ is -196 KJ
Calculating heat of reaction using Hess's law
From the question, we are to calculate the heat of reaction for the reaction
2H₂O₂ → 2H₂O +O₂ ∆H=?
Using Hess's law
Hess's Law of constant heat summation states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes
From the given equations,
2H₂ +O₂ → 2H₂O ∆H= -572 KJ ---------- (1)
2H₂O₂ → 2H₂ + 2O₂ ∆H= 376 KJ ----------- (2)
Adding equations (1) and (2), we get
2H₂O₂ → 2H₂O +O₂ ∆H= -572 KJ + 376 KJ
2H₂O₂ → 2H₂O +O₂ ∆H= -196 KJ
Hence, the heat of reaction for the reaction 2H₂O₂ → 2H₂O +O₂ is -196 KJ.
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