What is the standard heat of reaction (H0) for the combustion of ethane, C2H6(g), to form carbon dioxide gas and water?
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
Standard heats of formation (∆Hf⁰): C2H6 = –84.68 kJ
O2(g) = 0.00 kJ CO2(g) = –393.5 kJ H2O(l) = –285.5 kJ
dHrxn is delta H for the reaction as written and is the standard heat of reaction that the problem is asking for. n is the mols (the coefficients) in the balanced equation. dHo is delta Ho given to you in the problem.
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
So delta for the reaction is
(4*delta Hof + 6*dHo H2O) - (2*dHo C2H6 + 7*dHo O2)
I write dHo because it takes to long to type in delta Hof.
Hope this helps. BTW, if you are using multiple screen names, please don't. It helps us help you if you stick to the same name. Thanks. Another BTW. dHo for O2 is 0.
dHrxn = (n*dHo products) - (n*dHo reactants)
Substitute and solve for dH rxn. Post your work if you get stuck.
@DrBob222 what does dHrxn and n*dHo mean?
To find the standard heat of reaction (∆H₀) for the combustion of ethane, we can use the equation:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
We can calculate the standard heat of reaction by summing up the standard heat of formation (∆Hf⁰) of the products and subtracting the sum of the standard heat of formation of the reactants.
∆H₀ = ∑ (∆Hf⁰ of products) - ∑ (∆Hf⁰ of reactants)
From the given information:
∆H₀ = [4 × (∆Hf⁰ of CO2)] + [6 × (∆Hf⁰ of H2O)] - [2 × (∆Hf⁰ of C2H6)] - [7 × (∆Hf⁰ of O2)]
∆H₀ = [4 × (-393.5 kJ)] + [6 × (-285.5 kJ)] - [2 × (-84.68 kJ)] - [7 × (0.00 kJ)]
Calculating this expression:
∆H₀ = -1574 kJ + (-1713 kJ) + 169.36 kJ + 0 kJ
∆H₀ = -3118.64 kJ
Therefore, the standard heat of reaction (∆H₀) for the combustion of ethane is -3118.64 kJ.