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Calculate the heat of formation of butane (C4H10) using the following balanced equation. Write out the solution according to Hess's Law. C(s) + O2(g) -> CO2(g) ∆H^0f = -392.7 kJ/mol H2(g) + 1/2 O2(g) -> H2O(l) ∆H^0f = -280.7 kJ/mol 4 CO2(g) + 5 H2O(l) -> C4H10(g) + 13/2 O2(g) ∆H^0c = 2877.6 kJ/mol Answer choices are: -110.5 -96.7 -135.6 -146.9 -143.7 -159.2 -115 -127.1 -140.7 Answer in units of kJ
Added by Charles C.
Step 1
According to the balanced equation, 1 mole of butane is formed from 4 moles of carbon (C) and 5 moles of hydrogen (H2). Now, we can use Hess's Law to find the heat of formation for butane. Hess's Law states that the total enthalpy change for a reaction is the sum Show more…
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The heat of combustion of butane is 2877 kJ/mol. Use this value to find the heat of formation of butane. Use the data given below: Substance AHf (kJ/mol) 3935 -285.8 CO2(g) H2O(l) O2(g) Heat of formation (kJ/mol)
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The combustion reaction of butane is as follows: C4H10(g) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(l) Using Hess's law and the reaction enthalpies given below, find the change in enthalpy for this reaction. Reaction (1): C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ/mol Reaction (2): H2(g) + 1/2 O2(g) → H2O(l) ΔH = -285.8 kJ/mol Reaction (3): 4 C(s) + 5 H2(g) → C4H10(g) ΔH = -125.7 kJ/mol
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