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What is the standard enthalpy change for the reaction of 2-methylpropan-2-ol (C4H10O) with oxygen gas (O2) to form carbon dioxide (CO2) and water (H2O) at standard conditions?

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To calculate the standard enthalpy change for the reaction, we first need to write the balanced chemical equation for the combustion of 2-methylpropan-2-ol:C4H10O + 6O2  4CO2 + 5H2ONext, we need to find the standard enthalpies of formation  Hf  for each compound involved in the reaction. The values are as follows:Hf  C4H10O  = -267.3 kJ/molHf  O2  = 0 kJ/mol  since it is an element in its standard state Hf  CO2  = -393.5 kJ/molHf  H2O  = -285.8 kJ/molNow, we can use Hess's Law to calculate the standard enthalpy change for the reaction:H =  [Hf  products ] -  [Hf  reactants ]H = [4 -393.5  + 5 -285.8 ] - [ -267.3  + 6 0 ]H =  -1574 + -1429  -  -267.3 H = -3003 + 267.3H = -2735.7 kJ/molThe standard enthalpy change for the reaction of 2-methylpropan-2-ol with oxygen gas to form carbon dioxide and water at standard conditions is -2735.7 kJ/mol.

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