What is the enthalpy for reaction 1 reversed? reaction 1 reversed: 2co2+3h2o→c2h5oh+3o2?

Answer is: the enthalpy for reaction 1 reversed is 1370 kJ/mol.Balanced chemical reaction 1: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O  ΔrH= -1370 kJ/mol.Reaction 1 is exothermic reaction (heat is released), than reversed reaction is endothermic (chemical reactionthat absorbs more energy than it releases) and has same value for enthalpy, but positive sign.... Show More

The enthalpy for the reaction is .Further Explanation:This problem is based upon Hess’s Law. Hess’s law is utilized for calculating the enthalpy change of a reaction that can be obtained simply by summation of two or more reactions. In accordance with the Hess’s law, of the overall reaction is obtained by adding the enthalpy change for each individual step reaction involved to obtain the overall reaction. Enthalpy is defined as a state function and therefore its value depends upon the initial and final state of the system but not upon the path. This is the reason that the overall reaction can be simply obtained by adding or subtracting the enthalpy change of the individual steps utilized to get the final reaction.Combustion reactions:These are the reactions that take place when hydrocarbons are burnt in the presence of oxygen to form carbon dioxide and water. These are also referred to as burning.Example of combustion reactions are as follows: (a) (b) The given reaction is as follows:The value of is -1368 kJ.Step 1: The enthalpy change of the following reaction is .                                      …… (1)The reverse reaction for the given reaction is as follows:Step 2: The enthalpy change of the following reaction is .                                       …… (2)The reaction (2) can be obtained by reversing the reaction (1) so the value of can be obtained as follows:Learn more:1. Dissociation of ionic compounds: 2. Find the enthalpy of dissociation of 1 mole of MgO: Answer details:Grade: Senior schoolSubject: ChemistryChapter: ThermodynamicsKeywords: Hess’s Law, enthalpy, enthalpy change, CO2, H2O, O2, C2H5OH, 1368 kJ, -1368 kJ, state function, initial, final, reverse reaction, reversing, 2CO2, 3H2O.... Show More

The enthalpy of the given reaction is 1234.8kJ/mol.Explanation: Enthalpy change of the reaction is the amount of heat released or absorbed in a given chemical reaction. Mathematically, For the given reaction:Putting values in above equation, we get:... Show More

+283.0 kJ·mol⁻¹Explanation:If reaction 1 reversed isCO₂ ⟶ CO + ½O₂,then reaction 1 is CO + ½O₂ ⟶ CO₂Reaction 1 is the combustion of carbon monoxide, for whichΔH = -283.0 kJ·mol⁻¹.∴ The enthalpy change for reaction 1 reversed is +283.0 kJ·mol⁻¹.... Show More

The question is incomplete, here is a complete question.The given chemical reaction is:Reaction 1 :    ΔH = +109 kJ/molWhat is the enthalpy for reaction 1 reversed?Answer : The enthalpy of the reaction will be, -109 kJ/molExplanation :The given chemical reaction is:The enthalpy of given chemical reaction is, +109 kJ/mol.As we know that, when we are reversing the reaction then the sign of the enthalpy change of the reaction will also changed. That means, positive becomes negative and negative becomes positive.That means,    ΔH = -109 kJ/molHence, the enthalpy of the reaction will be, -109 kJ/mol... Show More

CO2→CO + 12O2 is not a balanced reaction, I think you meant to post " CO2→CO + 1/2O2"The simple answer is that the enthalphy for reaction 1 reversed is the same enthalpy as for the reaction as written but with the opposite sign.If A + B = AB with 30 kJ released thenAB = A + B with 30 kJ absorbed.We would write it asA + B = AB heat of reaction = -30 kJAB = A + B heat of reaction = +30 kJ... Show More

The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:CO2: -393.5 kJ/molCO: -99 kJ/molO2: 0 kJ/molAs observed O2 will not have enthalpy of formation as it is a pure substance. To calculate for the enthalpy of reaction,   enthalpy of formation of products - enthalpy of formation of reactants= (-99 kJ/mol) - (-393.5 kJ/mol)    = 294.5 kJ/molANSWER: 294.5 kJ/mol... Show More

The enthalpy of reaction for the given above and any other given can be calculated through the equation,     Hrxn = H(product) - H(reactants)From reliable sources, the values of H of the substances.  H(NO) = 90.25 kJ/mol  H(O2) = 0 kJ/mol  H(NO2) = 33.18 kJ/molSubstituting the known values,   Hrxn = (2)(33.18 kJ/mol) - (2)(90.25 kJ/mol)Simplifying,    Hrxn = -114.14 kJANSWER: -114.14 kJ... Show More