Calculate the enthalpy of formation (kJ/mol) of HCOOH( l ).The enthalpy of react

Question

Calculate the enthalpy of formation (kJ/mol) of HCOOH(l).The enthalpy of reaction for the equation as written is -540.66kJ/mol. If the answer is negative, enter the sign and then themagnitude.

2 HCOOH(l) + O2(g) 2CO2(g) + 2 H2O(l) .

the tolerance is +/-2%

Calculate the enthalpy of formation (kJ/mol) of HCOOH(l).The enthalpy of reaction for the equation as written is -540.66kJ/mol. If the answer is negative, enter the sign and then themagnitude.

2 HCOOH(l) + O2(g) 2CO2(g) + 2 H2O(l) .

Hof (kJ/mol) HCOOH(l) ? O2(g) 0 CO2(g) -393.5 H2O(l) -285.83

Explanation / Answer

          2HCOOH(l) + O2(g) 2CO2(g) + 2 H2O(l).               Hrxn = Hof (products)- Hof (reactants)                          =  [2Hof CO2 +2Hof H2O] - [Hof O2 +2Hof HCOOH ]    -540.66 kJ/mol = [ 2*(-393.5) +2*(-285.83) ] - [ 0 +2Hof HCOOH]                   Hof HCOOH = -409 kJ/mol

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