The equation fo Enthalpy of reaction
dHRxn = Hproducts- Hreactants
Products = CH3F and H2O
Reactants = CH3OH and HF
enthalpy of products =
CH3F = -223 kJ/mol
H2O = −285.830 kJ/mol
CH3OH =−238.7 kJ/mol
HF = −273.30
then..
dHRxn = Hproducts- Hreactants
dHRxn = Hproducts- Hreactants = ( -223 + -285.830 ) - (-238.7 + -273.30 ) = 3.17 kJ/mol
Need step by step solution, thank you Calculate Delta H^0 for the reaction. Enter your answer...
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