Determine the enthalpy change of the following reaction: CO + H2O -> H2 + CO2
2 answers:
I think A is correct:
deltaH=(Enthalpie of reactents)-(Enthalpie of products)=110.525+285.5-393.5=2.525~2.825
Answer:
A. The enthalpy change for the reaction is 2.825 kJ
Explanation:
The given reaction is:
CO + H2O → H2 + CO2
The enthalpy change for a reaction is given as:
where np and nr are the number of moles of products and reactants
ΔH⁰f are the standard enthalpies of formation of the respective reactants and products
Substituting the given enthalpy data:
ΔH = [1(0) + 1(-393.5)] - [1(-110.525) + 1(-285.8)] = 2.825 kJ
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