Determine the enthalpy change of the following reaction: co + h2o -> h2 + co2 given enthalpies: co: -110.525 kj/mol h2: 0 kj/mol h2o: -285.8 kj/mol co2: -393.5 kj/mol a. 2.825 kj/mol b. 789.825 kj/mol c. 1.007 kj/mol

I think A is correct:

deltaH=(Enthalpie of reactents)-(Enthalpie of products)=110.525+285.5-393.5=2.525~2.825

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

a is the answer to this question

explanation: