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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

the answer is c.the valence of the atoms combine .