Question

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

Answer 1

I think A is correct:

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

Answer 2

A. The enthalpy change for the reaction is 2.825 kJ

Step-by-step explanation:

The given reaction is:

CO + H2O → H2 + CO2

The enthalpy change for a reaction is given as:

`\Delta H = \sum n(p)\Delta H_(f)^(0)(products)-\sum n(r)\Delta H_(f)^(0)(reactants)`

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

`\Delta H = [1\Delta H_(f)^(0)(H2)+1\Delta H_(f)^(0)(CO2)]-[1\Delta H_(f)^(0)(CO)+1\Delta H_(f)^(0)(H2O)]`

Substituting the given enthalpy data:

ΔH = [1(0) + 1(-393.5)] - [1(-110.525) + 1(-285.8)] = 2.825 kJ