tonyhussam
contestada

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

Respuesta :

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:

[tex]\Delta H = \sum n(p)\Delta H_{f}^{0}(products)-\sum n(r)\Delta H_{f}^{0}(reactants)[/tex]

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

[tex]\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)][/tex]

Substituting the given enthalpy data:

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

Otras preguntas