Question

Determine the enthalpy change of the following reaction: CO + H2O -> H2 + CO2

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

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:

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

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