The standard enthalpy of formation for glucose [c6h12o6 (s) ] is - 1273.3 kj/mol. what is the correct formation equation corresponding to this δhof? the standard enthalpy of formation for glucose is. what is the correct formation equation corresponding to this? 6c (s, graphite) + 6h2o (l) →c6h12o6 (s, glucose) 6c (s, graphite) + 6h2 (l) + 3o2 (l) →c6h12o6 (s, glucose) 6c (s, graphite) + 6h2 (g) + 3o2 (g) →c6h12o6 (s, glucose) 6c (s, graphite) + 6h2o (g) →c6h12o6 (s, glucose)

The standard formation equation for glucose C6H12O6 (s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = - 1273.3 kJ/mol is

C (s) + H2 (g) + O2 (g) → C6H12O6 (s)

and the balanced chemical equation is

6C (s) + 6H2 (g) + 3O2 (g) → C6H12O6 (s)

Using the equation for the standard enthalpy change of formation

ΔHoreaction = ∑ΔHof (products) - ∑ΔHof (Reactants)

ΔHoreaction = ΔHfo[C6H12O6 (s) ] - {ΔHfo[C (s, graphite) + ΔHfo[H2 (g) ] + ΔHfo[O2 (g) ]}

C (s), H2 (g), and O2 (g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:

ΔHoreaction = [1*-1273.3] - [ (6*0) + (6*0) + (3*0) ]

= - 1273.3 - (0 + 0 + 0)

= - 1273.3