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

The answer is the first option just took the test

The correct answer is "false"