Question

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)

Answer 1

The correct formation equation for the standard enthalpy of formation for glucose is 6C (s, graphite) + 6H2 (g) + 3O2 (g) → C6H12O6 (s) and corresponds to a ΔH°f of −1273.3 kJ/mol.

Step-by-step explanation:

The correct formation equation corresponding to the standard enthalpy of formation for glucose, −1273.3 kJ/mol, involves combining elements in their most stable forms at 25°C to form glucose. According to the information provided, graphite is the most stable form of carbon at these conditions. Therefore, the correct reaction equation would be:

6C (s, graphite) + 6H2 (g) + 3O2 (g) → C6H12O6 (s), with a ΔH°f of −1273.3 kJ/mol.

The other options are incorrect because they do not use the stable forms of the elements at 25°C or appropriate states of matter. For example, H2 and O2 should be gases, not liquids, at standard conditions.

Answer 2

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