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
Answer:
SPECIFIC HEAT CAPACITY
Explanation:
Q = MC DELTA T
q = heat
c = specific heat
T = temperature ( final - initial )
Answer:
closer to F because fluorine has a higher electronegativity than carbon
Explanation:
Electronegativity refers to the ability of an atom in a bonding situation to draw the shared electrons of the bond closer to itself.
Electronegativity increases across the period and decreases down the group. A highly electronegative atom draws the shared electron pair of a bond towards itself.
When two atoms are bonded together, the electron pair is always drawn closer to the atom that has a higher electronegativity.
Hence, the electron pair in a C-F bond could be considered closer to F because fluorine has a higher electronegativity than carbon.
Hybridisation influences the bond length and bond enthalpy strength in organic compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp3 hybrid orbital.
