Use bond energies to calculate the enthalpy change for the following reaction. h2(g) + co2(g) ⟶ h2o(g) + co(g)

The given reaction is:

H2(g) + CO2(g) → H2O(g) + CO(g)

i.e.

H-H + O=C=O → H-O-H + C≡O

Reaction enthalpy (ΔH) is given as:

ΔH = ∑ΔH(Bonds broken) + ∑ΔH(Bonds formed)

     = [1*ΔH(H-H) + 2*ΔH(C=O)] + [2*ΔH(O-H) + 1*ΔH(C≡O)]

     =[+436 +2(799)] + [2(-463) +1(-1072)] = 36 kJ/mol

The enthalpy change for the given reaction is 36 kJ/mol

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890