Use bond energies to calculate the enthalpy change for the following reaction. H2(g) + CO2(g) → H2O(g) + CO(g)

Explanation:

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

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the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.