Hydrogen is one of only seven elements which exist as stable diatomic molecules at (or close to) room temperature and atmospheric pressure. Let’s investigate just how much more thermodynamically favorable diatomic hydrogen is compared to atomic hydrogen. Given the following reaction and associated data at T = 298.15 K.                2 H(g) ⇌ H"(g)    or equivalently    H(g) + H(g) ⇌ H2(g)             Δf?° (kJ mol-1)                                ?° (kJ K-1 mol-1) H(g)         218.0                                                          0.115 H2(g)           0                                                             0.131 Calculate ΔH, ΔS, and ΔG for the formation of H2(g) from H(g) at 298.15 K. Calculate KP for the reaction. Calculate the temperature at which the reverse reaction becomes favorable. Assume ΔH and ΔS do not change with temperature.

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Hydrogen is one of only seven elements which exist as stable diatomic molecules at (or close to) room temperature and atmospheric pressure. Let’s investigate just how much more thermodynamically favorable diatomic hydrogen is compared to atomic hydrogen.

Given the following reaction and associated data at T = 298.15 K.   

            2 H(g) ⇌ H"(g)    or equivalently    H(g) + H(g) ⇌ H2(g)

            Δf?° (kJ mol-1)                                ?° (kJ K-1 mol-1)

H(g)         218.0                                                          0.115

H2(g)           0                                                             0.131

  1. Calculate ΔH, ΔS, and ΔG for the formation of H2(g) from H(g) at 298.15 K.
  2. Calculate KP for the reaction.
  3. Calculate the temperature at which the reverse reaction becomes favorable. Assume ΔH and ΔS do not change with temperature.

 

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