Question

For each of the two reactions given below, calculate ∆H ⁰_rxn , ∆ S⁰_rxn , ∆T⁰_rxn   at 25⁰C, and state whether or not the reaction is spontaneous. If the reaction is non spontaneous , would a change in temperature make it spontaneous? If so, should the temperature be lowered or raised from 25 ⁰C?

( Use the Appendix in Any Gen Chem Book)

a) H₂ ( g)   +     CO₂ (g)   ------------>    H₂O (g) + CO (g)

b) 2 NH₃ (g) ---------> N₂H₄ (g)    +   H₂ (g)

Answer 1

a)

dH = Hproducts - Hreactants

dH = (H2O (g) + CO (g)) - (H2 ( g)   +     CO2 (g) ) = (-241.8 - 110.5) - (0+ -393.5) = 41.2

dS = (H2O (g) + CO (g)) - (H2 ( g)   +     CO2 (g) ) = (188.8 + 197.7) - (130.7+213.8) = 42

dG = dH - T*dS = 41.2*10^3 - 298*42 = 28684 J/mol

this is not spontaneous, also, it must have high T in order to become spontanoeus, so dG becomes negative.

b)

dH = Hproducts - Hreactants

dH = (N2H4 (g)    +   H2 (g)) - (2 NH3 (g)   ) = ( 95.395 + 0 ) - (2*-45.9) = 187.195

dS = (N2H4 (g)    +   H2 (g)) - (2 NH3 (g)   ) = ( 238.36 + 130.7) - (2*192.8) = -16.54

dG = dH - t*DS = 187.195*1000 - 298 * -16.54 = 192123.92J/mol.. not spontaneous as written

this is never spotnaneous, since dH is positive and dS negative