Question

Which of the following reactions is spontaneous as written?

C_(s) + 2H_{2 (g) =} CH_{4 (g)}           

2H_{2 (g)} + O_{2 (g)} = 2H₂O_(l)      

4HCN_(g) + 5O_2(g) = 2H₂O_(l) + 4CO_2(g) 2N_2(g)

NaI_(s) = Na⁺_(aq) + I^-_(aq)

Answer 1

In the first, second and third, the change in no. of gaseous moles (∆n) are 1-2 = -1, 0-3 = -3 and 6-9 = -3, respectively.

The negative ∆n values correspond to the negative value of a change in entropy (∆S).

If the value of ∆S is positive, then only the reaction will be spontaneous.

Hence, the 1^st, 2^nd and 3^rd reactions are nonspontaneous.

The increasing order of entropy for various phases is given by

solid (S_s) < liquid (S_l) < gas (S_g)

The fourth reaction is associated with a positive value of ∆S (S_l - S_s ), as the solid NaI (strong electrolyte) is completely dissolved in water to give the corresponding solution, i.e. liquid (where the NaI ionizes to corresponding Na⁺ and I^- ions).

Therefore, the fourth reaction is spontaneous.