Which equilibrium system least favors the formation of products?
2hgo(s) 2hg(l) + o2(g) keq = 1.2 x 10-22
ch3cooh(aq) + h2o(l) h3o+(aq) + ch3coo-(aq) keq = 1.8 x 10-5
2no(g) + o2(g) 2no2(g) keq = 6.5 x 105
2(g) + cl2(g) 2hcl(g) keq = 1.8 x 1033

The products are least favoured when the Keq value is the lowest it can be. Since the Keq value is calculated by [products]/[reactants], the higher the Keq value, the more the products are favoured. 

So, just by comparing all the Keq values, you can say that the decomposition of Mercury (II) Oxide has the lowest Keq value, and so has the system that least favours the formation of the products.

Hope I helped!! xx

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.