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

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.