asked 182k views
1 vote
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

asked
User Naor
by
7.8k points

2 Answers

3 votes

Answer:

its the first answer for plato users

Step-by-step explanation:

answered
User Dlev
by
7.8k points
3 votes
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
answered
User Raffael
by
8.8k points
Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.