At 100 ∘c the equilibrium constant for the reaction cocl2(g)←−→co(g)+cl2(g) has the value kc=2.19×10−10. are the following mixtures of cocl2, co, and cl2 at 100 ∘c at equilibrium? if not, indicate the direction that the reaction must proceed to achieve equilibrium. part a [cocl2]=2.00×10−3m, [co]=3.3×10−6m, [cl2]=6.62×10−6m , , the reaction is at equilibrium. the reaction will proceed right to attain equilibrium. the reaction will proceed left to attain equilibrium.

The reaction will proceed left to attain equilibrium.

Explanation:

Let´s consider the following reaction.

COCl₂(g)⇄CO(g)+Cl₂(g)     Kc = 2.19 × 10⁻¹⁰

To determine whether it is at equilibrium or not, we have to calculate the reaction quotient (Q).

Since Q > Kc, Q has to decrease its value to achieve equilibrium. To do so, it must decrease its numerator (products) and increase its denominator (reactants). All in all, the reaction will proceed left to attain equilibrium.

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.

samuel de champlain. in 1608, king henry iv sponsored pierre dugua, sieur de mons and samuel de champlain as founders of the city of quebec with 28 men. this was the second permanent french settlement in the colony of canada.