At high temperatures, carbon reacts with o2 to produce co as follows: c(s) o2(g) 2co(g). when 0.350 mol of o2 and excess carbon were placed in a 5.00-l container and heated, the equilibrium concentration of co was found to be 0.060 m. what is the equilibrium constant, kc, for this reaction?
a. 0.001
b. 0.072
c. 0.090
d. 1.2

Answer : The correct option is, (C) 0.090

Solution :  Given,

Initial moles of = 0.350 mole

volume of solution = 5.00 L

First we have to calculate the concentration .

The given equilibrium reaction is,

Initially                           0.07       0

At equilibrium             (0.07-x)    2x

The expression of will be,

As we are given the concentration of at equilibrium is, 0.060 M

That means,

2x = 0.060 M

x = 0.030 M

The concentration of at equilibrium = 0.07 - x = 0.07 - 0.03 = 0.04 M

Now put all the given values in the above expression, we get:

Therefore, the value of equilibrium constant for this reaction is, 0.090

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.

they are all solids.