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

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

< 3

i'm pretty sure it is letter c.) car