Question

Suppose a

250.mL

flask is filled with

1.0mol

of

NO2

,

0.40mol

of

CO

and

0.10mol

of

CO2

. The following reaction becomes possible:

+NO2gCOg

+NOgCO2g

The equilibrium constant

K

for this reaction is

7.16

at the temperature of the flask.

Calculate the equilibrium molarity of

NO2

. Round your answer to two decimal places.

Answer 1

The equilibrium molarity of NO2 is

2.55 M
MIcar - Given, equilibria, No29) + Cogs (Initially) 1.0 0.40 (change) - -x -x ( At equilibrium) (1-1) (0.4o.n) o.10 + + x (0.lotu since, equilibrium constant, k- Nonco | 1 NO,. hto 7:16. = uxlolot (1-x ) (0.40-2) - 7.16' Olon + 2? 0:40 - 1,4ox + x² - 2.864 - 10.024 x + 7.16x2 = 0.lont 6:16 x2 - 10:124 x + 21864 = 0 So, By quadratic formula :- x= -(-10.124) + (-10.124) ² - 4 X 6:16 X 2.86 2x 6.16 te:124 + 31:92 = 10.124 I 5.649 12:32 12:32 So, 2= 10.124 + 5-649 Ce x1 = 10:124-51649 12.32 12:32 1.280 x= 0.363 since, (x = 1.280 gives conchs negative. So, At equilibrium, [NO, I = 1-1 - 1-01363 = 2.548 M 250xlo 0.25