Question

Suppose a 250. mL flask is filled with 0.10 mol of Cl₂ and 1.4 mol of HCl. The following reaction becomes possible:

H₂(g)+Cl₂(g)=2HCl(g)

The equilibrium constant for this reaction is 0.414 at the temperature of the flask.

Calculate the equilibrium molarity of . Round your answer to two decimal places.

Answer 1

you of want H₂ cg to at find the molar concentration equilibrium. molar concentration of de a 0.10 0.40M 0.250 molar concentration of Hd = 1.4 5.60m 0.250 side b/c sital This duoction goes conco of Hq Lgs is left o

go H₂ & clog o 0.40m 2 Ha y 5.60m C + x + x - 2x E (x) (0.40+x) (5.60-2x) ke- [HAJ? [de] Hg 0,4 14 - - (5.60-2x e (0.404) (x2 10.4 14 = 31.36 + 4 x² - 22.48. 0.40x + x² 0.414 42 -0.1658 = 31.36+ 4x² – 22.48 3.5872 - 22.235X +31:36 x = 4.042 and 2.166 So, only 2.166 is right and were 4.04% is not right answer because if we but - value of x in concentration of Hel

I Hal = 5.60 - 2x = 5.60-9x4.042 = -ve. bo bo, it is negligible , then this value is not considered So the value of x = 2.166 [ H2] = 2.166