mr j
Lv 4

# how do i find the value of K_C when.......?

a 0.682 g sample of ICl (g) is place in a 625 mL reaction vessel at 682 K. when the equilibrium is reached between the ICl (g) and I_2 (g) and Cl_2 (g) is formed by its dissociation, 0.0383 g of I_2 (g) is present.

i already know that .682 g ICl is 110.728 mol and .0383 g I_2 is 9.72088 mol. in a 625 mL container they are 177.165 M and 15.5534 M respectively. how do i use this information to find the K_C value?

Relevance
• Dr.A
Lv 7

moles ICl = 0.682 g /162.357 g/mol= 0.00420

initial concentration ICl = 0.00420 mol / 0.625 L=0.00672 M

moles I2 = 0.0383 g /253.808 g/mol=0.000151

concentration I2 at equilibrium = 0.000151 / 0.625 L=0.000241 M

2 ICl <------> Cl2 + I2

start

0.00672

change

-2x. . . . .. .+x . .. .+x

x = 0.000241

2x = 0.000482

[ICl] at equilibrium = 0.00672 - 0.000482 =0.00624 M

Kc = [Cl2][I2] / [ICl]^2 = (0.000241)( 0.000241) / 0.00624 = 9.31 x 10^-6

• Chinese6 years agoReport

Kc = [ Cl2 ] [ I2 ] / [ ICl ]^2 = [0.000241] [0.000241] / [ 0.00624M ]^2 = 0.00149