A quantity of HI was sealed in a 2.0 L canister, heated to 425 degrees C, and he
ID: 499621 • Letter: A
Question
A quantity of HI was sealed in a 2.0 L canister, heated to 425 degrees C, and held a this temperature untik equillibrium as reached. The concentration of HI in the tube at equillibrium was founnd to be 0.0706 moles/ liter.
a. calculate the equillibrium concentration of H2 and I2 from the given information.
b. After the system was allowed to reach equillibrium and the concentrations calculated in (a), the equillobrium was disrupted by the addition of 71.5 g HI. Calculate the new equillibrium concentrations of HI, H2 and I2 from the given information.
thank you for the help!
Explanation / Answer
V = 2 L so:
[HI]initial = 0.0706 M
Kc = 54.6
Kc =[HI]^2 /[H2][I2]
a)
in equilibrium:
[HI] = 0.0706 - 2x
[H2] = 0 + x
[I2] 0 +x
substitute in KC
Kc =[HI]^2 /[H2][I2]
1/54.6 = (x*x) / (0.0706 - 2x)^2
sqrt(1/54.6) = x / (0.0706 - 2x)
0.1353 ( 0.0706-2x) = x
0.1353 ( 0.0706) - 2*0.1353x = x
(1+2*0.1353) x = 0.1353 ( 0.0706)
x = 0.00955218 / 1.2706
x =0.0075178
[HI] = 0.0706 - 2x = 0.0706 - 2*0.0075178 = 0.0555644
[H2] = 0 + x = 0.0075178
[I2] 0 +x = 0.0075178
Q = (0.0555644^2) / (0.0075178^2) = 54.62
b)
we add;
m = 71.5 g of HI
mol = mass/MW = 71.5/127.911 = 0.5589
M = mol/V = 0.5589 / 2 = 0.27945 M of HI added:
this will occur:
initially:
[HI] = 0.0555644 + 0.27945 = 0.3350144
[H2] = 0.0075178
[I2] = 0.0075178
after adding HI:
[HI] = 0.3350144 - 2x
[H2] = 0.0075178 + x
[I2] = 0.0075178 + x
substitute
1/kc = [H2][I2] / ([HI]
1/54.6 = x*x/( 0.3350144 - 2x)
sqrt(1/54.6) =x /( 0.3350144 - 2x)
0.13533299 * 0.3350144 -2*0.3350144 x = x
0.0453 = (1+ 0.6700288)x
x = 0.0453 / (1+ 0.6700288) = 0.02712
so
[HI] = 0.3350144 - 2x = 0.3350144 -2*0.02712 = 0.2807744
[H2] = 0.0075178 + x = 0.3350144 +0.02712 = 0.3621344
[I2] = 0.0075178 + x = 0.3350144 +0.02712 = 0.3621344