Homework 5l. 29

Jessica Booth 2F
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

Homework 5l. 29

At 25$^{\circ}C$ K = 3.2 x $10^{-34}$ for the reaction 2HCl$\rightleftharpoons$ H2 + Cl2. If a reaction vessel of volumme 1.0L is filled with HCl at .22 bar, what are the equilibrium partial pressures of HCl, H2, and Cl2.

Ruby Tang 2J
Posts: 102
Joined: Fri Aug 30, 2019 12:15 am

Re: Homework 5l. 29

For this problem, you can set up an ICE table, but instead of using the concentrations of the compounds, you use the partial pressures instead. You get:

2HCl (g) --> H2(g) + Cl2(g)
I 0.22 0 0
C -2x +x +x
E 0.22-2x x x

We know that Kp = PH2*PCl2/(PHCl)^2, which we can then plug our equilibrium expressions for each compound into: 3.2*10^-34 = x^2/(0.22-2x)^2. When you solve for x, you get +- 3.94*10^-18, but we know that x cannot be negative, because you cannot have a negative concentration of a compound. Therefore, if we plug in 3.94*10^-18 as x, then we get PHCl = 0.22 bar, PH2 = PCl2 = 3.9*10^-18 bar. Hope this helps!

Dina Marchenko 2J
Posts: 54
Joined: Thu Jul 25, 2019 12:16 am

Re: Homework 5l. 29

2HCl(g)->H2(g) + Cl2(g)
0.22 bar 0 0 initial
-2x +x +x change
0.22-2x +x +x final

Partial P(H2) x Partial P (Cl2)
K=------------------------------------
Partial P (HCl)^2
3.2x10^-34= x^2/(0.22-2x)^2 (square root both sides)
1.8x10^-17=x/0.22-2x (multiply both sides by denominator, distribute 1.8x10^-17 into parentheses and isolate x onto one side)
x=3.9x10^-18

plug in x into the above values (see final part of ice table)

Partial Pressure (H2)= 3.9x10^-18
Partial Pressure (Cl2)= 3.9x10^-18
Partial Pressure (HCl)= 0.22-2(3.9x10^-18)= 0.22