# Equilibrium, Kp constant, Partial Pressures

A system in equilibrium contains $$I_{2_(g)}\ \text{and} \ I_{(g)}$$ under pressures $$P_1=0,21 \,\text{atm. and}\ P_2=0,23 \,\text{atm.}$$ While its temperature remains constant, we reduce the system's volume to half its original value.

I have to determine $P_1'$ when the system's equilibrium is reestablished.

I know I can calculate $K_p$ and I know I somehow have to use $PV =NRT$, but from there on I am lost, I cannot figure out how to come up with enough equations to find all of my unknowns. Can anyone show me an explicit solution to the problem?

What I tried, but got really nowhere:

$$2I_{(g)} \leftrightarrow I_{2_{(g)}}, K_p=\frac{P_1}{P_2^2}=3,97.$$ $$P_t=P_1 + P_2 = 0,44 \,\text{atm.}$$

When the new equilibrium is established :

$$Kp=\frac{P_1'}{P_2'^2}=3,97$$

$$\frac{2P_1 }{P_1'}=\frac{n_1}{n_1'}$$ and same with $P_2, P_2', n_2, n_2'$, but I don't know where to go from there.

• Is that the entire question? Jan 22, 2018 at 5:00
• Yeah, that's the part where i am stuck. Nothing else is given that is of any interest, if that's what you're asking and i have to find a numerical value for P1'. Jan 22, 2018 at 5:01
• What are your equations and what are your unknowns? Jan 22, 2018 at 6:11
• Edited the question, hope you understand where my problem is now. Jan 22, 2018 at 6:29
• For your own understanding, please tell explicitly how many equations do you have, and how many unknowns do you have, and what are these unknowns. Jan 22, 2018 at 8:06

$$p_1'+p_2'=p_t\frac{V_t}{V_t'}=2p_t=0.88 \rm atm,$$
but this is invalid since $n_1+n_2\neq n_1'+n_2'$, thus $p_tV_t\neq p_t'V_t'$
The correct way is to use the fact that the number of iodine atoms is constant, from which we get, $$2n_1+n_2=2n_1'+n_2',$$ $$(2p_1+p_2)V_t=(2p_1'+p_2')V_t',$$ $$1.3 {\rm atm}=4p_1+2p_2=2p_1'+p_2',$$
Then you can go on by combining this with the equation you have already obtained from $K_p$ (listed in your question) and solve for $p_1'$ and $p_2'$.