[lilianag]

Initially a 2.00 L flask contains Cl2 with a partial pressure of 0.51 atm and Br2 wih a partial pressure of 0.34 atm. After equilibrium is established, the partial pressure of BrCl is 0.46 atm. Calculate the equilibrium partial pressures of Cl2 and Br2.
For this problem I did this:
kp=p(BrCl)\p(br2)p(cl2)

[kelvinLTR]

The number of moles of a gas is propotional to it's partial pressure.

Increase in partial pressure of BrCl=0.46atm

from the given equilibrium constant,

    Br₂+Cl₂  2BrCl can be derived.

therefore decrease in partial pressures of Cl₂ and Br₂=0.23atm (as one mole of Cl₂ and Br₂ produces two moles of BrCl)

new partial pressure of Cl₂=0.51atm-0.23atm=0.28atm
new partial pressure of Br₂=0.34atm-0.23atm=0.11atm

[lilianag]

so I am assuming no ICE chart is needed for this equation..

[Borek]

What kelvinLTR did is not different from what you could calculate using ICE table. You just don't have to do the final step and use Kp (which is not given) - everything is given so you can directly calculate Equilibrium pressures.

[lilianag]

if that 2.00 L flask contains 0.15 mol of each gas, what direction will the reaction proceed?
so what I did was....
Br2(g) + Cl2(g)--> 2BrCl
since its 1 to 2 moles the direction would go to the right