Midterm 2011 Q4 - Partial Pressures and K

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Melissa Trieu 1K
Posts: 6
Joined: Fri Sep 26, 2014 2:02 pm

Midterm 2011 Q4 - Partial Pressures and K

The question says that the partial pressures of each reactant (PCl3 and Cl2) is half that of the product (PCl5). In the book the partial pressures, after calculations is equal to 0.4 for the PCl5 and 0.2 for PCl3 and Cl2. Is it also right if I said that the partial pressure for PCl5 was 0.2 and 0.1 for PCl3 and Cl2?

Sarah H Brown 1L
Posts: 56
Joined: Fri Sep 26, 2014 2:02 pm

Re: Midterm 2011 Q4

It all depends on what your equation comes down to. In the case of Q4, it is
K = 10 = 2x/x2 when the reactant pressure is set to x and the product pressure is set to 2x.
10x2= 2x
An x can be crossed out from each side.
10x = 2
x = 0.2 bar. So, the reactant pressures are 0.2 bar and product pressure is 0.4 bar.
The answer would not be the same if the concentrations were 0.1 and 0.2. That would be
(0.2)/(0.1*0.1) which is equal to 20, not 10. K does not equal 20.
So, it all depends on the K you are setting your x expression equal to!