InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Consider the reaction `X(g) hArr Y(g)+Z(g)` When the system is at equilibrium at `100^(@)`, the concentrations are found to be `[X]=0.2 M, [Y]=[Z]=0.4 M` a. If the pressure of the container is suddenly halved at `100^(@)C`, find equilibrium concentration. b. If the pressure of the container is suddenly doubled at `100^(@)C`, find the equilibrium concentration |
|
Answer» `{:(X,hArr,Y,+,Z,),(0.2,,0.4,,0.4,"At equilibrium"):}` `K_(c )=(0.4xx0.4)/0.2=0.8` a. When pressure is halved, `(i.e. P rarr P/2)`, concentration is halved. `:. [X]=0.1 M, [Y]=[Z]=0.2 M` `Q_(c )=(0.2xx0.2)/(0.1)=0.4` `:. Q_(c ) lt K_(c )` (Reaction goes forward) `K_(c )=0.8=((0.2+x)(0.2+x))/((0.1-x))` solving for `x`: `x=(-0.4+sqrt(0.16+4xx0.04))/2=0.08` `:.` Equilibrium concentration: `[X]=0.1-x=0.1-0.08=0.02 M` `[Y]=[Z]=0.2+x=0.2+0.08=0.28 M` b. When pressure is doubled (i.e., `P rarr 2P)`, concentration is doubled. `:. [X]=0.4 M [Y]=[Z]=0.8 M` `Q_(c)=(0.8xx0.8)/0.4=1.6` `:. Q_(c) gt K_(c)` (Reaction goes backward) `K_(c)=0.8=((0.8-x)(0.8-x))/((0.4+x))` Solving for x: `x^(2)-0.4x+0.32=0` `x=(-(-0.4)+sqrt(0.16-1.28))/(2)~~0.73` `:.` Equilibrium concentration: `[X]=0.4+x=0.4+0.73=1.13 M` `[Y]=[Z]=0.8-x=0.8-0.73=0.07 M` |
|