Carbonylfluoride COF2 is an important intermediate used in theproduction of fluo
ID: 686583 • Letter: C
Question
Carbonylfluoride COF2 is an important intermediate used in theproduction of fluorine-containing compounds. For instance, it isused to make the refrigerant carbon tetrafluoride, CF4 via thereaction
2COF2(g) left and right arrows CO2(g) + CF4(g), Kc =6.30
If only COF2 ispresent initially at a concentration of 2.00 M, what concentrationof COF2 remains at equilibrium?
Part B
Consider thereaction
CO (g) + NH3 (g) left and right arrowsHCONH2(g), Kc =0.810
If a reactionvessel initially contains only CO and NH3 at concentrations of 1.00M and 2.00 M, respectively, what will the concentration of HCONH2be at equilibrium?
Expressthe molar concentration numerically.
Explanation / Answer
PATR - A --------------- 2COF2(g) <-----> CO2(g) + CF4(g), Kc= 6.30 initial conc.(in M) 2 0 0 atEquib. 2-2x x x Equilibrium constant , K = ( [CO2] [CF4] ) / [COF2]2 6.3 = ( x * x ) / ( 2-2x ) 2 solving for x we get x= 0.838 M concentration of COF2 remains at equilibrium = 2 - 2x = 2 -2*0.838 = 0.324 M PART-B ---------- CO (g) + NH3 (g) <------> HCONH2(g), Kc =0.810initial conc.(in M ) 1 2 0 atEquib. 1- x 2 -x x Equilibrium constant , K = ( [HCONH2 ] / ([CO] ) [NH3] 0.810= x / [ (1-x)( 2-x ) ] solving for x we get x= 0.429 M the concentration of HCONH2 be at equilibrium = 0.429M initial conc.(in M ) 1 2 0 atEquib. 1- x 2 -x x Equilibrium constant , K = ( [HCONH2 ] / ([CO] ) [NH3] 0.810= x / [ (1-x)( 2-x ) ] solving for x we get x= 0.429 M the concentration of HCONH2 be at equilibrium = 0.429M