The concentrations of reactants and products for a chemical reaction can be calc
ID: 1012785 • Letter: T
Question
The concentrations of reactants and products for a chemical reaction can be calculated if the equilibrium constant for the reaction and the starting concentrations of reactants and/or products are known
. Part A Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)CO2(g)+CF4(g), Kc=8.00 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Express the molar concentration numerically
[COF2] =
Part B Consider the reaction CO(g)+NH3(g)HCONH2(g), Kc=0.680 If a reaction vessel initially contains only CO and NH3 at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH2 be at equilibrium?
Express the molar concentration numerically to three significant figures.
[HCONH2] =
Explanation / Answer
1)
consider the given reaction
2 COF2 ----> C02 + CF4
using ICE table
initial conc of COF2 , CO2 , CF4 are 2 , 0 , 0
change in conc of COF2 , C02 , CF4 are -2x , +x , +x
equilibrium conc of COF2 , CO2 , CF4 are 2-2x , x , x
now
Kc = [C02] [CF4] / [COF2]^2
now
8 = [x] [x] / [2-2x]^2
8 = [x]^2 / [2-2x]^2
8 = [ x/ (2-2x)]^2
2.82843 = x / ( 2-2x)
x = 2.82843 * 2 ( 1-x)
x = 0.85
so
at equilibrium
[COF2] = 2 - 2x
[COF2] = 2 - (2 * 0.85)
[COF2] = 0.3
so
0.3 M of COF2 will remain at equilibrium
2)
consider the given reaction
CO + NH3 ---> HCONH2
using ICE table
initial conc of CO ,NH3 , HCONH2 are 1 , 2 , 0
change in conc of CO , NH3 , HCONH2 are -x , -x , +x
equilibrium conc of CO , NH3 , HCONH2 are 1-x , 2-x ,x
now
Kc = [HCONH2] / [CO] [NH3]
0.68 = [x] / [1-x] [2-x]
0.68 ( 2 - 3x + x2) = x
1.36 - 2.04x + 0.68x2 = x
0.68x2 -3.04x + 1.36 = 0
x = 0.504
so
at equilibrium
[HCONH2] = x = 0.504 M
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