CONSTANT EQUILIBRIUM GCHEM II LAB CONCEPT PROBLEMS
1) Consider following equilibrium equation: H2 (g) + I2(g) 2HI(g) at 450.0 K. If 5.00E-3 moles
of H2 and 2.00E-2 moles of I2 are placed in a 5.00 L flask and allowed to come to equilibrium. The flask
was analyzed and equilibrium concentration of the resulting HI was found to be 1.87E-3 M. Calculate
the Kc for the reaction.
R eaction
H2(g)
+
-3
I nitial (moles/L)
1.00 x 10
C hange (+/)
-9.350 x 10-4
E quilibrium
6.500 x 10-5
I2(g)
2HI(g)
-3
4.00 x 10
0
-9.350 x 10-4
+1.87 x 10-3
3.065 x 10-3
1.87 x 10-3
5.00 x 10-3 moles H2/5.00 L = 1.00 x 10-3 M H2
Kc = [HI]2
=
(1.87 x 10-3)2
[H2] [I2]
(6.50x10-5) (3.065x10-3)
at 450.0 K
and
2.00 x 10-2 moles I2/5.00 L = 4.00 x 10-3 M I2
= 3.4969 x 10-6
= 17.552
Kc = 17.6
1.99225 x 10-7
2) Consider the following equilibrium equation: H2(g) + Br2(g) 2HBr(g). A mixture of 1.374
grams of H2 and 70.31 grams of Br2 is heated in a 2.00 L flask to 700.0 K. At equilibrium the vessel is
found to contain 0.566 grams of H2. Calculate the Kc for the reaction.
R eaction
I nitial (moles/L)
C hange (+/)
E quilibrium
1.374 grams H2
+
Br2(g)
.21999
.2004
.01959
2HBr(g)
0
+ .4008
.4008
at 700.0 K
1 mole H2
. = .68154 moles H2/ 2.00 L = .34077 M
2.016 grams H2
1 mole Br2
. = .43998 moles Br2/ 2.00 L = .21999 M
159.8 grams Br2
1 mole H2
.
= .2807 moles H2/ 2.00 L = .14037 M
2.016 grams Hr2
70.31 grams Br2
.566 grams H2
Kc = [HBr]2
[H2] [Br2]
H2(g)
.34077
.2004
.14037
=
(.4008)2
=
(.14037) (.01959)
.16064
.002749
=
58.43
=
Kc = 58.4
3) A mixture of H2 and N2 gas is allowed to reach equilibrium with NH3 at 450.5 K and total pressure of
9.907 atm.. The equilibrium mixture was analyzed and found to contain 7.38 atm H2 and 2.46 atm N2.
Calculate the equilibrium constant.
R eaction
I nitial (moles/L)
C hange (+/)
E quilibrium
N2(g)
+
3H2(g)
2.46 atm
7.38 atm
2NH3
at 450.0 K
0.166 atm
P total = pPH2 + pP N2 + pP NH3
9.907 = 7.38 + 2.46
+ Pp NH3
[NH3]2
[N2] [H2]3
[.166]2
[2.46] [7.38]3
Kc =
=
Pp NH3 = 9.907 (7.38 + 2.46)
=
.02756
988.790
= 2.7872 x 10-5
PpNH3 = .166 atm
Kc = 2.7 X 10-5
4) Consider equilibrium reaction: H2(g) + I2 (g) 2HI (g) Kc = 2.25 at 450.0 o C. A mixture of
9.00 x 10 3 moles of H2 and I2 is placed in a 3.00 L flask and allowed to come to equilibrium. What is
equilibrium concentration reactants of and products ?
R eaction
H2
+
I2
2HI at 450.0 K
I nitial (moles/L)
9.00x10-3/3.00
9.00 x 10-3/3.00
0
C hange (+/)
x
x
+ 2x
E quilibrium
.00300 x
.00300 x
2x
9.00 x 10-3 moles H2/3.00 L = .003000 M H2 and
Kc =
9.00 x 10-3 moles I2/3.00 L = .003000 M I2
[HI]2 =
(2x)2
= 4x2
= 2.25 =
2.00x = 1.500
[H2] [I2] (.003000 x) (.003000 x) (.00300 x)2
(.00300 x)
2x = 2(0.001286) = 0.00257
x = 0.001286
[H2] = 0.00171, [I2] = 0.00171, [HI] = 0.00257
5) Consider equilibrium equation: Br2(g) + Cl2(g) 2 BrCl(g) Kc = 1.25E-6 at 400.0 o C. If 0.350
moles Br2 in a 2.00 L flask and 0.550 moles of Cl2 in a 4.00 L flask are combined into a 8.00 L flask and
allowed to reach equilibrium, what is the equilibrium concentration of the BrCl (g) ?
R eaction
Br2(g)
+
Cl2(g)
2BrCl(g)
at 400.0 K
I nitial (moles/L)
.04375
.06075
0
C hange (+/)
x
x
+2x
E quilibrium
.04375 x
.06075 x
2x
.350 moles Br2/8.00 L = .04375 M Br2 and .550 moles Cl2/8.00 L = .06875 M Cl2
Kc = 1.25 x10-4 = [ClBr]2
[Br2] [Cl2]
1.25 x 10-6 =
4x2
.003008
=
(2x)2
note: Kc (1.25 x 10-6) < 2 powers of
(.04375 x) (.06875 x)
of ten compared to [Br2] .04375
x2 = 9.400 x10-10
x = .0003066 [ClBr] = 2 (.0003066)
[BrCl] =6.13 x 10-4
6) Consider equation for deuterium a hydrogen isotope: D2(g) + H2(g) 2DH(g) Kc = 7.82 at 500.0 K.
Initial concentration of D2 is 1.08 M and H2 is 1.05 M . What is equilibrium concentration of DH(g) ?
R eaction
D2(g)
+
H2(g)
2DH(g)
at 500.0 K
I nitial (moles/L)
1.08
1.05
0
C hange (+/)
x
x
+2x
E quilibrium
1.08 x
1.05 x
2x
[DH]2
=
[2x]2
=
4x2
= 7.82
2
[D2] [H2] [1.08 x] [1.05 x]
x 2.13x + 1.134
7.82x2 16.656x + 8.867 = 4x2
3.82x2 16.656x + 8.967 = 0
(x = .6207 & ( x = 3.739 negative number) [DH] = 2(.6207) =
Kc =
[DH] = 1.24
7) Consider equation: 2 IBr(g) I2(g) + Br2(g) Kc = 2.80 at 250.5 o C. Suppose that 0.500 moles
BrI in a 4.00 L flask reaches equilibrium . What are equilibrium concentration of IBr and Br2 ?
R eaction
2 IBr(g)
I2 (g)
+ Br2 (g)
at 250.0 K
I nitial (moles/L)
.125
0
0
C hange (+/)
2x
+x
+x
E quilibrium
.125 2x
x
x
0.500 moles BrI/ 4.00 L = .125 M BrI
Kc = 2.80 = [I2] [Br2]
2.80 = (x * x)
1.673 =
x
1.673 (.125 2x) = x .209 3.346x = x
[IBr]2
(.125 - 2x)2
(.125 - 2x)
x = 0.04809
BrI = (.125 2(.04809)) = .02890
[IBr] = 0.0289 M
[Br2] = .0481 M
jpechunf07/2004