The ipR.O.B.O.T states
aA+bB⇌ cC+dD
the equilibrium constant is written as follows:
Kc=[C]c[D]d[A]a[B]b
The ICE Table
The easiest approach for calculating equilibrium concentrations is to use an ICE Table, which is an organized method to track which quantities are known and which need to be calculated. ICE stands for:
"I" is for the "initial" concentration or the initial amount
"C" is for the "change" in concentration or change in the amount from the initial state to equilibrium
"E" is for the "equilibrium" concentration or amount and represents the expression for the amounts at equilibrium.
For the gaseous hydrogenation reaction below, what is the concentration for each substance at equilibrium?
C2H4(g)+H2(g)⇌C2H6(g)(1)
with Kc=0.98 characterized from previous experiments and with the following initial concentrations:
[C2H4]0=0.33
[H2]0=0.53
SOLUTION
First the equilibrium expression is written for this reaction:
Kc=[C2H6][C2H4][H2]=0.98(2)
ICE Table
The concentrations for the reactants are added to the "Initial" row of the table. The initial amount of C2H6 is not mentioned, so it is given a value of 0. This amount will change over the course of the reaction.
ICE
C2H4
H2
C2H6
Initial
0.33
0.53
0
Change
Equilibrium
ICE
C2H4
H2
C2H6
Initial
0.33
0.53
0
Change
-x
-x
+x
Equilibrium
Equilibrium is determined by adding "Initial" and "Change together.
ICE
C2H4
H2
C2H6
Initial
0.33
0.53
0
Change
-x
-x
+x
Equilibrium
0.33-x
0.53-x
x
The expressions in the "Equilibrium" row are substituted into the equilibrium constant expression to find calculate the value of x. The equilibrium expression is simplified into a quadratic expression as shown:
0.98=x(0.33−x)(0.53−x)(3)
0.98=xx2−0.86x+0.1749(4)
0.98(x2−0.86x+0.1749)=x(5)
0.98x2−0.8428x+0.171402=x(6)
0.98x2−1.8428x+0.171402=0(7)
The quadratic formula can be used as follows to solve for x:
x=−b±b2−4ac−−−−−−−√2a(8)
x=−0.1572±(−0.1572)2−4(0.98)(0.171402)−−−−−−−−−−−−−−−−−−−−−−−−−√2(0.98)(9)
x=1.78 or0.098(10)
Because there are two possible solutions, each must be checked to determine which is the real solution. They are plugged into the expression in the "Equilibrium" row for [C2H4]Eq :
[C2H4]Eq=(0.33−1.78)=−1.45(11)
[C2H4]Eq=(0.33−0.098)=0.23(12)
If x=1.78 then [C2H4]Eq is negative, which is impossible, therefore, x must equal 0.098.
So:
[C2H4]Eq=0.23M(13)
[H2]Eq=(0.53−0.0981)=0.43M(14)
[C2H6]Eq=0.098M(15)
Problems
1. Find the concentration of iodine in the following reaction if the equilibrium constant is 3.76 X 103, and 2 mol of iodine are initially placed in a 2 L flask at 100 K.
I2(g)⇌2I−(aq)(16)
2. What is the concentration of silver ions in 1.00 L of solution with 0.020 mol of AgCl and 0.020 mol of Cl- in the following reaction? The equilibrium constant is 1.8 x 10-10.
AgCl(s)⇌Ag+(aq)+Cl−(aq)(17)
3. What are the equilibrium concentrations of the products and reactants for the following equilibrium reaction?
Initial concentrations: [HSO−4]0=0.4 [H3O+]0=0.01 [SO2−4]0=0.07 K=.012
HSO−4(aq)+H2O(l)⇌H3O+(aq)+SO2−4(aq)(18)
4. The initial concentration of HCO3 is 0.16 M in the following reaction. What is the H+ concentration at equilibrium? Kc=0.20.
H2CO3⇌H+(aq)+CO2−3(aq)(19)
5.The initial concentration of PCl5 is 0.200 moles per liter and there are no products in the system when the reaction starts. If the equilibrium constant is 0.030, calculate all the concentrations at equilibrium.
Solutions
1.
I2
I−
Initial
2mol/2L = 1 M
0
Change
−x
+2x
Equilibrium
1−x
2x
At equilibrium
Kc=[I−]2[I2]
3.76×103=(2x)21−x=4x21−x
cross multiply
4x2+3.76.103x−3.76×103=0
apply the quadratic formula:
−b±b2−4ac−−−−−−−√2a
with: a=4 , b=3.76×103 c=−3.76×103 .
The formula gives solutions of of x=0.999 and -940. The latter solution is unphysical (a negative concentration). Therefore, x=0.999 at equilibrium.
[I−]=2x=1.99M(20)
[I2]=1−x=1−.999=0.001M(21)
2.
Ag+
Cl−
Initial
0
0.02mol/1.00 L = 0.02 M
Change
+x
+x
Equilibrium
0.02+x
Kc=[Ag−][Cl−](22)
1.8×10−10=(x)(0.02+x)(23)
x2+0.02x−1.8×1010=0(24)
x=9×10−9(25)
[Ag−]=x=9×10−9(26)
[Cl−]=0.02+x=0.020(27)
3.
H2CO3
SO2−4
H3O+
Initial
0.4
0.01
0.07
Change
−x
Equilibrium
0.4−x
0.01+x
0.07+x
Kc=[SO2−4][H3O+]H2CO3(28)
0.012=(0.01+x)(0.07+x)0.4−x(29)
cross multiply and get:
x2+0.2x−0.0041=0(30)
apply the quadratic formula
x = 0.0328
[H2CO3]=0.4-x=0.4-0.0328=0.3672
[S042-]=0.01+x=0.01+0.0328=0.0428
[H30]=0.07+x=0.07+0.0328=0.1028
4.
H2CO3
H+
CO2−3
Initial
.16
0
Change
-x
Equilibrium
.16-x
apply the quadratic equation
x=0.1049
[H+]=x=0.1049
5. First write out the balanced equation:
PCl5(g)⇌PCl3(g)+Cl2(g)
PCl5
PCl3
Cl2
Initial
0.2
0
Change
-x
Equilibrium
0.2-x
Kc=[PC3][Cl2][PCl5](31)
0.30=x20.2−x(32)
Cross multiply:
x2+0.03x−0.006=0(33)
Apply the quadratic formula:
x=0.064
[PCl5]=0.2-x=0.136
[PCl3]=0.064
[Cl2]=0.064
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= 0.14mole of CF4