1. 2HI(g) <----> H2(g) + I2(g)
Hydrogen iodide decomposes at moderate temperatures according to the equation
2HI(g) <----> H2(g) I2(g)
The amount of I2 in the mixture can be determined from the intensity of the violet color of I2; the more intense, the more I2 in the reaction vessel. When 4.00 mol of HI were placed in a 5.00 L vessel at 458°C, the equilibrium mixture was found to contain mol I2. What is the value of K?
Initial:
4.00 mol
0.0 mol
0.0 mol
At Eq: ? ?
0.442 mol

2. 2HI(g) <----> H2(g) + I2(g)
Initial: mol mol mol
2HI(g) <----> H2(g) I2(g)
At Eq.: ? ? mol
0.442 mol
Which reaction proceeds initially? Why?
The forward…no products to make reverse go at first!
0.442 mol of I2 produced…how much H2?
0.442 mol of H2…one to one molar ratio and they both started out the same, zero

3. 2HI(g) <----> H2(g) + I2(g)
Initial: mol mol mol
2HI(g) <----> H2(g) I2(g)
At Eq.: ? ? mol
0.442 mol
Where did the I2 and the H2 come from?
The HI, obviously!
How much of that was used?
0.884 mol of HI…two to one molar ratio

4. 2HI(g) <----> H2(g) + I2(g)
Initial: mol mol mol
2HI(g) <----> H2(g) I2(g)
mol
At Eq.: ? ? mol
3.12 mol
0.442 mol
If we started out with 4.00 mol of HI initially, how much of it remains at equilibrium then?
3.12 mol of HI at eq.
The concentration of HI at equilibrium would be? (Remember the reaction vessel was 5.00 L)
0.624 M of HI

5. 2HI(g) <----> H2(g) + I2(g)
Initial: mol mol mol
2HI(g) <----> H2(g) I2(g)
At Eq.: ? ? mol
3.12 mol
0.442 mol
( M)
(0.624 M)
( M)
The [H2] = [I2] = ?
M = [H2] = [I2]
Now, determine the K value...
K = (0.0884M) (0.0884M)/ (0.624M)2 =

6. 2 IBr(g) <----> I2(g) + Br2(g)
For the reaction
2 IBr(g) <----> I2(g) + Br2(g)
K = 2.5 x 10-3 at 315 K
Calculate the equilibrium concentration of each species in a 4.0 L vessel starting with 0.30 mol of each substance. Also, what is the value of Kp.
IBr
Initial:
0.075 M M
0.075 M
4.0 L
+2x -x -x
Change:
At Eq:
0.075M + 2x ?
0.075M - x ?
0.075M - x ? I2
Br2
0.21M
0.010M
0.010M
(0.075M)(0.075M)
= Q reaction quotient
Kp=
= 1
(0.075M)2
K= (0.075-x)(0.075-x)
K= (0.075-x)2
Q>K (the reverse reaction is favored)
= 2.5 x 10-3
( x)2
0.075-x
= 0.050
x = 0.065 x

7. x = 0.00648 Consider the following reaction that takes place at 773 K
CO(g) + H2O(g) <----> CO2(g) H2 (g)
The reaction took place by injecting into a reaction vessel the materials shown below to produce the indicated initial conditions.
carbon monoxide: M water: M carbon dioxide: M hydrogen: M
Find the equilibrium concentrations of all the substances given the equilibrium constant, Kc, is equal to 4.00 at this temperature.
x =
students who answer correctly before the bell rings will receive 2 bonus points! (no sharing of info!)
Equilibrium Concentrations: [CO] = M [H2O]= [CO2]= M [H2] = M