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# Chemistry: Chemical Equilibrium

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Chemical Equilibrium

1- Determine the value of the equilibrium constant, K goal , for the reaction
N 2 (g)+H 2 O(g)⇌NO(g)+ 1 2 N 2 H 4 (g) , K goal =?
by making use of the following information:
1.N 2 (g)+O 2 (g)⇌2NO(g) ,
K 1 =
4.10×10 −31

2.N 2 (g)+2H 2 (g)⇌N 2 H 4 (g) ,
K 2 =
7.40×10 −26

3.2H 2 O(g)⇌2H 2 (g)+O 2 (g) ,
K 3 =
1.06×10 −10

2- Determine the equilibrium constant, K goal , for the reaction
4PCl 5 (g)⇌P 4 (s)+10Cl 2 (g), K goal =?
by making use of the following information:
1. P 4 (s)+6Cl 2 (g)⇌4PCl 3 (g), K 1 =2.00×10 19
2. PCl 5 (g)⇌PCl 3 (g)+Cl 2 (g), K 2 =1.13×10 −2
3- For the reaction 2A(g)+2B(g)⇌C(g)
K c = 89.2 at a temperature of 303 ∘ C .
Calculate the value of K p .

4- For the reaction X(g)+3Y(g)⇌2Z(g)
K p = 1.35×10−2 at a temperature of 265 ∘ C
Calculate the value of K c .

5- Why are pure solids and pure liquids not included in a Kc expression? Choose from:
Pure solids and pure liquids are not reactive enough to be included in an equilibrium expression.
Pure solids and pure liquids do not participate in reversible (equilibrium) reactions.
Pure solids and pure liquids do not change in volume over the course of a reaction.
The concentrations of pure solids and pure liquids do not change during reactions.

6- Part A :Phosgene (carbonyl chloride), COCl 2 , is an extremely toxic gas that is used in manufacturing certain dyes and plastics. Phosgene can be produced by reacting carbon monoxide and chlorine gas at high temperatures:
CO(g)+Cl 2 (g)⇌COCl 2 (g)
Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 465 ∘ C . At equilibrium, the concentrations were measured and the following results obtained:
Gas Partial Pressure
(atm )

CO
0.740
Cl 2
1.14
COCl 2
0.150
What is the equilibrium constant, K p , of this reaction?

7- In Part A, you were given the equilibrium pressures, which could be plugged directly into the formula for K . In Part B however, you will be given initial concentrations and only one equilibrium concentration. You must use this data to find all three equilibrium concentrations before you can apply the formula for K .
Part B
The following reaction was performed in a sealed vessel at 783 ∘ C :
H 2 (g)+I 2 (g)⇌2HI(g)
Initially, only H 2 and I 2 were present at concentrations of [H 2 ]=3.20M and [I 2 ]=2.00M .The equilibrium concentration of I 2is 0.0700M . What is the equilibrium constant, K c , for the reaction at this temperature?

8- How can Qc and Kc differ from one another despite having the same algebraic expression? Choose from:
Qc can include pure liquids and solids; Kc does not.
Because the two expressions have different units, they can never be equal unless the proper conversion factor is used.
Qc is calculated from the current concentrations of the species, while Kc only uses the equilibrium concentrations.
Because Qc represents the change from initial to equilibrium conditions, Kc represents the equilibrium conditions.
9- A mixture initially contains A, B, and C in the following concentrations: [A] = 0.700M , [B] = 1.05M, and [C] = 0.600M. The following reaction occurs and equilibrium is established:
A+2B⇌C
At equilibrium, [A]= 0.580M and [C] = 0.720M . Calculate the value of the equilibrium constant, K c .

https://brainmass.com/chemistry/general-chemistry/chemistry-chemical-equilibrium-552146

#### Solution Preview

Chemical Equilibrium

1- Determine the value of the equilibrium constant, K goal , for the reaction
N 2 (g)+H 2 O(g)⇌NO(g)+ 1 2 N 2 H 4 (g) , K goal =?
by making use of the following information:
1.N 2 (g)+O 2 (g)⇌2NO(g) ,
K 1 =
4.10×10 −31

2.N 2 (g)+2H 2 (g)⇌N 2 H 4 (g) ,
K 2 =
7.40×10 −26

3.2H 2 O(g)⇌2H 2 (g)+O 2 (g) ,
K 3 =
1.06×10 −10

½*reaction 2+1/2*reaction 1+1/2*reaction 3: we could obtain the destination reaction.
K(goal)=sqrt(k1*k2*k3)=sqrt(4.10*10^(-31)*7.40*10^(-26)*1.06*10^(-10))= 1.79*10^(-33)
2- Determine the equilibrium constant, K goal , for the reaction
4PCl 5 (g)⇌P 4 (s)+10Cl 2 (g), K goal =?
by making use of the following information:
1. P 4 (s)+6Cl 2 (g)⇌4PCl 3 (g), K 1 =2.00×10 19
2. PCl 5 (g)⇌PCl 3 (g)+Cl 2 (g), K 2 =1.13×10 −2