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**Part A:**Fill in the missing chemical formulae in the tables below.

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Conjugate acid of NH_{3 } =

Conjugate acid of CO_{3}^{-2} =

Conjugate base of HSO_{4}^{–}=

Conjugate base of H_{2}PO_{4}^{–} =

Conjugate acid of HSO_{4}^{–} =

Conjugate acid of CH_{3}NH_{2} =

**Part B:**At a certain temperature, the equilibrium constant K for the following reaction is 0.0017:

CO(g) + H_{2}O(g) ↔ CO_{2}(g) + H_{2}(g)

Use this information to complete the following :

1) Suppose at 6.0 L reaction vessel is filled with 1.3 mol of CO and 1.3 mol of H_{2}O. What can you say about the composition of the mixture in the vessel at equilibriu?

– There will be very little CO and H_{2}P

– There will be very little CO_{2} and H_{2}

– Neighter of the above is true

2) What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits.

CO_{2}(g) + H_{2}(g) ↔ CO (g)+ H_{2}O(g)

**Conjugate Acid-Base**

Conjugate acids can be made by letting a base species take proton. Conjugate acids- Bases are differed by a proton. A strong acid has weak conjugate bases and an acid that is weak has a strong base with a conjugate.

**Equilibrium Concentration of the Species**

A species’ equilibrium level involved in an reaction is determined by knowing the proper expression for the constant of equilibrium, the value of it and what the concentration at the beginning of the reaction. Then, we will design the RICE table based on the equilibrium reaction.

Part A:Conjugate acid of NH

_{3 }= NH_{4}+Conjugate acid of CO

_{3}^{-2}= HCO_{3}^{–}Conjugate base of HSO

_{4}^{− }= SO_{4}^{2−}Conjugate base of H

_{2}PO_{4}^{− }= HPO_{4}^{2−}Conjugate acid of HSO

_{4}^{−}=H_{2}SO_{4}Conjugate acid of CH

_{3}NH_{2}=CH_{3}NH^{+3}Part B:The following response:

CO(g) + H

_{2}O(g) ⇌ CO_{2}(g) + H_{2}(g) K=0.00171) We are given:

_{2}O=1.3 mol / 6.0 L=0.217MThen, we’ll create an RICE table to determine the equilibrium concentration of the species , as in the following manner:

_{2}O_{2}_{2}Solve for x:

K = [CO2][H2]/[CO][H2O]

0.0017=x×x / (0.217−x)(0.217−x)

x / (0.217−x) = 0.0412

x = 0.00893−0.0412

1.0412x = 0.00893

x = 0.00858

Therefore the equilibrium concentrations of the species is :

[CO]

_{eq }= 0.217−0.00858 = 0.208 M[H

_{2}O]_{eq }= 0.217−0.00858 = 0.208 M[CO

_{2}]_{eq }= 0.00858 M[H

_{2}]_{eq }= 0.00858 MIt will contain tiny amounts of CO_{2}and H_{2}at balance.2) The following response:

CO

_{2}(g) + H_{2}(g) ⇌ CO(g) + H_{2}O(g)The equilibrium constant is determined as follows:

K = ([CO]

_{eq}[H_{2}O]_{e}_{q}) / ([CO_{2}]_{eq}[H_{2}]_{eq})K = 0.208×0.208 / (0.00858×0.00858)

K = 5.9×10

^{2}