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Asked: 2022-04-11T18:46:33+00:00 In:

Fill in the missing chemical formulae in the tables below. Acid conjugate base base conjugate acid…

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

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Conjugate acid of NH3  =

Conjugate acid of  CO3-2 =

Conjugate base of  HSO4–=

Conjugate base of  H2PO4– =

Conjugate acid of  HSO4– =

Conjugate acid of  CH3NH2 =

Part B:

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

CO(g) + H2O(g) ↔ CO2(g) + H2(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 H2O. What can you say about the composition of the mixture in the vessel at equilibriu?

– There will be very little CO and H2P

– There will be very little CO2 and H2

– Neighter of the above is true

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

CO2(g) + H2(g) ↔ CO (g)+ H2O(g)

♦ Relevant knowledge
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.

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    Part A: 

    Conjugate acid of NH3 = NH4+

    Conjugate acid of CO3-2 = HCO3–

    Conjugate base of HSO4− = SO42−

    Conjugate base of H2PO4− = HPO42−

    Conjugate acid of HSO4−=H2SO4

    Conjugate acid of CH3NH2=CH3NH+3

    Part B: 

    The following response:

    CO(g) + H2O(g) ⇌ CO2(g) + H2(g) K=0.0017

    1) We are given:

    • Initial concentration of CO=1.3 mol / 6.0L=0.217M
    • Initial concentration of H2O=1.3 mol / 6.0 L=0.217M

    Then, we’ll create an RICE table to determine the equilibrium concentration of the species , as in the following manner:

    R CO H2O CO2 H2
    I 0.217 0.217 0 0
    C -x -x +x x
    E 0.217 – x 0.217 – x x x

    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

       [H2O]eq = 0.217−0.00858 = 0.208 M

       [CO2]eq = 0.00858 M

       [H2]eq = 0.00858 M

    It will contain tiny amounts of CO2 and H2 at balance.

    2) The following response:

    CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

    The equilibrium constant is determined as follows:

         K = ([CO]eq[H2O]eq) / ([CO2]eq[H2]eq) 

         K = 0.208×0.208 / (0.00858×0.00858)

         K = 5.9×102

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