Please answer F. Thank you very much! The equilibrium constant for the reaction

Question

Please answer F. Thank you very much!

The equilibrium constant for the reaction of gaseous carbon monoxide and water vapor to form carbon dioxide and hydrogen gas is K 10 at 287C. Given a mixture of 129.1 g of carbon monoxide, 207.3 g of water vapor, 55.4 g of carbon dioxide, and 38.6 g of hydrogen gas in a 3.25L vessel, answer the following questions. 2. a. Write the balanced chemical reaction, with phases b. Write the equilibrium expression for the reaction. d. Determine the reaction quotient. e. Determine the direction that the reaction will proceed to achieve equilibrium. LEFT / RIGHT NEITHER f. Determine the concentration and mass of all components at equilibrium.

Explanation / Answer

The reaction is CO(g)+ H2O(g) ------->CO2(g)+H2(g)

given data on masses need to be converted to moles first.

molar mass= sum of atomic weights of species

atomic weights :C= 12, O= 16 and H=1

molar masses : CO =12+16= 28, H2O= 2+16= 18, CO2= 12+2*16= 44 and H2= 2

moles= mass/molar mass and concentration = moles/volume,

mass of CO= 129.1 gm, moles of CO= mass/molar mass= 129.1/28= 4.62, concentration of CO= 4.62 moles/3.25 L=1.42 M

mass of H2O= 207.3 gm, moles of H2O= 207.3/18= 11.52, concentration = 11.52/3.25= 3.54M

mass of CO2= 55.4 gm, moles of CO2= 55.4/44= 1.26, concentration = 1.26/3.25=0.39M

moles of H2= 38.6/2= 19.3, concentration = 19.3/3.25= 5.94M

Q = reaction coefficient = [CO2][H2]/[CO]H2O]= 0.39*5.94/(1.42*3.54)= 0.46

if Q=K, the reaction will be at equilibrium, if Q>K, the reaction proceeds towards reactants side. if Q<K, the reaction proceeds towards products side.

since the reaction quotient, Q <KC, the reaction proceeds towards formation of more CO2 and H2.( right)

so let x= drop in concentration of CO to reach equilibrium, At equilibrium

[CO]= 1.42-x, [H2O]= 3.54-x, [CO2]= 0.39+x, [H2]= 5.94+x

KC= 10 = (0.39+x)*(5.94+x)/ (3.54-x)*(1.42-x)

when solved using excel, x= 1.0272

so at equilibrium, concentration , CO= 0.3928, H2O= 2.5128, CO2= 1.4172, H2= 6.9672M

so at equilibrum, moles : CO= 0.3928*3.25=1.28 , H2O= 2.5128*3.25= 8.2, CO2= 1.4272*3.25=4.6

H2= 6.9672*3.25= 22.64 moles

mass=moles* molar mass, mass (gm): CO= 1.28*28= 35.84, H2O= 8.2*18=147.6, CO2= 4.6*44=202.4 and H2= 22.64*2= 45.28 gm

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