Question

(6 marks) An easy way to analyse the amount of CO2 in a gaseous mixture of CO2 and CO is to pass the gaseous mixture through an aqueous solution that contains excess Ba(OH)2. The CO2 will react yielding a precipitate of BaCOs but the CO does not react (you will learn why in higher classes). The aforementioned method was used to analyse the equilibrium composition of the gas evolved when 1.77 g of CO2 reacted with 2.0 g of graphite (basically carbon) in a 1.0 L container at 1100 K. The analysis yielded 3.41 g of BaCOs. Use the data given to calculate Kp at 1100 K for the following reaction c) 2CO (g) CO2 (g) + C (s) (10 marks) (6 marks) An easy way to analyse the amount of CO2 in a gaseous mixture of CO2 and CO is to pass the gaseous mixture through an aqueous solution that contains excess Ba(OH)2. The CO2 will react yielding a precipitate of BaCOs but the CO does not react (you will learn why in higher classes). The aforementioned method was used to analyse the equilibrium composition of the gas evolved when 1.77 g of CO reacted with 2.0 g of graphite (basically carbon) in a 1.0 L container at 1100 K. The analysis yielded 3.41 g of BaCOs. Use the data given to calculate Kp at 1100 K for the following reaction c) 2CO (g) CO2 (g) + C (s) (10 marks)

Answer 1

Reaction of CO_{​​​​​2} with Ba(OH)₂

Ba(OH)2(aq) + CO2(g) → BaCO3(s)+ H2O(1)

Mass of BaCO₃ formed=3.41 g

Molar mass of BaCO₃=molar mass of Ba+molar mass of C+3xmolar mass of O=137.3 g/mol+12 g/mol+3x16 g/mol

=137.3 g/mol+12 g/mol+48 g/mol=197.3 g/mol

Number of moles of BaCO₃formed=given mass/molar mass=3.41 g/197.3 g/mol=0.017 mol

As per the balanced chemical equation, 1 mol BaCO₃ is formed from 1 mol CO_{​​​​​​2}

So 0.017 mol BaCO₃ is formed from 0.017 mol CO₂

Initial mass of CO₂=1.77 g

Molar mass of CO₂=molar mass of C+2xmolar mass of O

=12 g/mol+2x16 g/mol=12 g/mol+32 g/mol=44 g/mol

Initial number of moles of CO₂=initial mass of CO_{​​​​​2}/molar mass of CO₂=1.77 g/44 g/mol=0.04 mol

Initial concentration of CO₂=number of moles/volume (L)=0.04 mol/1.0 L=0.04 M

Number of moles of C=mass of C/molar mass of C=2.0 g/12 g/mol=0.17 mol

Here CO₂ is the limiting reactant as 0.04 mol CO₂ mol reacts with 0.04 mol=0.04 mol C as per the balanced chemical equation but we have 0.17 mol C which is in excess.

As 0.017 mol CO₂ react with Ba(OH)₂ to form 3.41 g BaCO₃

the equilibrium concentration of CO₂=[CO₂]=0.017 mol/1.0 L=0.017 M

Decrease in concentration of CO₂=0.04 M-0.017 M=0.023 M

Concentration of CO formed=[CO]=2x0.023 M=0.046 M

So Kc=[CO]²/[CO_{​​​​​2}]=(0.046 M)²/0.017 M=0.12 (C doesn't appear in Kc as concentration of pure solids remain almost unchanged in the reaction)

Kp=Kc(RT)^{$\Delta n$}

where $\Delta n$=number of moles of gaseous products-number of moles of gaseous reactants=1-1=0

Kp=0.12x(RT)⁰=0.12 x 1=0.12