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At temperatures above 100°C, hydrogen and carbon monoxide react to form methanol gas in this reversible reaction - HSC - SSCE Chemistry - Question 31 - 2015 - Paper 1
Question 31
At temperatures above 100°C, hydrogen and carbon monoxide react to form methanol gas in this reversible reaction. \[ 2H_2(g) + CO(g) \rightleftharpoons CH_3OH(g) \... show full transcript
Worked Solution & Example Answer:At temperatures above 100°C, hydrogen and carbon monoxide react to form methanol gas in this reversible reaction - HSC - SSCE Chemistry - Question 31 - 2015 - Paper 1
Step 1
The initial volume of the container is 1.00 L. Account for any changes in the concentration of hydrogen gas when the volume of the container is rapidly increased to 2.00 L.
Answer
When the volume of the container is increased from 1.00 L to 2.00 L, the concentration of hydrogen gas, along with that of the other reactants and products, decreases due to the increased volume. The concentration before the change is calculated as:
[ \text{Initial concentration of } H_2 = \frac{0.50 \text{ mol}}{1.00 \text{ L}} = 0.50 , \text{mol/L} ]
After the volume increase, the concentration becomes:
[ \text{New concentration of } H_2 = \frac{0.50 \text{ mol}}{2.00 \text{ L}} = 0.25 , \text{mol/L} ]
Such a decrease in concentration shifts the equilibrium towards the left according to Le Chatelier's principle, allowing more H2 to be formed from the reaction.
Step 2
Calculate the equilibrium constant for this reaction.
Answer
To calculate the equilibrium constant ( K ) for the reaction, we can use the equilibrium concentrations:
- Initial concentrations:
- [ [H_2] = 0.50 , \text{mol/L} ]
- [ [CO] = 1.00 , \text{mol/L} ]
- [ [CH_3OH] = 2.50 , \text{mol/L} ]
- Change in concentrations at equilibrium:
- [ [H_2] = 0.36 , \text{mol} ]
- [ [CH_3OH] = 2.32 , \text{mol} ]
- Find the concentration of [CO]: Since the volume also doubles, the moles need to be adjusted for the new volume:
[ ext{Equilibrium concentrations:} ]
[ [H_2] = \frac{0.36}{2.00} = 0.18 , \text{mol/L} ]
[ [CO] = (0.50 - 0.18) = 0.32 , \text{mol/L} ]
[ [CH_3OH] = \frac{2.32}{2.00} = 1.16 \text{mol/L} ]
- The equilibrium constant is calculated as:
[ K = \frac{[CH_3OH]}{[H_2]^2[CO]} = \frac{1.16}{(0.18)^2 \times 0.32} = 20.0 \text{ (approximately)} ]
Step 3
Describe how saponification can be safely carried out as part of a first-hand investigation.
Answer
To safely carry out saponification:
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Materials: Use appropriate materials such as sodium hydroxide (NaOH), fats/oils, and water in a well-ventilated area.
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Apparatus: Set up using a round-bottom flask, a heating mantle, and a condenser.
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Handling caustics: Wear gloves, goggles, and a lab coat to protect against NaOH spills.
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Procedure:
- Dissolve NaOH in water slowly.
- Heat the fats until melted, add the NaOH solution, and stir gently.
- After the reaction, allow the mixture to cool before neutralizing.
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Disposal: Disposal of any caustic chemicals should follow safety and environmental regulations.
Step 4
Explain the chemistry related to the cleaning properties of the product of saponification.
Answer
Saponification results in the formation of soap, which is a salt of fatty acids. The chemistry behind its cleaning properties involves its molecular structure:
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Amphiphilic Nature: Soap molecules have a hydrophilic (water-attracting) "head" and a hydrophobic (water-repelling) "tail." This allows soap to interact with both water and oils.
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Emulsification: When soap is mixed with water and grease, the hydrophobic tails attach to the grease while the hydrophilic heads remain in the water, effectively allowing the oil to be lifted away from surfaces.
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Cleaning Process: The soap allows for the dispersion of grease and dirt in water, facilitating cleaning as it breaks large grease droplets into smaller ones, which can be rinsed away.
Step 5
Outline the chemistry of the production of sodium carbonate in the process shown.
Answer
The production of sodium carbonate in the Solvay process involves several key reactions:
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Ammoniation of Brine: [ NH_3(aq) + H_2O(l) + CO_2(g) \rightleftharpoons NH_4^+(aq) + HCO_3^-(aq) ]
where carbon dioxide reacts with ammonium hydroxide from brine, forming bicarbonate. -
Precipitation of Sodium Bicarbonate: [ Na^+(aq) + HCO_3^-(aq) \rightleftharpoons NaHCO_3(s) ]
Sodium ions in brine react with bicarbonate to precipitate sodium bicarbonate, which is then filtered. -
Formation of Sodium Carbonate: [ 2NaHCO_3(s) \rightarrow Na_2CO_3(s) + H_2O(g) + CO_2(g) ]
Heating sodium bicarbonate decomposes it into sodium carbonate.
Step 6
By making specific reference to the diagram, justify the requirements for the location of a Solvay process plant.
Answer
For the Solvay process plant location:
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Raw Material Availability: The plant needs proximity to purified brine and a reliable supply of carbon dioxide from limestone calcination.
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Water Source: An ample water source is required for the reactions and cooling processes.
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Environmental Considerations: The plant should be situated away from residential areas to minimize pollution and health risks associated with ammonia and carbon dioxide emissions.
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Waste Management: Facilities for handling and disposing of calcium chloride, a by-product, must be in place to prevent environmental contamination.
Step 7
Compare the membrane cell method with ONE other method used in the industrial production of sodium hydroxide in terms of technical and environmental issues.
Answer
The membrane cell method for producing sodium hydroxide:
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Technical Efficiency: This method offers higher purity of sodium hydroxide and lower energy consumption compared to the diaphragm cell method, which has higher by-product contamination.
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Environmental Impact: Membrane cells reduce the emission of chlorine gas, minimizing air pollution, whereas diaphragm cells can cause issues with gaseous chlorine release, requiring more extensive control measures.
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Waste Management: The membrane method produces fewer hazardous wastes compared to the mercury cell method, which is being phased out due to environmental hazards from mercury contamination.