8. Methanol (CH₃OH) is an important chemical in industry - Scottish Highers Chemistry - Question 8 - 2016
Question 8
8. Methanol (CH₃OH) is an important chemical in industry.
(a) Methanol is produced from methane in a two-step process.
In step 1, methane is reacted with steam as s... show full transcript
Worked Solution & Example Answer:8. Methanol (CH₃OH) is an important chemical in industry - Scottish Highers Chemistry - Question 8 - 2016
Step 1
Complete the table to show the most favourable conditions to maximise the yield for each step.
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Temperature (High/Low)
Pressure (High/Low)
Step 1
High
Low
Step 2
Low
High
Step 2
Suggest a structure for compound X.
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Answer
The structure for compound X can be proposed as 2-methylpropene:
H H
| |
H—C—C
| /
H—C—C
|
H
This structure corresponds to 2-methylpropene, which allows the formation of methyl tertiary-butyl ether with methanol.
Step 3
The atom economy of this reaction is 100%. Explain what this means.
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An atom economy of 100% indicates that all atoms from the reactants are converted into the desired product. This means there are no by-products or waste products generated in the reaction, ensuring maximum efficiency in using the original materials.
Step 4
Using bond enthalpy and mean bond enthalpy values from the data booklet, calculate the enthalpy change, in kJ mol⁻¹, for the reaction.
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To calculate the enrthalpy change for the reaction, we can use the bond enthalpy values:
Breaking bonds:
412 (C-H) x 4 = 1648 kJ
360 (C-O) = 360 kJ
463 (O-H) x 2 = 926 kJ
Total bond breaking = 1648 + 360 + 926 = 2934 kJ
Forming bonds:
743 (C=O) = 743 kJ
412 (C-H) x 2 = 824 kJ
463 (O-H) x 2 = 926 kJ
Total bond forming = 743 + 824 + 926 = 2493 kJ
Enthalpy change (ΔH) = Total bond breaking - Total bond forming
ΔH = 2934 - 2493 = 441 kJ
Thus, the enthalpy change for this reaction is 441 kJ mol⁻¹.
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