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8. Methanol (CH₃OH) is an important chemical in industry - Scottish Highers Chemistry - Question 8 - 2016

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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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Answer

Temperature (High/Low)Pressure (High/Low)
Step 1HighLow
Step 2LowHigh

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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Answer

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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