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

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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 show... show full transcript

Worked Solution & Example Answer: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

TemperaturePressure
Step 1High
Step 2Low

Step 2

Suggest a structure for compound X.

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Answer

The structure for compound X is 2-methylpropene, which can be represented as:

     H   H
      \ /
       C
      / \
   H-C   C-H
      \ / 
       H

This structure allows for the addition reaction with methanol.

Step 3

Explain what this means.

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An atom economy of 100% means that all the atoms from the reactants are converted into the desired product without any waste. In this reaction, there are no by-products; therefore, every part of the reactants contributes directly to the formation of the target compound.

Step 4

Calculate the enthalpy change, in kJ mol⁻¹, for the reaction.

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To calculate the enthalpy change:

  1. Bond Breaking:

    • C-H bonds: 4 bonds x 412 kJ/mol = 1648 kJ
    • O-H bonds: 1 bond x 463 kJ/mol = 463 kJ
    • C-O bonds: 1 bond x 360 kJ/mol = 360 kJ
    • Total bond breaking = 1648 + 463 + 360 = 2471 kJ
  2. Bond Formation:

    • C=O bonds: 1 bond x 743 kJ/mol = 743 kJ
    • H-H bonds: 2 bonds x 436 kJ/mol = 872 kJ
    • Total bond formation = 743 + 872 = 1615 kJ
  3. Enthalpy Change:

    extEnthalpyChange=extTotalbondbreakingextTotalbondformation=24711615=856extkJ/mol ext{Enthalpy Change} = ext{Total bond breaking} - ext{Total bond formation} = 2471 - 1615 = 856 ext{ kJ/mol}

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