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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 sh... show full transcript

Worked Solution & Example Answer:Methanol (CH₃OH) is an important chemical in industry - Scottish Highers Chemistry - Question 8 - 2016

Step 1

Step 1: Complete the table to show the most favourable conditions to maximise the yield for step 1.

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Answer

For the first step, the reaction is endothermic as indicated by the positive ΔH value (+210 kJ mol⁻¹). To favour the forward reaction, high temperature is needed. Since the reaction produces more molecules of gas (3 moles of H₂ and 1 mole of CO from 1 mole of CH₄ and 1 mole of H₂O), low pressure is favoured to enhance yield.

Therefore:

  • Temperature: High
  • Pressure: Low

Step 2

Step 2: Complete the table to show the most favourable conditions to maximise the yield for step 2.

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Answer

The second step is exothermic, indicated by the negative ΔH value (-91 kJ mol⁻¹). To favour the forward reaction under these conditions, low temperature is preferred. Additionally, since the reaction results in fewer gas molecules (2 moles of reactants to 1 mole of product), high pressure is desirable to maximise the yield.

Thus:

  • Temperature: Low
  • Pressure: High

Step 3

(b)(i) Suggest a structure for compound X.

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The structure for compound X is:

     H
     |
H₃C−C−CH₃
     | 
     H

This represents 2-methylpropene, which has the formula C₄H₈.

Step 4

(b)(ii) The atom economy of this reaction is 100%. Explain what this means.

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An atom economy of 100% means that all the atoms of the reactants are converted into the desired product without generating any by-products. In other words, the mass of the desired product is equal to the total mass of the reactants used, ensuring maximum efficiency in material use.

Step 5

(c) 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 enthalpy change, we need to account for bond breaking and bond forming.

The total bond breaking enthalpies are:

  • C−H: 4 bonds (412 kJ/mol each)
  • C−C: 1 bond (346 kJ/mol)
  • C=O: 1 bond (743 kJ/mol)
  • H−H: 1 bond (436 kJ/mol)

Total for bond breaking:

extTotalbondbreaking=4(412)+1(346)+1(743)+1(436)=743+412+412+346+436=2059extkJ/mol ext{Total bond breaking} = 4(412) + 1(346) + 1(743) + 1(436) = 743 + 412 + 412 + 346 + 436 = 2059 ext{ kJ/mol}

The bond formation enthalpies are:

  • C=O: 1 bond (743 kJ/mol)
  • H−H: 2 bonds (436 kJ/mol each)

Total for bond formation:

extTotalbondforming=1(743)+4(412)=743+1648=2391extkJ/mol ext{Total bond forming} = 1(743) + 4(412) = 743 + 1648 = 2391 ext{ kJ/mol}

Thus, the enthalpy change is:

extΔH=extBondbreakingextBondforming=20592391=332extkJ/mol ext{ΔH} = ext{Bond breaking} - ext{Bond forming} = 2059 - 2391 = -332 ext{ kJ/mol}

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