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Question: Compute the enthalpy of formation of liquid methyl alcohol in kJ mol-¹, using the following data. E...

Compute the enthalpy of formation of liquid methyl alcohol in kJ mol-¹, using the following data.

Enthalpy of vaporisation of liquid CH3OH = 38 kJ/mol.

Enthalpy of formation of gaseous atoms from the elements in their standard states are H→218 kJ/mol; C→715 kJ/mol; O→249 kJ/mol.

Bond Enthalpies C-H→415 kJ/mol; C-0→356 kJ/mol; O-H→463 kJ/mol

Answer

-266 kJ mol⁻¹

Explanation

Solution

To compute the enthalpy of formation of liquid methyl alcohol (CH3OH\text{CH}_3\text{OH}), we can use Hess's Law by breaking down the overall formation process into several steps involving atomization and bond formation.

The target reaction is: C(graphite)+2H2(g)+12O2(g)→CH3OH(l)\text{C(graphite)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}

We can follow these steps:

Step 1: Atomization of reactants

Convert the elements in their standard states to gaseous atoms. C(graphite)→C(g)\text{C(graphite)} \rightarrow \text{C(g)}; ΔHa(C)=715 kJ/mol\Delta H_a(\text{C}) = 715 \text{ kJ/mol} 2H2(g)→4H(g)2\text{H}_2\text{(g)} \rightarrow 4\text{H(g)}; Since H2(g)→2H(g)\text{H}_2\text{(g)} \rightarrow 2\text{H(g)} requires 2×218 kJ/mol2 \times 218 \text{ kJ/mol}, for 2H2(g)2\text{H}_2\text{(g)} it will be 2×(2×218)=4×218=872 kJ/mol2 \times (2 \times 218) = 4 \times 218 = 872 \text{ kJ/mol}. 12O2(g)→O(g)\frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{O(g)}; ΔHa(O)=249 kJ/mol\Delta H_a(\text{O}) = 249 \text{ kJ/mol}

The total enthalpy change for atomization of reactants (ΔH1\Delta H_1) is: ΔH1=ΔHa(C)+4×ΔHa(H)+ΔHa(O)\Delta H_1 = \Delta H_a(\text{C}) + 4 \times \Delta H_a(\text{H}) + \Delta H_a(\text{O}) ΔH1=715 kJ/mol+872 kJ/mol+249 kJ/mol\Delta H_1 = 715 \text{ kJ/mol} + 872 \text{ kJ/mol} + 249 \text{ kJ/mol} ΔH1=1836 kJ/mol\Delta H_1 = 1836 \text{ kJ/mol}

Step 2: Formation of gaseous methyl alcohol from gaseous atoms

This step involves the formation of bonds. The structure of CH3OH\text{CH}_3\text{OH} is:

  H
  |
H-C-O-H
  |
  H

It contains:

  • 3 C-H bonds
  • 1 C-O bond
  • 1 O-H bond

When bonds are formed, energy is released (exothermic, negative enthalpy change). Enthalpy change for bond formation (ΔH2\Delta H_2): ΔH2=−[3×BE(C-H)+1×BE(C-O)+1×BE(O-H)]\Delta H_2 = - [3 \times \text{BE(C-H)} + 1 \times \text{BE(C-O)} + 1 \times \text{BE(O-H)}] ΔH2=−[3×415 kJ/mol+1×356 kJ/mol+1×463 kJ/mol]\Delta H_2 = - [3 \times 415 \text{ kJ/mol} + 1 \times 356 \text{ kJ/mol} + 1 \times 463 \text{ kJ/mol}] ΔH2=−[1245+356+463] kJ/mol\Delta H_2 = - [1245 + 356 + 463] \text{ kJ/mol} ΔH2=−2064 kJ/mol\Delta H_2 = - 2064 \text{ kJ/mol}

The enthalpy of formation of gaseous methyl alcohol (ΔHf∘(CH3OH(g))\Delta H_f^\circ(\text{CH}_3\text{OH(g)})) from its elements is the sum of ΔH1\Delta H_1 and ΔH2\Delta H_2: ΔHf∘(CH3OH(g))=ΔH1+ΔH2=1836 kJ/mol+(−2064 kJ/mol)\Delta H_f^\circ(\text{CH}_3\text{OH(g)}) = \Delta H_1 + \Delta H_2 = 1836 \text{ kJ/mol} + (-2064 \text{ kJ/mol}) ΔHf∘(CH3OH(g))=−228 kJ/mol\Delta H_f^\circ(\text{CH}_3\text{OH(g)}) = -228 \text{ kJ/mol}

Step 3: Condensation of gaseous methyl alcohol to liquid methyl alcohol

This is the reverse of vaporization. CH3OH(g)→CH3OH(l)\text{CH}_3\text{OH(g)} \rightarrow \text{CH}_3\text{OH(l)} The enthalpy change for this step (ΔH3\Delta H_3) is the negative of the enthalpy of vaporization: ΔH3=−ΔHvap(CH3OH(l))=−38 kJ/mol\Delta H_3 = - \Delta H_{vap}(\text{CH}_3\text{OH(l)}) = -38 \text{ kJ/mol}

Overall Enthalpy of Formation of Liquid Methyl Alcohol

The total enthalpy of formation of liquid methyl alcohol (ΔHf∘(CH3OH(l))\Delta H_f^\circ(\text{CH}_3\text{OH(l)})) is the sum of the enthalpy changes of all steps: ΔHf∘(CH3OH(l))=ΔHf∘(CH3OH(g))+ΔH3\Delta H_f^\circ(\text{CH}_3\text{OH(l)}) = \Delta H_f^\circ(\text{CH}_3\text{OH(g)}) + \Delta H_3 ΔHf∘(CH3OH(l))=−228 kJ/mol+(−38 kJ/mol)\Delta H_f^\circ(\text{CH}_3\text{OH(l)}) = -228 \text{ kJ/mol} + (-38 \text{ kJ/mol}) ΔHf∘(CH3OH(l))=−266 kJ/mol\Delta H_f^\circ(\text{CH}_3\text{OH(l)}) = -266 \text{ kJ/mol}

The final answer is −266 kJ mol−1-266 \text{ kJ mol}^{-1}.

Explanation of the solution:

  1. Calculate the energy required to convert elements in their standard states to gaseous atoms (atomization energy). ΔHatomization=ΔHf(C(g))+4×ΔHf(H(g))+ΔHf(O(g))=715+4(218)+249=1836 kJ/mol\Delta H_{\text{atomization}} = \Delta H_f(\text{C(g)}) + 4 \times \Delta H_f(\text{H(g)}) + \Delta H_f(\text{O(g)}) = 715 + 4(218) + 249 = 1836 \text{ kJ/mol}.
  2. Calculate the energy released when gaseous atoms form gaseous methyl alcohol by forming bonds. ΔHbond formation=−(3×BE(C-H)+1×BE(C-O)+1×BE(O-H))=−(3×415+356+463)=−2064 kJ/mol\Delta H_{\text{bond formation}} = - (3 \times \text{BE(C-H)} + 1 \times \text{BE(C-O)} + 1 \times \text{BE(O-H)}) = - (3 \times 415 + 356 + 463) = -2064 \text{ kJ/mol}.
  3. The enthalpy of formation of gaseous methyl alcohol is the sum of these two steps: ΔHf(CH3OH(g))=1836−2064=−228 kJ/mol\Delta H_f(\text{CH}_3\text{OH(g)}) = 1836 - 2064 = -228 \text{ kJ/mol}.
  4. Finally, convert gaseous methyl alcohol to liquid methyl alcohol using the enthalpy of vaporization. Since we are going from gas to liquid, it's the negative of vaporization enthalpy. ΔHcondensation=−ΔHvaporization=−38 kJ/mol\Delta H_{\text{condensation}} = - \Delta H_{\text{vaporization}} = -38 \text{ kJ/mol}.
  5. The enthalpy of formation of liquid methyl alcohol is the sum of the enthalpy of formation of gaseous methyl alcohol and the enthalpy of condensation: ΔHf(CH3OH(l))=−228−38=−266 kJ/mol\Delta H_f(\text{CH}_3\text{OH(l)}) = -228 - 38 = -266 \text{ kJ/mol}.

Answer: −266 kJ mol−1-266 \text{ kJ mol}^{-1}