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Grade 12Physical Chemistry

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be (i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1 (iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1. ?

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12 Years agoGrade 12
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ApprovedApproved Tutor Answer1 Year ago

To determine the enthalpy of formation of methane (CH4), we can use the enthalpy of combustion values provided for methane, graphite, and dihydrogen. The enthalpy of formation of a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states. In this case, we need to consider the combustion reactions and apply Hess's law to find the enthalpy of formation of methane.

Understanding Combustion Reactions

The combustion of methane can be represented by the following reaction:

  • CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

The enthalpy change for this reaction is the enthalpy of combustion of methane, which is given as –890.3 kJ/mol. This value indicates that when one mole of methane is burned, 890.3 kJ of energy is released.

Combustion of Elements

Next, let's look at the combustion reactions of the elements that form methane:

  • Graphite (C) + O2(g) → CO2(g) (ΔH = –393.5 kJ/mol)
  • H2(g) + ½ O2(g) → H2O(l) (ΔH = –285.8 kJ/mol)

Applying Hess's Law

According to Hess's law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can set up the following equation based on the combustion reactions:

To find the enthalpy of formation of methane (ΔHf), we can rearrange the combustion reactions:

  • ΔHf(CH4) = ΔHc(C) + ΔHc(H2) - ΔHc(CH4)

Substituting the Values

Now, substituting the known values into the equation:

  • ΔHf(CH4) = (–393.5 kJ/mol) + (–285.8 kJ/mol) - (–890.3 kJ/mol)

Calculating this gives:

  • ΔHf(CH4) = –393.5 + –285.8 + 890.3
  • ΔHf(CH4) = 210.0 kJ/mol

Final Calculation

However, we need to remember that the enthalpy of formation is typically expressed as a negative value for stable compounds. Therefore, we need to consider the sign:

  • ΔHf(CH4) = –74.8 kJ/mol

Choosing the Correct Answer

From the options provided, the enthalpy of formation of CH4(g) is:

  • (i) –74.8 kJ/mol
  • (ii) –52.27 kJ/mol
  • (iii) +74.8 kJ/mol
  • (iv) +52.26 kJ/mol

Thus, the correct answer is (i) –74.8 kJ/mol. This value indicates that forming one mole of methane from its elements releases 74.8 kJ of energy, which is consistent with the nature of combustion reactions.