Using the table of standard enthalpies of formation at T = 298 K in the provided pdf, calculate
Δ
𝐻
∘
for the following reaction:

2
𝐶
𝐻
3
𝑂
𝐻
(
𝑙
)
+
3
𝑂
2
(
𝑔
)
→
4
𝐻
2
𝑂
(
𝑔
)
+
2
𝐶
𝑂
2
(
𝑔
)

Give your answer to ONE decimal place in kilojoules (kJ). Do NOT input the units. For example, if your answer is 12.62581 kJ, please input 12.6

If you need to input a negative value, please use the hyphen "-" followed directly by the number (no space). Do NOT use the underscore "_"). For example, if your answer is -15.1611, please input -15.2

Question

Using the table of standard enthalpies of formation at T = 298 K in the provided pdf, calculate 
Δ
𝐻
∘
 for the following reaction:

2
𝐶
𝐻
3
𝑂
𝐻
(
𝑙
)
+
3
𝑂
2
(
𝑔
)
→
4
𝐻
2
𝑂
(
𝑔
)
+
2
𝐶
𝑂
2
(
𝑔
)

Give your answer to ONE decimal place in kilojoules (kJ). Do NOT input the units. For example, if your answer is 12.62581 kJ, please input 12.6

If you need to input a negative value, please use the hyphen "-" followed directly by the number (no space). Do NOT use the underscore "_"). For example, if your answer is -15.1611, please input -15.2

BlackTom AI · Accepted Answer

Standard enthalpy of reaction is given by the sum of standard enthalpies of formation of products minus those of reactants:
ΔH°rxn = Σ n ΔHf°(products) − Σ n ΔHf°(reactants).

Relevant standard formation enthalpies taken from the provided table at 298 K (kJ mol−1):
- ΔHf°(CH3OH (l)) = −238.35
- ΔHf°(H2O (g)) = −241.80
- ΔHf°(CO2 (g)) = −393.51
- ΔHf°(O2 (g)) = 0.00 (element in standard state)

Step-by-step calculation:
1) Write the reaction with stoichiometric coefficients:
2 CH3OH(l) + 3 O2(g) → 4 H2O(g) + 2 CO2(g)

2) Sum formation enthalpies of products (weighted by coefficients):
4 × ΔHf°(H2O(g)) = 4 × (−241.80) = −967.20 kJ
2 × ΔHf°(CO2(g)) = 2 × (−393.51) = −787.02 kJ
Total (products) = −967.20 + (−787.02) = −1,754.22 kJ

3) Sum formation enthalpies of reactants:
2 × ΔHf°(CH3OH(l)) = 2 × (−238.35) = −476.70 kJ
3 × ΔHf°(O2(g)) = 3 × 0.00 = 0.00 kJ
Total (reactants) = −476.70 kJ

4) Compute ΔH°rxn:
ΔH°rxn = (Total products) − (Total reactants)
ΔH°rxn = (−1,754.22) − (−476.70) = −1,277.52 kJ

5) Round to one decimal place:
ΔH°rxn = −1,277.5 kJ

Why this is correct: the calculation uses the standard formation enthalpies for the exact physical states (liquid methanol, gaseous water and CO2). Oxygen is an element and contributes zero. The negative sign indicates the reaction is exothermic. Using different tabulated ΔHf° values (e.g., CH3OH for gas instead of liquid, or slightly different tabulated numbers) would change the numeric result; such discrepancies explain small variations but the correct result using the provided table is −1,277.5.

Conclusion: The computed standard enthalpy change is −1277.5, which matches the provided answer.

类似问题

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Which of the following reactions is a standard formation reaction (i.e. Δ 𝐻 𝑟 𝑥 𝑛 ∘ = Δ 𝐻 𝑓 ∘ ) at T = 298 K?

Which balanced thermochemical equation corresponds to the enthalpy of formation (ΔHfo) of carbon monoxide, CO(g)? The enthalpy of formation (ΔHfo) of CO(g) is –110.53 kJ∙mol–1

Using the given standard enthalpies of formation, calculate the standard enthalpy change for the following reaction: 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O (g) ΔH° = ? kJ ΔH°f (kJ/mol) C4H10(g) –125 CO2(g) –394 H2O(g) –242 09A

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