Using this information together with the standard enthalpies of formation of O2(g), CO2(g), and H2O(l) from Appendix C, calculat

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Anton

2 months ago

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Using this information together with the standard enthalpies of formation of O2(g), CO2(g), and H2O(l) from Appendix C, calculat

e the standard enthalpy of formation of acetone.

Chemistry

1 answer:

Tems11 [2.7K] · Answer 1 · 2026-03-10T11:24:30+03:00

The question is not fully stated; here is the full version:

Using this data alongside the standard enthalpies of formation for $O_2(g)$ , $CO_2(g)$ , and $H_2O(l)$ found in Appendix C, determine the standard enthalpy of formation for acetone.

The complete combustion of 1 mole of acetone $(C_3H_6O)$ releases 1790 kJ:

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l);\Delta H^o=-1790kJ$

Answer: The standard enthalpy of formation for $CO_2(g)$ is calculated to be -247.9 kJ/mol

Explanation:

Enthalpy change represents the variation in enthalpy for all products and reactants based on their respective mole counts. This is denoted as $\Delta H^o$

The enthalpy change calculation for a chemical reaction follows this equation:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

Concerning the chemical reaction in question:

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l)$

The equation reflecting the enthalpy change for this reaction is:

$\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(3\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_3H_6O(l))})+(4\times \Delta H^o_f_{(O_2(g))})]$

Provided data includes:

$\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-1790kJ$

Substituting values from the equation gives us:

$-1790=[(3\times {(-393.5)})+(3\times (-285.8))]-[(1\times \Delta H^o_f_{(C_3H_6O(g))})+(4\times (0))]\\\\\Delta H^o_f_{(C_3H_6O(g))}=-247.9kJ/mol$

Thus, the enthalpy of formation of $C_3H_6O(g)$ computes to -247.9 kJ/mol.