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Anton
29 days ago
12

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 [2.6K]29 days ago
4 0

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.

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