The standard enthalpies of formation of CO2 (g) and H2O (l) are -393.5 kJ/mol and -285.8 kJ/mol....
Question:
The standard enthalpies of formation of CO2 (g) and H2O (l) are -393.5 kJ/mol and -285.8 kJ/mol. Given the standard enthalpy change of the following reaction, calculate the standard enthalpy of formation of acetone C3H6O: C3H6O(l) + 4 O2(g) --> 3 CO2(g) + 3 H2O(l) ?H^? = -1790 kJ
Using Standard Enthalpies of Formation :
The enthalpy change for any reaction can be obtained by invoking the Hess's law of constant heat summation, provided we have access to the standard enthalpies of formation of all the reactants and products. The simple rule is:
{eq}\Delta H_{rxn} = \Sigma \Delta H_f(products) - \Sigma \Delta H_f(reactants) {/eq}.
Note that the standard enthalpy of formation of a pure element in its standard state is zero.
Answer and Explanation: 1
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View this answerThe equation of interest is the combustion of acetone:
{eq}C_3H_6O(l) + 4O_2(g) \rightarrow 3CO_2(g) + 3H_2O(l) {/eq}.
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