State Hess's law and give a simple example.

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Multiple Choice

State Hess's law and give a simple example.

Explanation:
Hess's law is about adding up enthalpy changes along a path because enthalpy is a state function—the overall enthalpy change depends only on the starting and ending states, not on how you get from one to the other. That means you can break a reaction into steps and sum their ΔH values to obtain the total ΔHrxn. A convenient way to express this is that the enthalpy change of a reaction equals the difference between the sum of standard enthalpies of formation of the products and that of the reactants: ΔHrxn = ΣΔHf(products) − ΣΔHf(reactants). A simple example is the combustion of methane: CH4 + 2 O2 → CO2 + 2 H2O(l). Using standard enthalpies of formation, ΔHrxn = [ΔHf(CO2) + 2ΔHf(H2O)] − [ΔHf(CH4) + 2ΔHf(O2)]. Since the enthalpy of formation of O2 is zero, you can plug in the values and find a negative ΔHrxn (about −890 kJ per mole of CH4), illustrating a large release of energy. This shows why the statement about the enthalpy change being the sum of the enthalpy changes of steps is the correct description of Hess's law, and it also reflects why enthalpy is considered independent of the path taken. The other ideas—such as enthalpy being simply the product of enthalpies or equating enthalpy change to the energy of the surroundings—do not capture Hess's law.

Hess's law is about adding up enthalpy changes along a path because enthalpy is a state function—the overall enthalpy change depends only on the starting and ending states, not on how you get from one to the other. That means you can break a reaction into steps and sum their ΔH values to obtain the total ΔHrxn. A convenient way to express this is that the enthalpy change of a reaction equals the difference between the sum of standard enthalpies of formation of the products and that of the reactants: ΔHrxn = ΣΔHf(products) − ΣΔHf(reactants).

A simple example is the combustion of methane: CH4 + 2 O2 → CO2 + 2 H2O(l). Using standard enthalpies of formation, ΔHrxn = [ΔHf(CO2) + 2ΔHf(H2O)] − [ΔHf(CH4) + 2ΔHf(O2)]. Since the enthalpy of formation of O2 is zero, you can plug in the values and find a negative ΔHrxn (about −890 kJ per mole of CH4), illustrating a large release of energy.

This shows why the statement about the enthalpy change being the sum of the enthalpy changes of steps is the correct description of Hess's law, and it also reflects why enthalpy is considered independent of the path taken. The other ideas—such as enthalpy being simply the product of enthalpies or equating enthalpy change to the energy of the surroundings—do not capture Hess's law.

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