how much heat is produced by the combustion of 125 g of acetylene c2h2
What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%?

We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. Energy is transferred into a system when it absorbs heat (q) from the surroundings or when the surroundings do work (w) on the system. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. 1. Writing out these reactions, and noting their relationships to the values for these compounds (from Appendix G), we have: Summing these reaction equations gives the reaction we are interested in: Summing their enthalpy changes gives the value we want to determine: So the standard enthalpy change for this reaction is ΔH° = −138.4 kJ. Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. In a thermochemical equation, the enthalpy change of a reaction is shown as a ΔH value following the equation for the reaction. For example, when 1 mole of hydrogen gas and [latex]\frac{1}{2}[/latex] mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. }\hfill \end{array}[/latex], [latex]\Delta{H}_{\text{reaction}}^{\textdegree }=\sum n\times \Delta{H}_{\text{f}}^{\textdegree }\text{(products)}-\sum n\times \Delta{H}_{\text{f}}^{\textdegree }\left(\text{reactants}\right)[/latex], [latex]\begin{array}{l}\\ =\left[2\cancel{\text{mol}{\text{HNO}}_{3}}\times \frac{-207.4\text{kJ}}{\cancel{\text{mol}{\text{HNO}}_{3}\left(aq\right)}}+1\cancel{\text{mol}\text{NO}\left(g\right)}\times \frac{\text{+90.2}\text{kJ}}{\cancel{\text{mol}\text{NO}\left(g\right)}}\right]\\ -\left[3\cancel{\text{mol}{\text{NO}}_{2}\left(g\right)}\times \frac{\text{+33.2}\text{kJ}}{\cancel{\text{mol}{\text{NO}}_{2}\left(g\right)}}+1\cancel{\text{mol}{\text{H}}_{2}\text{O}\left(l\right)}\times \frac{-285.8\text{kJ}}{\cancel{\text{mol}{\text{H}}_{2}\text{O}\left(l\right)}}\right]\\ =2\left(-207.4\text{kJ}\right)+1\left(\text{+90.2}\text{kJ}\right)-3\left(\text{+33.2}\text{kJ}\right)-1\left(-285.8\text{kJ}\right)\\ =-138.4\text{kJ}\end{array}[/latex], One mole under the same conditions would require, [latex]\frac{80.0432\cancel{\text{g}}{\text{mol}}^{-1}}{3.21\cancel{\text{g}}}\times 1000\text{J}=25\text{kJ}{\text{mol}}^{-1}. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. (Note that this is similar to determining the intensive property specific heat from the extensive property heat capacity, as seen previously.). (credit a: modification of work by Micah Sittig; credit b: modification of work by Robert Kerton; credit c: modification of work by John F. Williams). We can apply the data from the experimental enthalpies of combustion in Table 2 to find the enthalpy change of the entire reaction from its two steps: The result is shown in Figure 6. This is usually rearranged slightly to be written as follows, with ∑ representing “the sum of” and n standing for the stoichiometric coefficients: The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest. Here is a less straightforward example that illustrates the thought process involved in solving many Hess’s law problems. Is this a typo?

Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline.

The relationship between internal energy, heat, and work can be represented by the equation: as shown in Figure 1. Standard enthalpy of combustion [latex]\left(\Delta{H}_{C}^{\textdegree }\right)[/latex] is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called “heat of combustion.” For example, the enthalpy of combustion of ethanol, -1366.8 kJ/mol, is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 °C and 1 atmosphere pressure, yielding products also at 25 °C and 1 atm. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the enthalpy change (ΔH) is: The mathematical product PΔV represents work (w), namely, expansion or pressure-volume work as noted.


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