# 0.8 Energetics of chemical reactions  (Page 3/7)

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Second, and perhaps more importantly for our purposes, we can use the known specific heat of water to measurethe heat released in any chemical reaction. To analyze a previous example, we observed that the combustion of 1.0g of methane gasreleased sufficient heat to increase the temperature of 1000g of water by 13.3°C. The heat capacity of 1000g of water must be $1000g4.184\frac{J}{g°C}=4184\frac{J}{°C}$ . Therefore, by , elevating the temperature of 1000g of water by 13.3°C must require $55,650J=55.65\mathrm{kJ}$ of heat. Therefore, burning 1.0g of methane gas produces exactly 55.65kJ of heat.

The method of measuring reaction energies by capturing the heat evolved in a water bath and measuring thetemperature rise produced in that water bath is called calorimetry . This method is dependent on the equivalence of heat and work as transfers of energy, and on thelaw of conservation of energy. Following this procedure, we can straightforwardly measure the heat released or absorbed in anyeasily performed chemical reaction. For reactions which are difficult to initiate or which occur only under restrictedconditions or which are exceedingly slow, we will require alternative methods.

## Observation 2: hess' law of reaction energies

Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction ofcarbon (coal) and water:

$C\left(s\right)+2{H}_{2}O\left(g\right)\to C{O}_{2}\left(g\right)+2{H}_{2}\left(g\right)$

Calorimetry reveals that this reaction requires the input of 90.1kJ of heat for every mole of $C\left(s\right)$ consumed. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: inchemical terms, $q> 0$ for an endothermic reaction. When heat is evolved, the reaction is exothermic and $q< 0$ by convention.

It is interesting to ask where this input energy goes when the reaction occurs. One way to answer thisquestion is to consider the fact that the reaction converts one fuel, $C\left(s\right)$ , into another, ${H}_{2}\left(g\right)$ . To compare the energy available in each fuel, we can measure theheat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that

$C\left(s\right)+{O}_{2}\left(g\right)\to C{O}_{2}\left(g\right)$

produces 393.5kJ for one mole of carbon burned; hence $q=-393.5\mathrm{kJ}$ . The reaction

$2{H}_{2}\left(g\right)+{O}_{2}\left(g\right)\to 2{H}_{2}O\left(g\right)$

produces 483.6kJ for two moles of hydrogen gas burned, so $q=-483.6\mathrm{kJ}$ . It is evident that more energy is available from combustion of thehydrogen fuel than from combustion of the carbon fuel, so it is not surprising that conversion of the carbon fuel to hydrogen fuelrequires the input of energy.

Of considerable importance is the observation that the heat input in , 90.1kJ, is exactly equal to the difference between the heat evolved, -393.5kJ, in the combustion of carbon and the heat evolved, -483.6kJ, in the combustion of hydrogen . This is not a coincidence: if we take the combustion of carbon and add to it the reverse of the combustion of hydrogen , we get $C\left(s\right)+{O}_{2}\left(g\right)\to C{O}_{2}\left(g\right)$ $2{H}_{2}O\left(g\right)\to 2{H}_{2}\left(g\right)+{O}_{2}\left(g\right)$

$C\left(s\right)+{O}_{2}\left(g\right)+2{H}_{2}O\left(g\right)\to C{O}_{2}\left(g\right)+2{H}_{2}\left(g\right)+{O}_{2}\left(g\right)$

Canceling the ${O}_{2}\left(g\right)$ from both sides, since it is net neither a reactant nor product, is equivalent to . Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen (or more accurately, adding the reverse of the combustion of hydrogen ) yields . And, the heat of the combustion of carbon minus the heat of the combustion of hydrogen equals the heat of .

#### Questions & Answers

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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Sherica
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Sherica
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Tamia
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Uday
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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China
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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I'm interested in nanotube
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
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Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
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Uday
I'm interested in Nanotube
Uday
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Prasenjit
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