# 5.4 Thermochemistry: calorimetry  (Page 4/14)

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${q}_{\text{reaction}}+{q}_{\text{solution}}=0$

This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:

${q}_{\text{reaction}}=\text{−}{q}_{\text{solution}}$

This concept lies at the heart of all calorimetry problems and calculations.

## Heat produced by an exothermic reaction

When 50.0 mL of 0.10 M HCl( aq ) and 50.0 mL of 0.10 M NaOH( aq ), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction?

$\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NaCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)$

## Solution

To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C.

The heat given off by the reaction is equal to that taken in by the solution. Therefore:

${q}_{\text{reaction}}=\text{−}{q}_{\text{solution}}$

(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and its surroundings.)

Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:

${q}_{\text{solution}}={\left(c\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}m\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}T\right)}_{\text{solution}}$

To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 $×$ 10 2 g (two significant figures). The specific heat of water is approximately 4.18 J/g °C, so we use that for the specific heat of the solution. Substituting these values gives:

${q}_{\text{solution}}=\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(28.9\phantom{\rule{0.2em}{0ex}}\text{°C}-22.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)=2.89\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}$

Finally, since we are trying to find the heat of the reaction, we have:

${q}_{\text{reaction}}=\text{−}{q}_{\text{solution}}=-2.89\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}$

The negative sign indicates that the reaction is exothermic. It produces 2.89 kJ of heat.

When 100 mL of 0.200 M NaCl( aq ) and 100 mL of 0.200 M AgNO 3 ( aq ), both at 21.9 °C, are mixed in a coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?

1.34 $×$ 10 3 J; assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water

## Thermochemistry of hand warmers

When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands ( [link] ). A common reusable hand warmer contains a supersaturated solution of NaC 2 H 3 O 2 (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable NaC 2 H 3 O 2 quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).

The process ${\text{NaC}}_{2}{\text{H}}_{3}{\text{O}}_{2}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NaC}}_{2}{\text{H}}_{3}{\text{O}}_{2}\left(s\right)$ is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the NaC 2 H 3 O 2 redissolves and can be reused.

Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is $2\text{Fe}\left(s\right)+\frac{3}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Fe}}_{2}{\text{O}}_{3}\left(s\right).$ Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.

what's the easiest and fastest way to the synthesize AgNP?
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Azam
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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