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A student dissolves 10.7 g of lithium chloride (LiCl) in 300. g of water in a well-insulated open cup. He then observes the temperature of the water rise from 22.0 °C to 28.6 °C over the course of 3.8 minutes. Use this data, and any information you need from the ALEKS Data resource:
LiCl(s) rightarrow Li+(aq) + Cl-(aq)
You can make any reasonable assumptions about the physical properties of the solution. Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and published values for this reaction.
1) Is this reaction exothermic, endothermic, or neither?
2) If you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in this case.
3) Calculate the reaction enthalpy delta Hrxn per mole of LiCl.

Sagot :

sebassandin

Answer:

1) Exothermic.

2) [tex]Q_{rxn}=-8580J[/tex]

3) [tex]\Delta _rH=-121.0kJ/mol[/tex]

Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature increases the reaction is exothermic because it is releasing heat to solution, that is why the temperature goes from 22.0 °C to 28.6 °C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is absorbed by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

[tex]Q_{rxn}=-m_{Total}C(T_2-T_1)\\\\Q_{rxn}=-(300g+10.7g)*4.184 \frac{J}{g\°C} (28.6\°C-22.0\°C)\\\\Q_{rxn}=-8580J[/tex]

3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case LiCl, we proceed as follows:

[tex]\Delta _rH=\frac{Q_{rxn}}{n_{LiCl}} \\\\\Delta _rH=\frac{-8580J}{10.7g*\frac{1mol}{150.91g} }*\frac{1kJ}{1000J} \\\\\Delta _rH=-121.0kJ/mol[/tex]

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