Answered

Find answers to your questions and expand your knowledge with IDNLearn.com. Get the information you need from our community of experts, who provide detailed and trustworthy answers.

A student dissolves 15.0 g of ammonium chloride(NH4Cl) in 250. 0 g of water in a well-insulated open cup. She then observes the temperature of the water fall from 20.0 oC to 16.0 oC over the course of minutes. Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction:
NH4Cl(s) rightarrow NH4+(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 deltaHrxn per mole of NH4CI.

Sagot :

Sebassandinidn

Answer:  

1) Endothermic.  

2) [tex]Q_{rxn}=4435.04J[/tex]  

3) [tex]\Delta _rH=15.8kJ/mol[/tex]

Explanation:  

Hello there!  

1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.  

2) Now, for the total heat released by the reaction, we first need to assume that all of it is released 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}=-(15.0g+250.0g)*4.184\frac{J}{g\°C}(16.0-20.0)\°C\\\\ Q_{rxn}=4435.04J[/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 NH4Cl, we proceed as follows:

[tex]\Delta _rH=\frac{ Q_{rxn}}{n}\\\\\Delta _rH= \frac{ 4435.04J}{15.0g*\frac{1mol}{53.49g} } *\frac{1kJ}{1000J} \\\\\Delta _rH=15.8kJ/mol[/tex]

Best regards!  

Best regards!

We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.