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lana66690 [7]
3 years ago
10

A student dissolves 14.9 g of ammonium chloride (NH4Cl in 250. g of water in a well-insulated open cup. She then observes the te

mperature of the water fall from 20.0 °C to 16.0 °C over the course of 5.2 minutes. Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction: NH,Cl(s) – NH (aq) + Cl (aq) You can make any reasonable assumptions about the physical properties of the solution. Be sure answers you calculate using measured data are rounded to 2 significant digits. 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. O exothermic Is this reaction exothermic, endothermic, or neither? endothermic x I ? neither 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. kJ
Calculate the reaction enthalpy AH xn per mole of NHACI.
Chemistry
1 answer:
ololo11 [35]3 years ago
6 0

Answer:

Endothermic

4.433kJ

15.91 kJ / mol

Explanation:

In water, ammonium chloride dissolves thus:

NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq)

As the student observed the temperature decreased, the reaction is <em>endothermic </em>because absorbs heat from surroundigns when occurs.

It is possible to find the absorbed heat using coffee cup calorimeter equation:

Q = m×C×ΔT

<em>Where Q is heat, m is mass of solution (250g + 14.9g = 264.9g), C is specific heat of solution (It is possible to assume specific heat of pure water, 4.184J/g°C), and ΔT is change in temperature (20.0°C-16.0°C = 4.0°C)</em>

Replacing:

Q = 264.9g×4.184J/g°C×4.0°C

Q = 4433 J = <em>4.433kJ</em>

<em></em>

As enthalpy is the change in heat per mole of reaction, moles of ammonium chloride that reacted were:

14.9g NH₄Cl × (1mol / 53.491g) = <em>0.2786 moles</em>

As heat produced per 0.2786moles were 4.433kJ, heat per mole of ammonium chloride is:

4.433kJ / 0.2786mol = <em>15.91 kJ / mol</em>

<em />

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