Question

a student dissolves 11.8 g of sodium chloride (nacl) in 250. g of water in a well-insulated open cup. he then observes the temperature of the water fall from 21.0 °c to 19.5 °c over the course of 3.3 minutes.use this data, and any information you need from the aleks data resource, to answer the questions below about this reaction:nacl(s) → na⁺(aq) + cl⁻(aq)you can make any reasonable assumptions about the physical properties of the solution.note for advanced students: its 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?○ exothermic○ endothermic○ neither2. 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.round your answer to 2 significant digits.____ kj3. calculate the reaction enthalpy $delta h_{rxn}$ per mole of nacl.round your answer to 2 significant digits.$\frac{\text{kj}}{\text{mol}}$

Explanation:

Step1: Identify temp change, reaction type

The water temperature decreases, so the reaction absorbs heat from water → endothermic.
$\Delta T = T_{final} - T_{initial} = 19.5\ ^\circ\text{C} - 21.0\ ^\circ\text{C} = -1.5\ ^\circ\text{C}$

Step2: Calculate heat absorbed by water

Assume solution has same density/specific heat as water: $c=4.184\ \text{J/g}^\circ\text{C}$, mass = 250. g.
$q_{water} = m \cdot c \cdot \Delta T = 250.\ \text{g} \times 4.184\ \text{J/g}^\circ\text{C} \times (-1.5\ ^\circ\text{C}) = -1570.5\ \text{J} = -1.6\ \text{kJ}$

Step3: Find heat absorbed by reaction

Heat lost by water = heat gained by reaction: $q_{rxn} = -q_{water} = 1.6\ \text{kJ}$

Step4: Moles of NaCl used

Molar mass of NaCl: $M=22.99+35.45=58.44\ \text{g/mol}$
$n = \frac{m}{M} = \frac{11.8\ \text{g}}{58.44\ \text{g/mol}} = 0.202\ \text{mol}$

Step5: Calculate enthalpy per mole

$\Delta H_{rxn} = \frac{q_{rxn}}{n} = \frac{1.6\ \text{kJ}}{0.202\ \text{mol}} = 7.9\ \text{kJ/mol}$

Answer:

• Reaction type: endothermic
• Heat absorbed by reaction: $1.6$ kJ
• Reaction enthalpy: $\boldsymbol{7.9\ \frac{\text{kJ}}{\text{mol}}}$