Question

a student dissolves 11.7 g of ammonium chloride ($\text{nh}_4\text{cl}$) in 300. g of water in a well-insulated open cup. she then observes the temperature of the water fall from 21.0 $^\circ$c to 18.7 $^\circ$c over the course of 8.4 minutes.
use this data, and any information you need from the aleks data resource, to answer the questions below about this reaction:
$\text{nh}_4\text{cl}(s) \
ightarrow \text{nh}_4^+(aq) + \text{cl}^-(aq)$
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
○ neither

1. 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.
$square$ kj

1. calculate the reaction enthalpy $\delta h_{\text{rxn}}$ per mole of $\text{nh}_4\text{cl}$.

round your answer to 2 significant digits.
$square$ $\frac{\text{kj}}{\text{mol}}$

Explanation:

Step1: Classify reaction type

The water temperature decreases, so the reaction absorbs heat from the surroundings, meaning it is endothermic.

Step2: Calculate heat absorbed by reaction

Assume the solution has the same specific heat as water ($c = 4.184\ \frac{\text{J}}{\text{g}^\circ\text{C}}$), and total mass is $300.\ \text{g} + 11.7\ \text{g} = 311.7\ \text{g}$. Temperature change $\Delta T = 18.7^\circ\text{C} - 21.0^\circ\text{C} = -2.3^\circ\text{C}$.
The heat lost by the solution is:
$$q_{\text{soln}} = mc\Delta T = 311.7\ \text{g} \times 4.184\ \frac{\text{J}}{\text{g}^\circ\text{C}} \times (-2.3^\circ\text{C})$$
$$q_{\text{soln}} \approx -2990\ \text{J} = -3.0\ \text{kJ}$$
Heat absorbed by the reaction: $q_{\text{rxn}} = -q_{\text{soln}} = 3.0\ \text{kJ}$

Step3: Calculate moles of $\text{NH}_4\text{Cl}$

Molar mass of $\text{NH}_4\text{Cl}$ is $14.01 + 4\times1.008 + 35.45 = 53.492\ \frac{\text{g}}{\text{mol}}$.
$$n = \frac{11.7\ \text{g}}{53.492\ \frac{\text{g}}{\text{mol}}} \approx 0.2187\ \text{mol}$$

Step4: Calculate $\Delta H_{\text{rxn}}$ per mole

$$\Delta H_{\text{rxn}} = \frac{q_{\text{rxn}}}{n} = \frac{3.0\ \text{kJ}}{0.2187\ \text{mol}} \approx 14\ \frac{\text{kJ}}{\text{mol}}$$

Answer:

Is this reaction exothermic, endothermic, or neither?
$\circ$ exothermic
$\bullet$ endothermic
$\circ$ neither

Amount of heat absorbed: $3.0$ kJ

Reaction enthalpy $\Delta H_{\text{rxn}}$: $14\ \frac{\text{kJ}}{\text{mol}}$