Question

a student dissolves 13.6 g of ammonium nitrate ($\text{nh}_4\text{no}_3$) in 250. g of water in a well-insulated open cup. she then observes the temperature of the water fall from 23.0 °c to 17.6 °c over the course of 5.3 minutes.
use this data, and any information you need from the aleks data resource, to answer the questions below about this reaction:
$\text{nh}_4\text{no}_3(s) \
ightarrow \text{nh}_4^+(aq) + \text{no}_3^-(aq)$
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 $\text{nh}_4\text{no}_3$.

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

Explanation:

Step1: Identify reaction type

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

Step2: Calculate heat change of solution

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 $250. + 13.6 = 263.6\ \text{g}$.
Temperature change: $\Delta T = 17.6^\circ\text{C} - 23.0^\circ\text{C} = -5.4^\circ\text{C}$
Heat change of solution: $q_{\text{soln}} = mc\Delta T = 263.6\ \text{g} \times 4.184\ \frac{\text{J}}{\text{g}^\circ\text{C}} \times (-5.4^\circ\text{C}) = -5980\ \text{J} = -5.98\ \text{kJ}$

Step3: Find heat absorbed by reaction

For an insulated system, $q_{\text{rxn}} = -q_{\text{soln}} = -(-5.98\ \text{kJ}) = 6.0\ \text{kJ}$ (rounded to 2 sig figs)

Step4: Calculate moles of $\text{NH}_4\text{NO}_3$

Molar mass of $\text{NH}_4\text{NO}_3$: $14.01 + 4\times1.008 + 14.01 + 3\times16.00 = 80.052\ \frac{\text{g}}{\text{mol}}$
Moles: $n = \frac{13.6\ \text{g}}{80.052\ \frac{\text{g}}{\text{mol}}} = 0.1699\ \text{mol}$

Step5: Calculate $\Delta H_{\text{rxn}}$ per mole

$\Delta H_{\text{rxn}} = \frac{q_{\text{rxn}}}{n} = \frac{6.0\ \text{kJ}}{0.1699\ \text{mol}} = 35\ \frac{\text{kJ}}{\text{mol}}$ (rounded to 2 sig figs)

Answer:

Is this reaction exothermic, endothermic, or neither?
endothermic

Amount of heat absorbed by the reaction:
6.0 kJ

Reaction enthalpy $\Delta H_{\text{rxn}}$ per mole of $\text{NH}_4\text{NO}_3$:
35 $\frac{\text{kJ}}{\text{mol}}$