Question

a student dissolves 10.7 g of potassium hydroxide (koh) in 200. g of water in a well-insulated open cup. he then observes the temperature of the water rise from 21.0 °c to 30.1 °c over the course of 6.8 minutes.
use this data, and any information you need from the aleks data resource, to answer the questions below about this reaction:
$ce{koh(s) -> k^{+}(aq) + oh^{-}(aq)}$
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 $delta h_{rxn}$ per mole of koh.

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

Explanation:

Step1: Classify reaction type

The temperature of the solution increases, so the reaction releases heat, meaning it is exothermic.

Step2: Calculate temperature change

$\Delta T = T_{final} - T_{initial} = 30.1\ ^\circ\text{C} - 21.0\ ^\circ\text{C} = 9.1\ ^\circ\text{C}$

Step3: Calculate mass of solution

Assume the solution has the same density as water ($1\ \text{g/mL}$), so mass of solution $m = 200.\ \text{g} + 10.7\ \text{g} = 210.7\ \text{g}$

Step4: Calculate heat absorbed by solution

Use $q = mc\Delta T$, where $c = 4.184\ \text{J/(g·}^\circ\text{C)}$ (specific heat of water, assumed for solution):
$q = 210.7\ \text{g} \times 4.184\ \text{J/(g·}^\circ\text{C)} \times 9.1\ ^\circ\text{C} \approx 7990\ \text{J} = 8.0\ \text{kJ}$
This is the heat released by the reaction, so $q_{rxn} = -8.0\ \text{kJ}$

Step5: Calculate moles of KOH

Molar mass of $\text{KOH} = 39.10 + 16.00 + 1.008 = 56.108\ \text{g/mol}$
$n = \frac{10.7\ \text{g}}{56.108\ \text{g/mol}} \approx 0.1907\ \text{mol}$

Step6: Calculate reaction enthalpy

$\Delta H_{rxn} = \frac{q_{rxn}}{n} = \frac{-8.0\ \text{kJ}}{0.1907\ \text{mol}} \approx -42\ \text{kJ/mol}$

Answer:

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

Amount of heat released: $8.0$ kJ

Reaction enthalpy $\Delta H_{rxn}$: $\boldsymbol{-42\ \frac{\text{kJ}}{\text{mol}}}$