Question

a student dissolves 11.1 g of potassium hydroxide (koh) in 300. g of water in a well - insulated open cup. he then observes the temperature of the water rise from 20.0°c to 29.3°c over the course of 3.2 minutes.
use this data, and any information you need from the aleks data resource, to answer the questions below about this reaction:
koh(s)→k+(aq)+oh−(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 the correct number of significant digits.
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.
is this reaction exothermic, endothermic, or neither? exothermic endothermic 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 δhrxn per mole of koh. kj/mol

Explanation:

Step1: Determine if the reaction is exothermic or endothermic

Since the temperature of the water rises, heat is released by the reaction. So the reaction is exothermic.

Step2: Calculate the mass of the solution

Assume the density of water is \(1\space g/mL\). The mass of water \(m_{water}=300\space g\), mass of \(KOH\) \(m_{KOH} = 11.1\space g\). The mass of the solution \(m=m_{water}+m_{KOH}=300 + 11.1=311.1\space g\)

Step3: Calculate the heat absorbed by the solution

Use the formula \(q = mc\Delta T\), where \(c = 4.184\space J/(g\cdot^{\circ}C)\) (specific - heat capacity of water), \(\Delta T=T_{final}-T_{initial}=29.3 - 20.0 = 9.3^{\circ}C\)
\(q=(311.1\space g)\times(4.184\space J/(g\cdot^{\circ}C))\times(9.3^{\circ}C)\)
\(q = 311.1\times4.184\times9.3\space J\)
\(q=311.1\times38.9112\space J\)
\(q = 12015.27432\space J\approx12.0\space kJ\) (rounded to three significant figures)
Since the reaction is exothermic, \(q_{reaction}=-q_{solution}=- 12.0\space kJ\)

Step4: Calculate the number of moles of \(KOH\)

The molar mass of \(KOH\) is \(M = 39.1+16 + 1.01=56.11\space g/mol\)
\(n=\frac{m}{M}=\frac{11.1\space g}{56.11\space g/mol}\approx0.198\space mol\)

Step5: Calculate the reaction enthalpy \(\Delta H_{rxn}\) per mole of \(KOH\)

\(\Delta H_{rxn}=\frac{q_{reaction}}{n}=\frac{-12.0\space kJ}{0.198\space mol}\approx - 60.6\space kJ/mol\)

Answer:

• Is this reaction exothermic, endothermic, or neither? exothermic
• 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. \(-12.0\space kJ\)
• Calculate the reaction enthalpy \(\Delta H_{rxn}\) per mole of \(KOH\). \(-60.6\space kJ/mol\)