Respuesta :
Answer : The enthalpy change for the solution is 42.8 kJ/mol
Explanation :
Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water
[tex]q=[q_1+q_2][/tex]
[tex]q=[c_1\times \Delta T+m\times c_2\times \Delta T][/tex]
where,
q = heat released by the reaction
[tex]q_1[/tex] = heat absorbed by the calorimeter
[tex]q_2[/tex] = heat absorbed by the water
[tex]c_1[/tex] = specific heat of calorimeter = [tex]15.8J/^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m[/tex] = mass of water = 100.0 g
[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(32.0-23.9)=8.1^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)][/tex]
[tex]q=3513.8J=3.5138kJ[/tex] (1 kJ = 1000 J)
Now we have to calculate the enthalpy change for the solution.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
q = heat released = 3.5138 kJ
m = mass of NaOH = 3.25 g
Molar mass of NaOH = 40 g/mole
[tex]\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole[/tex]
Now,
[tex]\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol[/tex]
Therefore, the enthalpy change for the solution is 42.8 kJ/mol
The [tex]\rm \Delta H[/tex] for the reaction is 42.8 kJ/mol.
[tex]\rm \Delta H\;=\;\frac{q}{n}[/tex]
q = heat released by the reaction
= heat absorbed by calorimeter + heat absorbed by water
= heat capacity of calorimeter [tex]\times[/tex] change in temperature + mass of water [tex]\times[/tex] heat capacity of water [tex]\times[/tex] change in temperature
= 15.8 J[tex]\rm /^\circ C[/tex] [tex]\times[/tex] (32 - 29 [tex]\rm ^\circ C[/tex] ) + 100 g [tex]\times[/tex] 4.18 J[tex]\rm /^\circ C[/tex] [tex]\times[/tex] (32 - 29 [tex]\rm ^\circ C[/tex])
= 3513.8 J
q = 3.5138 kJ
n = moles of NaOH
moles of NaOH = [tex]\rm \frac{weight}{molecular\;weight}[/tex]
moles of NaOH = [tex]\rm \frac{3.25}{40}[/tex]
n = 0.0812 moles
[tex]\rm \Delta H\;=\;\frac{q}{n}[/tex]
[tex]\rm \Delta H\;=\;\frac{3.15}{0.08}[/tex] kJ/mol
[tex]\rm \Delta[/tex]H = 42.8 kJ/mol.
For more information about calorimeter, refer the link:
https://brainly.com/question/10987564?referrer=searchResults
