Respuesta :
According to the reaction equation:
4C6H5NH2 +35 O2 → 24CO2(g) + 14 H2O(g) + 4NO2(g) ΔH = -1.28x10^4 KJ
from this equation, we can need 4 mol of aniline to give 1.28 x 10^4 KJ ΔH
first, we need to get moles of aniline = mass/molar mass
= 6.55 g / 93.13g/mol
=0.07 mol
the heat generated = moles of aniline / 4 mol * ΔH
= 0.07mol / 4 mol * 1.28 x 10^4KJ
= 224 KJ
when ΔT = Q/C
when we have C the heat capacity = 14.25 KJ/°C
ΔT = 224 / 14.25
= 15.7°C
∴ Tf = Ti + ΔT
= 32.9 °C + 15.7°C
= 48.6 °C
4C6H5NH2 +35 O2 → 24CO2(g) + 14 H2O(g) + 4NO2(g) ΔH = -1.28x10^4 KJ
from this equation, we can need 4 mol of aniline to give 1.28 x 10^4 KJ ΔH
first, we need to get moles of aniline = mass/molar mass
= 6.55 g / 93.13g/mol
=0.07 mol
the heat generated = moles of aniline / 4 mol * ΔH
= 0.07mol / 4 mol * 1.28 x 10^4KJ
= 224 KJ
when ΔT = Q/C
when we have C the heat capacity = 14.25 KJ/°C
ΔT = 224 / 14.25
= 15.7°C
∴ Tf = Ti + ΔT
= 32.9 °C + 15.7°C
= 48.6 °C
The final temperature of the calorimeter will be 48.619 [tex]\rm ^\circ C[/tex].
The complete question has given the enthalpy for the reaction, [tex]\rm \Delta H^\circ _r_x_n[/tex] = 1.28 [tex]\rm \times\;10^4[/tex] kJ.
The complete reaction is:
[tex]\rm 4\;C_6H_5NH_2\;+\;35\;O_2\;\rightarrow\;24\;CO_2\;+\;14\;H_2O\;+\;4\;NO_2[/tex]
The moles of the aniline combusted in the bomb calorimeter:
weight of aniline = 6.55 g
molecular weight = 93.13 g/mol
[tex]\rm moles\;=\;\dfrac{weight}{molecular\;weight}[/tex]
[tex]\rm moles\;=\;\dfrac{6.55}{93.13}[/tex]
moles = 0.07 moles.
The equation utilized that, for 4 moles of aniline enthalpy is 1.28 [tex]\rm \times\;10^4[/tex] kJ.
For 0.07 moles of aniline, [tex]\rm \Delta H^\circ _r_x_n[/tex] = [tex]\rm \dfrac{1.28\;\times\;10^4}{4}\;\times\;0.07[/tex]
[tex]\rm \Delta H^\circ _r_x_n[/tex] = 224 kJ.
Given, The heat capacity = 14.25 [tex]\rm kJ/^\circ C[/tex].
Initial temperature = [tex]\rm 32.9 \;^\circ C[/tex].
Heat capacity of reaction = [tex]\rm \dfrac{total\;amount\;of\;heat}{temperature\;change}[/tex]
heat capacity of the reaction = [tex]\rm \dfrac{total\;amount\;of\;heat}{final\;temperature\;-\;initial\;temperature}[/tex]
final temperature - initial temperature = [tex]\rm \dfrac{enthalpy\;of\;reaction}{heat\;capacity}[/tex]
Final temperature - [tex]\rm 32.9 \;^\circ C[/tex] = [tex]\rm \dfrac{224}{14.25}[/tex] [tex]\rm kJ/^\circ C[/tex].
Final temperature - [tex]\rm 32.9 \;^\circ C[/tex] = 15.71 [tex]\rm kJ/^\circ C[/tex]
Final temperature = 15.71 + 32.9 [tex]\rm ^\circ C[/tex]
Final temperature = 48.619 [tex]\rm ^\circ C[/tex].
The final temperature of the calorimeter will be 48.619 [tex]\rm ^\circ C[/tex].
For more information, refer to the link:
https://brainly.com/question/16200365?referrer=searchResults
