Chemistry, 25.07.2019 14:00 wyattjefferds05

A6.55 g sample of aniline (c6h5nh2, molar mass = 93.13 g/mol) was combusted in a bomb calorimeter with a heat capacity of 14.25 kj/°c. if the initial temperature was 32.9°c, use the information below to determine the value of the final temperature of the calorimeter. 4 c6h5nh2(l) + 35 o2(g) → 24 co2(g) + 14 h2o(g) + 4 no2(g)

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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

Answer from: july00

The final temperature of the calorimeter will be 48.619 $\rm ^\circ C$ .

The complete question has given the enthalpy for the reaction, $\rm \Delta H^\circ _r_x_n$ = 1.28 $\rm \times\;10^4$ kJ.

The complete reaction is:

$\rm 4\;C_6H_5NH_2\;+\;35\;O_2\;\rightarrow\;24\;CO_2\;+\;14\;H_2O\;+\;4\;NO_2$

The moles of the aniline combusted in the bomb calorimeter:

weight of aniline = 6.55 g

molecular weight = 93.13 g/mol

$\rm moles\;=\;\dfrac{weight}{molecular\;weight}$

$\rm moles\;=\;\dfrac{6.55}{93.13}$

moles = 0.07 moles.

The equation utilized that, for 4 moles of aniline enthalpy is 1.28 $\rm \times\;10^4$ kJ.

For 0.07 moles of aniline, $\rm \Delta H^\circ _r_x_n$ = $\rm \dfrac{1.28\;\times\;10^4}{4}\;\times\;0.07$

$\rm \Delta H^\circ _r_x_n$ = 224 kJ.

Given, The heat capacity = 14.25 $\rm kJ/^\circ C$ .

Initial temperature = $\rm 32.9 \;^\circ C$ .

Heat capacity of reaction = $\rm \dfrac{total\;amount\;of\;heat}{temperature\;change}$

heat capacity of the reaction = $\rm \dfrac{total\;amount\;of\;heat}{final\;temperature\;-\;initial\;temperature}$

final temperature - initial temperature = $\rm \dfrac{enthalpy\;of\;reaction}{heat\;capacity}$

Final temperature - $\rm 32.9 \;^\circ C$ = $\rm \dfrac{224}{14.25}$ $\rm kJ/^\circ C$ .

Final temperature - $\rm 32.9 \;^\circ C$ = 15.71 $\rm kJ/^\circ C$

Final temperature = 15.71 + 32.9 $\rm ^\circ C$

Final temperature = 48.619 $\rm ^\circ C$ .

The final temperature of the calorimeter will be 48.619 $\rm ^\circ C$ .

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Answer from: Quest

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A6.55 g sample of aniline (c6h5nh2, molar mass = 93.13 g/mol) was combusted in a bomb calorimeter wi...

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