Question

A 5.0L bomb calorimeter, with a heat capacity of 14.25 kJ/°C, is filled to a pressure of 5.0 atm with O2 gas at 32.9°C. A 6.55 g sample of aniline (C6H5NH2, molar mass = 93.13 g/mol) was then combusted in this calorimeter.

Determine the value of the final temperature of the calorimeter.

4 C6H5NH2(l) + 35 O2(g) → 24 CO2(g) + 7 H2O(g) + 2 NO2(g)

ΔH°rxn= -1.28 x 104kJ/mol

Answer 1

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Given that,

Volume,$V= 5 \;L=5\times10^{-3}\;m^{3}$

Pressure, $P= 5 \;atm=5\times10^{5}\;Pa$

Temperature, $T=32.9+273\;K=304.9\;K$

Hence, no.of moles of oxygen gas is given by,

$n=\frac{PV}{RT}=\frac{5\times10^{5}\;Pa\times5\times10^{-3}\;m^{3}}{8.314\;J/mol.K\times304.9\;K}$

$\therefore n=0.983\;mol$

Also, no.of moles of aniline is given by,

$n_{1}=\frac{w}{M}=\frac{6.55\;g}{93.13\;g/mol}=0.0703\;mol$

Combustion reaction of aniline is given by,

$4 C_{6}H_{5}NH_{2}_{(l)} + 35 O_{2}_{(g) }\rightarrow 24 CO_{2}_{(g)} + 7 H_{2}O_{(g)} + 2 NO_{2}_{(g)}$

From the above balanced reaction we have, 4 moles of aniline reacts with 35 moles of O₂.

Hence, 1 mole of aniline reacts with 35/4 moles of O₂.

Therefore, 0.0703 moles of aniline reacts with,

$n'=0.0703\;mol\times\frac{35}{4}$

$\therefore n'=0.6154\;mol$

Hence, energy released for this reaction is given by,

$H=\frac{1.28\times10^{4}\;kJ/mol}{35}\times n'=\frac{1.28\times10^{4}\;kJ/mol}{35}\times0.6154\;mol$

$\mathbf{\therefore H=225.062\;kJ}$

Hence, Change/increase in temperature of the calorimeter is given by,

$\Delta T=\frac{H}{C}=\frac{225.062\;J}{14.25\;kJ/^{o}C}$

$\therefore \Delta T=15.79^{o}C}$

Hence,Final temperature of the calorimeter is given by,

$\therefore T_{final}=T+\Delta T=32.9^{o}C+15.79^{o}C}$

$\mathbf{\therefore T_{final}=48.69^{o}C}}$