Question

the decomposition of trinitrotoluene (tnt, c₇h₅n₃o₆) was studied in a bomb calorimeter: 6.1891 g of tnt (molar mass = 227.1311 gmol⁻¹) was placed in the calorimeter and detonated. the temperature of the calorimeter rose from 22.905°c to 25.879°c. the heat capacity of the bomb calorimeter was 9.625 kj °c⁻¹. the decomposition of tnt occurs according to the following reaction: c₇h₅n₃o₆(s) → 3/2n₂(g) + 5/2h₂o(l) + 7/2co(g) + 7/2c(s, graphite) determine δᵣu and δᵣh (at 298 k) for the decomposition reaction.

Explanation:

Step1: Calculate the heat absorbed by the calorimeter

The heat absorbed by the calorimeter $q_{cal}$ is given by the formula $q_{cal}=C_{cal}\Delta T$, where $C_{cal}$ is the heat - capacity of the calorimeter and $\Delta T$ is the change in temperature.
$\Delta T=T_2 - T_1=25.879^{\circ}C - 22.905^{\circ}C = 2.974^{\circ}C$
$C_{cal}=9.625\ kJ\ ^{\circ}C^{-1}$
$q_{cal}=9.625\ kJ\ ^{\circ}C^{-1}\times2.974^{\circ}C = 28.62575\ kJ$

Step2: Calculate the number of moles of TNT

The number of moles $n$ of TNT is calculated using the formula $n=\frac{m}{M}$, where $m$ is the mass and $M$ is the molar mass.
$m = 6.1891\ g$ and $M = 227.1311\ g\ mol^{-1}$
$n=\frac{6.1891\ g}{227.1311\ g\ mol^{-1}}=0.02725\ mol$

Step3: Calculate $\Delta_{r}U$

For a bomb - calorimeter, the heat of reaction at constant volume $\Delta_{r}U$ is equal to the heat absorbed by the calorimeter per mole of reactant.
$\Delta_{r}U=\frac{q_{cal}}{n}=\frac{28.62575\ kJ}{0.02725\ mol}=1050.486\ kJ\ mol^{-1}$

Step4: Calculate $\Delta n_{gas}$

From the balanced chemical equation $C_{7}H_{5}N_{3}O_{6}(s)
ightarrow\frac{3}{2}N_{2}(g)+\frac{5}{2}H_{2}O(l)+\frac{7}{2}CO(g)+\frac{7}{2}C(s, graphite)$, the number of moles of gaseous products minus the number of moles of gaseous reactants $\Delta n_{gas}=\frac{3 + 7}{2}-0 = 5$

Step5: Calculate $\Delta_{r}H$

The relationship between $\Delta_{r}H$ and $\Delta_{r}U$ is $\Delta_{r}H=\Delta_{r}U+\Delta n_{gas}RT$.
$R = 8.314\ J\ K^{-1}\ mol^{-1}=8.314\times10^{- 3}\ kJ\ K^{-1}\ mol^{-1}$, $T = 298\ K$
$\Delta_{r}H=1050.486\ kJ\ mol^{-1}+5\times8.314\times10^{-3}\ kJ\ K^{-1}\ mol^{-1}\times298\ K$
$\Delta_{r}H=1050.486\ kJ\ mol^{-1}+12.395\ kJ\ mol^{-1}=1062.881\ kJ\ mol^{-1}$

Answer:

$\Delta_{r}U = 1050.49\ kJ\ mol^{-1}$, $\Delta_{r}H = 1062.88\ kJ\ mol^{-1}$