73. when 0.5141 g of biphenyl (c₁₂h₁₀) undergoes combustion in a bomb calorimeter, the temperature rises…

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. $\Delta T=T_2 - T_1=29.419^{\circ}C - 25.823^{\circ}C = 3.596^{\circ}C$ $C_{cal}=5.861\ kJ^{\circ}C^{-1}$ $q_{cal}=C_{cal}\Delta T=5.861\ kJ^{\circ}C^{-1}\times3.596^{\circ}C = 21.076\ kJ$

Step2: Calculate the number of moles of biphenyl

The molar mass of biphenyl ($C_{12}H_{10}$) is $M=(12\times12.01 + 10\times1.01)\ g/mol=154.22\ g/mol$ The mass of biphenyl $m = 0.5141\ g$ The number of moles $n=\frac{m}{M}=\frac{0.5141\ g}{154.22\ g/mol}=0.003334\ mol$

Step3: Calculate $\Delta_{r}U$

For a bomb - calorimeter, $\Delta_{r}U = \frac{q_{cal}}{n}$ $\Delta_{r}U=\frac{21.076\ kJ}{0.003334\ mol}=6321.5\ kJ/mol$

Step4: Calculate $\Delta n_{gas}$ for the combustion reaction

The combustion reaction of biphenyl is $C_{12}H_{10}(s)+\frac{29}{2}O_{2}(g)\rightarrow12CO_{2}(g)+5H_{2}O(l)$ $\Delta n_{gas}=n_{products,gases}-n_{reactants,gases}=12-\frac{29}{2}=-\frac{5}{2}$

Step5: Calculate $\Delta_{r}H$

We know the relationship $\Delta_{r}H=\Delta_{r}U+\Delta n_{gas}RT$ $T = 298\ K$, $R=8.314\ J\ K^{-1}\ mol^{-1}=8.314\times10^{-3}\ kJ\ K^{-1}\ mol^{-1}$ $\Delta_{r}H=\Delta_{r}U+\Delta n_{gas}RT$ $\Delta_{r}H = 6321.5\ kJ/mol+(-\frac{5}{2})\times8.314\times10^{-3}\ kJ\ K^{-1}\ mol^{-1}\times298\ K$ $\Delta_{r}H = 6321.5\ kJ/mol-6.2\ kJ/mol=6315.3\ kJ/mol$

Answer:

$\Delta_{r}U = 6321.5\ kJ/mol$, $\Delta_{r}H = 6315.3\ kJ/mol$