Question

1. when 0.5141 g of biphenyl (c₁₂h₁₀) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823°c to 29.419°c. find δᵣu and δᵣh for the combustion of biphenyl in kjmol⁻¹ at 298 k. the heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.861 kj°c⁻¹.

Explanation:

Step1: Calculate the heat absorbed by the calorimeter ($q_{cal}$)

$q_{cal}=C_{cal}\Delta T$, where $C_{cal} = 5.861\ kJ/^{\circ}C$ and $\Delta T=T_2 - T_1$.
$\Delta T=29.419^{\circ}C - 25.823^{\circ}C=3.596^{\circ}C$.
$q_{cal}=5.861\ kJ/^{\circ}C\times3.596^{\circ}C = 21.076\ kJ$.
Since the heat released by the combustion of biphenyl is equal in magnitude to the heat absorbed by the calorimeter, $q_{rxn}=-q_{cal}=- 21.076\ kJ$.

Step2: Calculate the number of moles of biphenyl ($n$)

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

Step3: Calculate $\Delta_{r}U$

$\Delta_{r}U=\frac{q_{rxn}}{n}$.
$\Delta_{r}U=\frac{-21.076\ kJ}{0.003334\ mol}=- 6321.5\ kJ/mol$.

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

The combustion reaction of biphenyl is $C_{12}H_{10}(s)+\frac{29}{2}O_{2}(g)
ightarrow12CO_{2}(g)+5H_{2}O(l)$.
$\Delta n_{gas}=n_{products,g}-n_{reactants,g}=12-\frac{29}{2}=-\frac{5}{2}$.

Step5: Calculate $\Delta_{r}H$

We know that $\Delta_{r}H=\Delta_{r}U+\Delta n_{gas}RT$.
At $T = 298\ K$ and $R=8.314\ J/(mol\cdot K)=8.314\times10^{- 3}\ kJ/(mol\cdot K)$.
$\Delta n_{gas}RT=-\frac{5}{2}\times8.314\times10^{-3}\ kJ/(mol\cdot K)\times298\ K=-6.2\ kJ/mol$.
$\Delta_{r}H=\Delta_{r}U+\Delta n_{gas}RT=-6321.5\ kJ/mol-6.2\ kJ/mol=-6327.7\ kJ/mol$.

Answer:

$\Delta_{r}U=-6321.5\ kJ/mol$, $\Delta_{r}H=-6327.7\ kJ/mol$