Question

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 kJ mol−1 at 298 K. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.861 kJ °C−1.

Answer 1

Answer:

$\Delta_{r}U$ of the reaction is -6313 kJ/mol

$\Delta_{r}H$ of the reaction is -6312 kJ/mol

Explanation:

$Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C$

$q_{cal}= C \times \Delta T$

$=5.861 \times 3.596 = 21.076\,kJ$

$q_{rxn}= -q_{cal}= -21.076\,kJ$

$\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol$

Therefore, $\Delta_{r}U$ of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

$C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)$

$Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5$

$\Delta_{r}H=\Delta E+ \Delta n. RT$

$=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol$

Therefore, $\Delta_{r}H$ of the reaction is -6312 kJ/mol.