Question

Find δrh for the combustion of biphenyl in kj mol−1 at 298 k.

Answer 1

The $\mathbf{\Delta _rH }$ of combustion of biphenyl is -6312.8 kJ/mol.

What is the enthalpy change of a reaction?

The enthalpy change of a reaction is denoted by ΔHrxn and it is the disparities in the enthalpy between the product vs the reactant.

Since there is limited information in the question, Let's assume that:

• When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C.

The heat gained by the calorimeter:

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

where;

• C = heat capacity of the bomb calorimeter = 5.861 KJ/ °C

$\mathbf{q_{cal} = 5.861 \times (29.419 -25.823)}$

$\mathbf{q_{cal}= 21.076 \ kJ}$

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

The change in the internal energy is:

$\mathbf{\Delta U_r = -21.076 \times \dfrac{154}{0.5141}} \\ \\ \\ \mathbf{\Delta U_r = -6313 \ kJ/mol}$

In a Bomb calorimeter, the chemical reaction of biphenyl can be computed as:

$\mathbf{C_{12}H_{10}_{(s)} + \dfrac{27}{2}O_{2(g)} \to 12 CO_{2 (g)} + 5 H_2O_{(g)}}$

The number of moles (n) in the chemical reaction is:

$\mathbf{= (12+5) - \dfrac{27}{2}}$

= 3.5

Therefore, $\mathbf{\Delta _rH = \Delta E + \Delta nRT}$

$\mathbf{\Delta _rH = (-6313 +3.5 \times 8.314 \times 10^{-3} \times 3.596 )}$

$\mathbf{\Delta _rH = -6312.8\ kJ/mol}$

Therefore, we can conclude that the $\mathbf{\Delta _rH }$ of the combustion of biphenyl is -6312.8 kJ/mol.

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