The [tex]\mathbf{\Delta _rH }[/tex] of combustion of biphenyl is -6312.8 kJ/mol.
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:
The heat gained by the calorimeter:
[tex]\mathbf{q_{cal} = C \times \Delta T}[/tex]
where;
[tex]\mathbf{q_{cal} = 5.861 \times (29.419 -25.823)}[/tex]
[tex]\mathbf{q_{cal}= 21.076 \ kJ}[/tex]
[tex]\mathbf{q_{rxn}= -q_{cal} = -21.076 \ kJ}[/tex]
The change in the internal energy is:
[tex]\mathbf{\Delta U_r = -21.076 \times \dfrac{154}{0.5141}} \\ \\ \\ \mathbf{\Delta U_r = -6313 \ kJ/mol}[/tex]
In a Bomb calorimeter, the chemical reaction of biphenyl can be computed as:
[tex]\mathbf{C_{12}H_{10}_{(s)} + \dfrac{27}{2}O_{2(g)} \to 12 CO_{2 (g)} + 5 H_2O_{(g)}}[/tex]
The number of moles (n) in the chemical reaction is:
[tex]\mathbf{= (12+5) - \dfrac{27}{2}}[/tex]
= 3.5
Therefore, [tex]\mathbf{\Delta _rH = \Delta E + \Delta nRT}[/tex]
[tex]\mathbf{\Delta _rH = (-6313 +3.5 \times 8.314 \times 10^{-3} \times 3.596 )}[/tex]
[tex]\mathbf{\Delta _rH = -6312.8\ kJ/mol}[/tex]
Therefore, we can conclude that the [tex]\mathbf{\Delta _rH }[/tex] of the combustion of biphenyl is -6312.8 kJ/mol.
Learn more about enthalpy change here:
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