Answer : The molar heat of combustion of benzene is 1.95 kJ/mol
Explanation :
Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water
[tex]q=[q_1+q_2][/tex]
[tex]q=[c_1\times \Delta T+m_2\times c_2\times \Delta T][/tex]
where,
q = heat released by the reaction
[tex]q_1[/tex] = heat absorbed by the calorimeter
[tex]q_2[/tex] = heat absorbed by the water
[tex]c_1[/tex] = specific heat of calorimeter = [tex]800J/^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_2[/tex] = mass of water = 100 g
[tex]\Delta T[/tex] = change in temperature = [tex]4^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=[(800J/^oC\times 4^oC)+(100g\times 4.18J/g^oC\times 4^oC)][/tex]
[tex]q=4872J[/tex]
Now we have to determine the molar heat of combustion of benzene in kJ/mol.
As, 2.50 mole of benzene produces heat on combustion = 4872 J
So, 1 mole of benzene produces heat on combustion = [tex]\frac{4872}{2.50}=1948.8J/mol=1.95kJ/mol[/tex]
conversion used : (1 kJ = 1000 J)
Therefore, the molar heat of combustion of benzene is 1.95 kJ/mol
The molar heat of combustion of benzene is 1.95 kJ/mol
A calorimeter is a device applied when measuring the change of heat taking place during a chemical or physical change.
In a calorimeter, the total heat absorbed by the reaction is equal to the heat absorbed by the calorimeter in addition to the heat absorbed by water,
Total heat absorbed by the reaction = C_cΔT+ mcΔT
where;
= (800 J/C° × 4° C) + (100 × 4.18 × 4) J
= 3200 J + 1672 J
= 4872 J
Now, in a 2.5 mol sample of benzene, the molar heat of combustion can be computed as:
[tex]\mathbf{\Delta H = \dfrac{q}{n}}[/tex]
[tex]\mathbf{\Delta H = \dfrac{4872}{2.5}}[/tex]
ΔH = 1948.8 J/mol
ΔH = 1.95 kJ/mol
Learn more about molar enthalpy of combustion here:
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