Answer:
[tex]-8.032\ \text{kJ/mol}[/tex]
Explanation:
m = Mass of mixture = 62.6 g
n = Number of mole of reactant = 0.15 mol
[tex]\Delta T[/tex] = Change in temperature = [tex]4.6^{\circ}\text{C}[/tex]
c = Specific heat of mixture = [tex]4.184\ \text{J/g}^{\circ}\text{C}[/tex] (assumed to be the same as water)
Heat generated
[tex]Q=mc\Delta T\\\Rightarrow Q=62.6\times 4.184\times 4.6\\\Rightarrow Q=1204.82\ \text{J}=1.20482\ \text{kJ}[/tex]
Heat of reaction is given by
[tex]\Delta H=-\dfrac{Q}{n}\\\Rightarrow \Delta H=-\dfrac{1.20482}{0.15}\\\Rightarrow \Delta H=-8.032\ \text{kJ/mol}[/tex]
The heat of the reaction is [tex]-8.032\ \text{kJ/mol}[/tex]
A 62.6 g-reaction mixture that contains 0.15 moles of reactant A and causes a temperature increase of 4.6 °C in a calorimeter, has a heat of reaction of -8.0 kJ/mol.
We have a reaction mixture in a calorimeter that causes a temperature increase of 4.6 °C (ΔT). We can calculate the heat absorbed by the solution (Qs) using the following expression.
[tex]Qs = c \times m \times \Delta T = \frac{4.184J}{g.\° C} \times 62.6 g \times 4.6 \° C \times \frac{1kJ}{1000J} = 1.2 kJ[/tex]
where,
According to the law of conservation of energy, the sum of the heat absorbed by the solution and the heat released by the reaction (Qr) is zero.
[tex]Qs + Qr = 0\\\\Qr = -Qs = -1.2 kJ[/tex]
1.2 kJ are released by the reaction of 0.15 moles of reactant A. The heat of the reaction, in kJ/mol, is:
[tex]\frac{-1.2kJ}{0.15mol} = -8.0 kJ/mol[/tex]
A 62.6 g-reaction mixture that contains 0.15 moles of reactant A and causes a temperature increase of 4.6 °C in a calorimeter, has a heat of reaction of -8.0 kJ/mol.
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