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When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution (dissolving) can be determined using a coffee cup calorimeter. In the laboratory a general chemistry student finds that when 5.15 g of NH4Br(s) are dissolved in 105.00 g of water, the temperature of the solution drops from 23.82 to 21.63 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.76 J/°C. Based on the student's observation, calculate the enthalpy of dissolution of NH4Br(s) in kJ/mol.

Answer :

Answer : The enthalpy change of dissolution of [tex]NH_4Br[/tex] is 18.4 kJ/mole

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]1.76J/^oC[/tex]

[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]

[tex]m_2[/tex] = mass of water = 105.00 g

[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(23.82-21.63)=2.19^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]q=[(1.76J/^oC\times 2.19^oC)+(105.00g\times 4.184J/g^oC\times 2.19^oC)][/tex]

[tex]q=965.9652J=0.966kJ[/tex]

Now we have to calculate the enthalpy change of dissolution of [tex]NH_4Br[/tex]

[tex]\Delta H=\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy change = ?

q = heat released = 0.966 kJ

m = mass of [tex]NH_4Br[/tex] = 5.15 g

Molar mass of [tex]NH_4Br[/tex] = 97.94 g/mol

[tex]\text{Moles of }NH_4Br=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{5.15g}{97.94g/mole}=0.0526mole[/tex]

[tex]\Delta H=\frac{0.966kJ}{0.0526mole}=18.4kJ/mole[/tex]

Therefore, the enthalpy change of dissolution of [tex]NH_4Br[/tex] is 18.4 kJ/mole

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