A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both solutions was 20.2 degree celsious before mixing and 26.3 degree celsious after mixing . The heat capacity of the calorimeter is 12.1 J/K. From the data, calculate delta Hfor the process
CuSO4(1 M)+2KOH(2 M)----------------------Cu(OH)2(s)+K2SO4(0.5 M)
Assume that the specific heat and density of the solution after mixing are the same as those of pure water and that the volumes are additive.

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Alleei Alleei · Answer 1 · 2019-11-06T09:55:51+01:00

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

[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 solution

[tex]c_1[/tex] = specific heat of calorimeter = [tex]12.1J/^oC[/tex]

[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_2[/tex] = mass of water or solution = [tex]Density\times Volume=1/mL\times 100.0mL=100.0g[/tex]

[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(26.3-20.2)^oC=6.1^oC[/tex]

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

[tex]q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)][/tex]

[tex]q=2623.61J[/tex]

Now we have to calculate the enthalpy change for the process.

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

where,

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

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = [tex]Concentration\times Volume=1M\times 0.050L=0.050mole[/tex]

[tex]\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole[/tex]

Therefore, the enthalpy change for the process is 52.5 kJ/mole.