Question

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.

Answer 1

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

`q=[q_1+q_2]`

`q=[c_1* \Delta T+m_2* c_2* \Delta T]`

where,

q = heat released by the reaction

`q_1` = heat absorbed by the calorimeter

`q_2` = heat absorbed by the solution

`c_1` = specific heat of calorimeter =
`12.1J/^oC`

`c_2` = specific heat of water =
`4.18J/g^oC`

`m_2` = mass of water or solution =
`Density* Volume=1/mL* 100.0mL=100.0g`

`\Delta T` = change in temperature =
`T_2-T_1=(26.3-20.2)^oC=6.1^oC`

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

`q=[(12.1J/^oC* 6.1^oC)+(100.0g* 4.18J/g^oC* 6.1^oC)]`

`q=2623.61J`

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

`\Delta H=(q)/(n)`

where,

`\Delta H` = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used =
`Concentration* Volume=1M* 0.050L=0.050mole`

`\Delta H=(2623.61J)/(0.050mole)=52472.2J/mole=52.5kJ/mole`

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