Question

a 50 ml sample of a 1.00 M solution of CuSO4 is mixed with 50 ml of 2.00 M KOH in a calorimeter. The temperature of both solutions was 20.2 degrees Celsius before mixing and 26.3 degrees celsius after mixing. The heat capacity of the calorimeter is 12.1 J/K. From these data calculate delta H for the process. Assume the specific heat and density of the solution after mixing are the same as those of pure water.

Answer 1

The enthalpy change for the process can be calculated using the equation ΔH = q / n, where q is the heat absorbed or released by the system and n is the number of moles of the substance involved in the reaction. Then, we can calculate the heat absorbed by the system using the equation q = C * ΔT, where C is the heat capacity of the calorimeter and ΔT is the change in temperature. Finally, we can calculate the enthalpy change using the formula ΔH = q / (n(CuSO4) + n(KOH)).

Step-by-step explanation:

The enthalpy change, ΔH, for a process can be calculated using the equation:

ΔH = q / n

where q is the heat absorbed or released by the system and n is the number of moles of the substance involved in the reaction.

In this case, we can determine the heat absorbed by the system using the equation:

q = C * ΔT

where C is the heat capacity of the calorimeter and ΔT is the change in temperature.

Substituting the given values, we can calculate the heat absorbed by the system:

q = 12.1 J/K * (26.3°C - 20.2°C)

Next, we can calculate the number of moles of CuSO4 and KOH:

n(CuSO4) = 1.00 M * 0.050 L

n(KOH) = 2.00 M * 0.050 L

Finally, we can calculate the enthalpy change:

ΔH = q / (n(CuSO4) + n(KOH))

Answer 2

Explanation

Solution:- Total volume of the solution = 5.0 mL + 50.0 mL = 100.0 mL

Density of solution is same as of water that is 1.0 g per mL. So, the mass of solution, m =

m = 100.0 g

change in temperature, = 26.3 - 20.2 = 6.1 degree C

specific heat, s of water is 4.184 J per g per degree C.

where q is the heat energy. let's plug in the values in the equation:

q = 2552.24 J

energy of calorimeter = change in temperature*heat capacity of calorimeter

energy of calorimeter =

= 73.81 J

Total heat = 2552.24 J + 73.81 J = 2626.05 J

Let's convert it to kJ and for this we divide by 1000 since, 1000 J = 1 kJ

= 2.626 kJ

0.05 moles of copper sulfate are used so the of the reaction =

=

Since, there is an increase in temperature, the heat is released and so the sign of will be negative.

Hence, = 52.52 jk