Back again with another problem :)
How do you calculate the enthalpy change, in kJ/mol for the reaction of copper(II) sulfate?
0.0210 mol of copper sulfate was added to 25 cm3 of water. The temperature increased by 14°C. The specific heat capacity of water is 4.18 J k-1 g-1
What I did was get heat released with mcΔT, which was 25 x 4.18 x 14 = 1463
I then divided this by 0.0210 to get 69.7 kJ/mol.
Is this correct? Thanks for checking my answer!
Also, why can't we measure the value of ΔH for the reaction of turning anhydrous copper sulfate to hydrated copper sulfate directly? Why are we forced to make it into copper sulfate solution instead?
you had 25 grams of water, not 25kg. Rework your mc*deltaT with units.
Then divide by .0210
on the second reaction, how would you measure the heat given off? In a solution of water, it is easy, you use a balance and a thermometer.
To calculate the enthalpy change for the reaction of copper(II) sulfate, you need to use the equation:
ΔH = q / n
Where:
- ΔH is the enthalpy change in kJ/mol;
- q is the heat released or absorbed in J;
- n is the number of moles of the substance involved in the reaction.
For your calculation:
The heat released, q, can be found using the equation:
q = m × c × ΔT
Where:
- m is the mass of the water in grams;
- c is the specific heat capacity of water;
- ΔT is the change in temperature.
To convert the volume of water to mass, you can use the density of water, which is 1 g/cm^3.
Given that 25 cm^3 of water were used and using the density of water, the mass of water, m, is equal to 25 g.
The change in temperature, ΔT, is 14°C.
The specific heat capacity of water, c, is 4.18 J k^-1 g^-1.
So, calculating the heat released, q:
q = 25 g × 4.18 J k^-1 g^-1 × 14°C
q = 1465 J
Now, to calculate the enthalpy change, ΔH:
ΔH = q / n
ΔH = 1465 J / 0.0210 mol
ΔH ≈ 69.8 kJ/mol (rounded to 3 significant figures)
Your calculation is correct, the enthalpy change for the reaction of copper(II) sulfate is approximately 69.8 kJ/mol.
Regarding your second question, the reason why we use a copper sulfate solution instead of directly measuring the enthalpy change for turning anhydrous copper sulfate to hydrated copper sulfate is that the reaction between anhydrous copper sulfate and water (hydration) is highly exothermic and rapid. It releases a significant amount of heat and can be difficult to control and measure accurately. By using a solution, we can more easily control the reaction and measure the enthalpy change.
To calculate the enthalpy change, ΔH, for the reaction of copper(II) sulfate, you need to consider the heat absorbed by the water and the molar quantity of the reactant.
First, let's calculate the heat absorbed by the water using the formula: q = mcΔT, where q is the heat absorbed, m is the mass of the water in grams, c is the specific heat capacity of water, and ΔT is the change in temperature.
In this case, you have 25 cm3 of water, which is equivalent to 25 grams since the density of water is approximately 1 g/cm3.
q = (25 g) x (4.18 J g-1 °C-1) x (14 °C) = 1465 J
Now, to convert the heat absorbed to kJ/mol, you divide the heat absorbed by the number of moles of copper sulfate:
Molar mass of copper sulfate (CuSO4) = atomic mass of Cu + 4 x (atomic mass of S) + 6 x (atomic mass of O)
= (63.55 g/mol) + 4 x (32.06 g/mol) + 6 x (16.00 g/mol)
= 159.61 g/mol
Number of moles of copper sulfate = mass / molar mass
= 0.0210 mol
So, ΔH = q / n = (1465 J) / (0.0210 mol) = 69.8 kJ/mol
Your calculation is correct!
Now, moving on to your second question. The reason we cannot measure the value of ΔH directly for the reaction of turning anhydrous copper sulfate to hydrated copper sulfate is that this reaction involves the absorption or release of water molecules. The amount of water that is absorbed or released depends on the reaction conditions, such as temperature, pressure, and concentration.
To measure the enthalpy change accurately, we can conduct the reaction in a closed system where the water vapor can be condensed and collected. This allows us to measure the actual amount of water absorbed or released during the reaction.
By making the anhydrous copper sulfate into a solution, we ensure that the reaction takes place in the presence of excess water. In this case, all the water required for the reaction is available, and the enthalpy change can be measured accurately.