How to calculate the quantity of heat gained or lost in the following change:

3.50 mol of water freezes at 0 degrees C?

Would it be 6.01kJ/ 3.50 mol ?

Not quite.

6.01 kJ/mol x 3.50 mol = ??
Note that your arrangement gives kJ/mol; my arrangement the mol cancel to give kJ as the answer and that is the quantity of heat.

-21.0 KJ (21.0 KJ lost)

21kj lost

Well, calculating the quantity of heat gained or lost during a phase change like freezing can be quite chilling. In this case, we're dealing with the freezing of 3.50 mol of water at 0 degrees C. To find the quantity of heat, we must use the formula Q = nΔH, where Q is the quantity of heat, n is the number of moles, and ΔH is the enthalpy change.

For water freezing at 0 degrees C, the enthalpy change (ΔH) is approximately -6.01 kJ/mol. So, to determine the total quantity of heat, we multiply the number of moles (3.50 mol) by the enthalpy change:

Q = (3.50 mol) * (-6.01 kJ/mol)

Hold your melting heart, the calculation will give you -21.035 kJ of heat lost during the freezing process. But remember, the negative sign indicates that heat is lost, not gained. Stay cool!

To calculate the quantity of heat gained or lost during a phase change, you need to use the heat of fusion (ΔHfus) or heat of solidification (ΔHsolid).

For water, the heat of fusion (ΔHfus) is 6.01 kJ/mol.

To calculate the heat gained or lost in the given change, where 3.50 mol of water freezes at 0 degrees Celsius, you need to multiply the number of moles of water by the heat of fusion.

So the calculation would be:

Heat gained or lost = number of moles × heat of fusion
= 3.50 mol × 6.01 kJ/mol

The correct answer would be:
Heat gained or lost = 21.035 kJ

Therefore, the answer is not 6.01 kJ/3.50 mol, but rather 21.035 kJ.