What is the value of q when 8.21g of water vaporizes at 373 K? The enthalpy of condensation of water at 373 K is -40.7 kJ/mol.
I tried using the equation q= m x Cs x Delta T, but I never got the right answer which is 18.5 kJ apparently, so I need to find out what I did wrong.
You used the correct formula for calculating q WITHIN a phase; however, the formula to use at a phase change is
q = mass x heat vap for boiling/condensing.
q = mass x heat fusion for freezing/melting.
git gud
To find the value of q when water vaporizes at 373 K, you need to use the equation:
q = n x ΔH
Where:
q is the heat absorbed or released (in this case, it will be absorbed as water vaporizes)
n is the number of moles
ΔH is the enthalpy change (in this case, the enthalpy of condensation of water)
First, you need to calculate the number of moles of water by using the formula:
n = m / M
Where:
m is the mass of water
M is the molar mass of water
The molar mass of water is approximately 18.015 g/mol. Therefore:
m = 8.21 g
M = 18.015 g/mol
n = 8.21 g / 18.015 g/mol
n ≈ 0.455 mol (rounded to three decimal places)
Now, substitute the values of n and ΔH into the equation:
q = 0.455 mol x (-40.7 kJ/mol)
q ≈ -18.5275 kJ
Therefore, the value of q when 8.21g of water vaporizes at 373 K is approximately -18.5 kJ.
To calculate the value of q when water vaporizes, you need to consider the enthalpy of vaporization, not the enthalpy of condensation. The enthalpy of vaporization represents the energy required to convert a substance from the liquid phase to the gas phase at its boiling point.
Since the enthalpy of condensation is given (-40.7 kJ/mol), but we need the enthalpy of vaporization, we can use the fact that the reverse process has the opposite sign. Hence, the enthalpy of vaporization at 373 K is +40.7 kJ/mol.
To find out the value of q, we first need to calculate the number of moles of water that vaporizes. For that, use the molar mass of water (H2O), which is approximately 18.02 g/mol.
Number of moles (n) = Mass (m) / Molar mass (M)
= 8.21 g / 18.02 g/mol
Next, we can use the formula:
q = n x ΔHvap
where q is the heat energy transferred, n is the number of moles, and ΔHvap is the enthalpy of vaporization.
q = (8.21 g / 18.02 g/mol) x (40.7 kJ/mol)
Now, let's calculate q:
q = (0.455 mol) x (40.7 kJ/mol)
q ≈ 18.52 kJ
Therefore, the approximate value of q when 8.21 g of water vaporizes at 373 K is 18.52 kJ.