Given the following reaction and enthalpy change, calculate the amount of heat for converting 100.4g of liquid water (18.02g/mole) to a vapor?
delta H= +44kJ
q = mass H2O x deltaHvap. You list delta Hvap for H2O as 44 kJ but "per WHAT"
If that is per grams use grams for mass H2O. If that is per mole, use moles for mass H2O.
To calculate the amount of heat required to convert a given amount of a substance from a liquid state to a vapor state, you can use the equation:
q = m * delta H
where:
q is the heat (in joules or calories),
m is the mass of the substance (in grams), and
delta H is the enthalpy change (in joules or calories).
First, convert the mass of water from grams to moles:
moles of water = mass / molar mass
= 100.4 g / 18.02 g/mol
= 5.57 mol
Now, you can calculate the amount of heat using the equation:
q = m * delta H
= 5.57 mol * 44 kJ/mol
= 244.28 kJ
Therefore, the amount of heat required to convert 100.4g of liquid water to vapor is 244.28 kJ.
To calculate the amount of heat required to convert a certain amount of liquid water to vapor, we can use the equation:
q = m * ΔH
Where:
q = amount of heat (in Joules)
m = mass of the substance (in grams)
ΔH = enthalpy change (in Joules/gram)
First, we need to convert the given mass of water from grams to moles using the molar mass of water, which is 18.02 g/mol.
moles = mass / molar mass
moles = 100.4 g / 18.02 g/mol
moles ≈ 5.569 mol (rounded to three decimal places)
Now we can calculate the amount of heat required:
q = m * ΔH
q = 5.569 mol * (+44 kJ/mol) (convert ΔH from kJ to J by multiplying by 1000)
q = 5.569 mol * (+44,000 J/mol)
q ≈ 244,076 J (rounded to the nearest whole number)
Therefore, the amount of heat needed to convert 100.4 g of liquid water to vapor is approximately 244,076 Joules.