calcualte the enthalpy change for the process in which 6.00g of steam at 100C is coverted to liquid water at a temperature of 35.0
To calculate the enthalpy change for the given process, we need to consider two steps:
Step 1: Heating the steam from 100°C to its boiling point at 100°C.
Step 2: Condensing the steam to liquid water at 100°C and then cooling it further to 35.0°C.
Step 1: Heating the steam (q1)
The equation we will use is q1 = m * C * ΔT, where:
- q1 is the heat absorbed by the steam,
- m is the mass of the steam (6.00g),
- C is the specific heat capacity of steam (2.03 J/g°C),
- ΔT is the change in temperature (100°C - 100°C = 0°C).
Since the temperature change is zero, there is no enthalpy change or heat absorbed in this step.
Step 2: Condensation and cooling (q2)
Now, we need to calculate the heat released during condensation and cooling.
The equation is q2 = -m * ΔH, where:
- q2 is the heat released by the steam,
- m is the mass of the steam (6.00g),
- ΔH is the enthalpy of vaporization/condensation of water (40.7 kJ/mol).
To find the enthalpy change, we need to convert grams of steam to moles of water.
- The molar mass of water (H₂O) is 18.015 g/mol.
- Moles of water = mass of steam / molar mass of water.
moles = 6.00g / 18.015 g/mol = 0.333 mol
Now, the heat released during condensation and cooling can be calculated:
q2 = - (0.333 mol) * (40.7 kJ/mol) = -13.5 kJ.
Since q1 = 0 J and q2 = -13.5 kJ, the total enthalpy change (ΔH) is given by:
ΔH = q1 + q2 = 0 J - 13.5 kJ = -13.5 kJ.
Therefore, the enthalpy change for the process of converting 6.00g of steam at 100°C to liquid water at 35.0°C is -13.5 kJ. The negative sign indicates that heat is released during the process.