Calculate the enthalpy change (in kJ) associated with the conversion of 25.0 grams of ice at -4.00 °C to water vapor at 109.0 °C.
The specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively.
For H2O, ΔHfus = 6.01 kJ/mol and ΔHvap = 40.67 kJ/mol.
Well, let's break it down step by step, shall we?
First, we need to calculate the heat required to raise the temperature of the ice from -4.00 °C to 0 °C. The specific heat of ice is 2.09 J/g-K, and we have 25.0 grams of ice. So the heat required for this step is:
Q1 = (2.09 J/g-K) * (25.0 g) * (0 °C - (-4.00 °C))
= (2.09 J/g-K) * (25.0 g) * (4.00 °C)
= 209.00 J
Next, we need to calculate the heat required to melt the ice at 0 °C and turn it into water at 0 °C. The enthalpy of fusion for water is 6.01 kJ/mol. To convert from grams to moles, we can use the molar mass of water (18.02 g/mol):
moles of H2O = (25.0 g) / (18.02 g/mol)
= 1.386 mol
So the heat required for this step is:
Q2 = (6.01 kJ/mol) * (1.386 mol)
= 8.323 kJ
After that, we need to calculate the heat required to raise the temperature of the water from 0 °C to 109.0 °C. The specific heat of water is 4.18 J/g-K, and we have 25.0 grams of water. So the heat required for this step is:
Q3 = (4.18 J/g-K) * (25.0 g) * (109.0 °C - 0 °C)
= (4.18 J/g-K) * (25.0 g) * (109.0 °C)
= 11,076.50 J
Finally, we need to calculate the heat required to convert the water vapor at 109.0 °C to steam at 109.0 °C. The enthalpy of vaporization for water is 40.67 kJ/mol. Using the same mole calculation as before, we get:
moles of H2O = (25.0 g) / (18.02 g/mol)
= 1.386 mol
So the heat required for this step is:
Q4 = (40.67 kJ/mol) * (1.386 mol)
= 56.40 kJ
Now, we just need to add up all the heat values:
ΔH = Q1 + Q2 + Q3 + Q4
= 209.00 J + 8.323 kJ + 11,076.50 J + 56.40 kJ
And don't forget to convert all the units to kJ:
ΔH = (209.00 J / 1000) + 8.323 kJ + (11,076.50 J / 1000) + 56.40 kJ
= 0.209 kJ + 8.323 kJ + 11.077 kJ + 56.40 kJ
= 76.009 kJ
So, the enthalpy change associated with the conversion of 25.0 grams of ice at -4.00 °C to water vapor at 109.0 °C is approximately 76.009 kJ. Hope this answer didn't leave you cold!
To calculate the enthalpy change associated with the conversion of ice to water vapor, we need to consider the following steps:
Step 1: Heating the ice from -4.00 °C to 0 °C
Step 2: Melting the ice at 0 °C
Step 3: Heating the water from 0 °C to 100 °C
Step 4: Boiling the water at 100 °C
Step 5: Heating the steam from 100 °C to 109 °C
Let's calculate the enthalpy change for each step:
Step 1: Heating the ice from -4.00 °C to 0 °C
To heat the ice, we need to use the specific heat of ice:
q1 = m * c * ΔT
where,
m = mass of ice = 25.0 g
c = specific heat of ice = 2.09 J/g-K
ΔT = change in temperature = (0 °C - (-4.00 °C)) = 4.00 °C
q1 = (25.0 g) * (2.09 J/g-K) * (4.00 °C)
q1 = 209 J
Step 2: Melting the ice at 0 °C
To melt the ice, we need to use the enthalpy of fusion:
q2 = m * ΔHfus
where,
ΔHfus = enthalpy of fusion of water = 6.01 kJ/mol (given)
To convert grams to moles, we need the molar mass of water:
MM(H2O) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
moles of ice = (25.0 g) / (18.02 g/mol)
moles of ice ≈ 1.3868 mol
q2 = (1.3868 mol) * (6.01 kJ/mol) * (1000 J/kJ)
q2 = 83.43 kJ
Step 3: Heating the water from 0 °C to 100 °C
To heat the water, we need to use the specific heat of water:
q3 = m * c * ΔT
where,
m = mass of water = 25.0 g
c = specific heat of water = 4.18 J/g-K
ΔT = change in temperature = (100 °C - 0 °C) = 100 °C
q3 = (25.0 g) * (4.18 J/g-K) * (100 °C)
q3 = 10450 J
Step 4: Boiling the water at 100 °C
To boil the water, we need to use the enthalpy of vaporization:
q4 = m * ΔHvap
where,
ΔHvap = enthalpy of vaporization of water = 40.67 kJ/mol (given)
q4 = (1.3868 mol) * (40.67 kJ/mol) * (1000 J/kJ)
q4 = 56508 J
Step 5: Heating the steam from 100 °C to 109 °C
To heat the steam, we need to use the specific heat of steam:
q5 = m * c * ΔT
where,
m = mass of steam = 25.0 g
c = specific heat of steam = 1.84 J/g-K
ΔT = change in temperature = (109 °C - 100 °C) = 9 °C
q5 = (25.0 g) * (1.84 J/g-K) * (9 °C)
q5 = 414 J
Finally, to find the total enthalpy change, we add up all the individual enthalpy changes:
ΔH = q1 + q2 + q3 + q4 + q5
ΔH = 209 J + 83.43 kJ + 10450 J + 56508 J + 414 J
ΔH ≈ 65.59 kJ
Therefore, the enthalpy change associated with the conversion of 25.0 grams of ice at -4.00 °C to water vapor at 109.0 °C is approximately 65.59 kJ.
To calculate the enthalpy change associated with the conversion of ice at -4.00 °C to water vapor at 109.0 °C, we need to consider the different steps involved in the process.
Step 1: Heating the ice from -4.00 °C to 0.00 °C
To find the heat required to raise the temperature of the ice, we use the equation:
q = m * c * ΔT
where q is the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we are calculating the heat necessary to raise the temperature of ice (m = 25.0 g) from -4.00 °C to 0.00 °C. The specific heat capacity of ice is 2.09 J/g-K. The change in temperature is ΔT = 0.00 °C - (-4.00 °C) = 4.00 °C.
So, for Step 1, the heat absorbed is given by:
q1 = (25.0 g) * (2.09 J/g-K) * (4.00 °C)
q1 = 209 J
Step 2: Melting the ice at 0.00 °C into water at 0.00 °C
To find the heat required for the phase change from ice to water (melting), we use the equation:
q = m * ΔHfus
where ΔHfus is the heat of fusion or enthalpy change of fusion. In this case, ΔHfus is given as 6.01 kJ/mol.
The mass of ice is 25.0 g, so the number of moles of ice can be calculated using the molar mass of water (18.02 g/mol):
moles = mass / molar mass = (25.0 g) / (18.02 g/mol) = 1.387 mol
The heat absorbed for Step 2 is then:
q2 = (1.387 mol) * (6.01 kJ/mol)
q2 = 8.328 kJ
Step 3: Heating the water from 0.00 °C to 100.0 °C
To find the heat required to raise the temperature of the water, we use the equation:
q = m * c * ΔT
In this case, we are calculating the heat necessary to raise the temperature of water (m = 25.0 g) from 0.00 °C to 100.0 °C. The specific heat capacity of water is 4.18 J/g-K. The change in temperature is ΔT = 100.0 °C - 0.00 °C = 100.0 °C.
So, for Step 3, the heat absorbed is given by:
q3 = (25.0 g) * (4.18 J/g-K) * (100.0 °C)
q3 = 10450 J
Step 4: Vaporizing the water at 100.0 °C into steam at 100.0 °C
To find the heat required for the phase change from water to steam (vaporization), we use the equation:
q = m * ΔHvap
where ΔHvap is the heat of vaporization or enthalpy change of vaporization. In this case, ΔHvap is given as 40.67 kJ/mol.
Using the number of moles calculated earlier:
q4 = (1.387 mol) * (40.67 kJ/mol)
q4 = 56.45 kJ
Step 5: Heating the steam from 100.0 °C to 109.0 °C
To find the heat required to raise the temperature of the steam, we use the equation:
q = m * c * ΔT
In this case, we are calculating the heat necessary to raise the temperature of steam (m = 25.0 g) from 100.0 °C to 109.0 °C. The specific heat capacity of steam is 1.84 J/g-K. The change in temperature is ΔT = 109.0 °C - 100.0 °C = 9.0 °C.
So, for Step 5, the heat absorbed is given by:
q5 = (25.0 g) * (1.84 J/g-K) * (9.0 °C)
q5 = 414 J
Now, add up the heat absorbed in each step to get the total enthalpy change:
Total enthalpy change = q1 + q2 + q3 + q4 + q5
Total enthalpy change = 209 J + 8.328 kJ + 10450 J + 56.45 kJ + 414 J
Converting everything to kJ:
Total enthalpy change = (209 J + 8.328 kJ + 10450 J + 56.45 kJ + 414 J) / 1000
Total enthalpy change = 66.987 kJ
A. heat to move ice from -4.00 to zero C.
q = mass ice x specific heat ice x (Tfinal-Tinitial)
B. heat to melt ice at zero C to liquid water at zero C.
q = mass ice x heat fusion.
C. heat to move liquid water from zero C to 100 C.
q = mass water x specific heat H2O x (Tfinal-Tinitial)
D. heat to convert H2O @ 100 C to steam @ 100 C.
q = mass water x heat vaporization
E. heat to move steam @ 100 C to 109 C.
q = mass steam x specific heat steam x (Tfinal-Tinitial)
Total q = delta H = q1 + q2 + q3 + q4 + q5.
Note that specific heats are given in J/g*K but heat fusion and heat vaporization are given in kJ/mol. I would suggest you change kJ/mol to J/g so as to keep all of the units the same.
Post your work if you get stuck. I'll check your answer and your work but only if you show your work.