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Enthalpy of Phase Transitions
When a substance changes from solid to liquid, liquid to gas or solid to gas, there are specific enthalpies involved in these changes.
The latent heat of melting describes the transition from solid to liquid (the reverse is minus this value and called the enthalpy of fusion), the enthalpy of vaporization describes the transition from liquid to gas (and the opposite is condensation) and the enthalpy of sublimation describes the transition from solid to gas (the reverse is again called the enthalpy of condensation).
For water, the enthalpy of melting is ∆Hmelting = 6.007 kJ/mol. Imagine that you heat ice from 250 Kelvin until it melts, and then heat the water to 300 K. The enthalpy change for the heating parts is just the heat required, so you can find it using:
∆H = nC∆T
Where (n) is the number of moles, (∆T) is the change in temperatue and (C) is the specific heat. The specific heat of ice is 38.1 J/K mol and the specific heat of water is 75.4 J/K mol. So the calculation takes place in a few parts. First, the ice has to be heated from 250 K to 273 K (i.e., −23 °C to 0°C). For 5 moles of ice, this is:
∆H = nC∆T
= 5 mol × 38.1 J/K mol × 23 K
= 4.382 kJ
Now multiply the enthalpy of melting by the number of moles:
∆H = n ∆Hmelting
= 5 mol × 6.007 kJ/mol
= 30.035 kJ
Calculations for vaporization are the same, except with the vaporization enthalpy in place of the melting one. Finally, calculate the final heating phase (from 273 to 300 K) in the same way as the first:
∆H = nC∆T
= 5 mol × 75.4 J/K mol × 27 K
= 10.179 kJ
Sum these parts to find the total change in enthalpy for the reaction:
∆Htotal = 10.179 kJ + 30.035 kJ + 4.382 kJ
= 44.596 kJ
Hess’s Law
Hess’s law is useful for when the reaction you’re considering has two or more parts and you want to find the overall change in enthalpy. It states that the enthalpy change for a reaction or process is independent of the route through which it occurs. This means that if reaction transforms on substance into another, it doesn’t matter if the reaction occurs in one step (reactants become products immediately) or whether it goes through many steps (reactants become intermediaries and then become products), the resulting enthalpy change is the same in both cases.
It usually helps to draw a diagram (see Resources) to help you use this law. One example is if you start with six moles of carbon combined with three of hydrogen, they combust to combine with oxygen as an intermediary step and then form benzene as an end-product.
Hess’ law states that the change in enthalpy of the reaction is the sum of the changes in enthalpy of both parts. In this case, the combustion of one mole of carbon has ∆H = −394 kJ/mol (this happens six times in the reaction), the change in enthalpy for the combustion of one mole of hydrogen gas is ∆H = −286 kJ/mol (this happens three times) and the carbon dioxide and water intermediaries become benzene with an enthalpy change of ∆H = +3,267 kJ/mol.
Take the sum of these changes to find the total enthalpy change, remembering to multiply each by the number of moles needed in the first stage of the reaction:
∆Htotal = 6×(−394) + 3×(−286) +3,267
= 3,267 − 2,364 – 858
= 45 kJ/mol
The enthalpy change is negative because the reaction releases heat to the surroundings, resulting in an increase in temperature of the water.
Sample Problem:
Calculate the value of heat if the temperature of 200g of water was increased by 10.0oc when 0.46g of ethanol was burned in the air.
The specific heat capacity of water is 4.18J.K-1.g-1
Solution
So we can simply plug the given values in the equation
Q= C.m. = 4.18 X 200 X 8 = 6688J
In such problem, you might also be asked to provide the value of calculated heat in kJ/mol. This would be the enthalpy change of a reaction. In such case, you will simply convert the mass of ethanol to moles. To do so, divide the mass of ethanol by the molar mass of ethanol.
For this sample problem, we would get the following:
n (ethanol) = mM= 0.46g46g/mol = 0.01mol
Now, we convert 6688J to kJ:
1kJ = 1000J
6688J = 6.688kJ
And finally, we find the value of heat in kJ/mol by dividing 6.688kJ by 0.01mol of ethanol and we get the following:
668.8kJ/mol
Since the reaction is exothermic, we should have minus in front of 668.8kJ/mol. So the final answer would be the following:
∆H = -668.8kJ/mol
NOTE:
Since the problem gives the temperature change, there is no need to convert the value to Kelvin. The reason is that if we are given 2 temperature values (initial and final) in Celsius and we convert both values to Kelvin, the difference between the two values will be the same since.
In case if we are not looking for the temperature change, the temperature value provided in Celsius must be converted into Kelvin to get the proper result.
Sample Problem:
Calculate the enthalpy change of the exothermic reaction if 0.036mol of anhydrous copper (II) sulphate was added to 30cm of DI water and the temperature was increased by 16OC.
Specific heat capacity of water is 4.18J.K-1.g-1
Question
A solution was made by dissolving a spatula of potassium nitrate into 50 cm3 of water. The temperature changed from 20.4˚C to 18.7˚C. Calculate the enthalpy change for this reaction.
To calculate the enthalpy change (ΔH) we must know the values for c, m and ΔT. The specific heat capacity (c) is a constant, with a value of 4.18. Since 50 cm3 of water have been used, the mass of water (m) is 0.05 kg.
From the question we can see that the temperature has decreased by 1.7 ˚C. This means that the reaction is endothermic (so ΔH will be positive).
ΔH=cmΔT
= 4.18 x 0.05 x 1.7
= 0.3553 kJ
Heat Transfer between Substances at Different Temperatures
A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron, and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).
Solution
The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or:
qrebar=−qwater
Since we know how heat is related to other measurable quantities, we have:
(c×m×ΔT)rebar=−(c×m×ΔT)water
Letting f = final and i = initial, in expanded form, this becomes:
Crebar×mrebar×(Tf,rebar−Ti,rebar)=−cwater×mwater×(Tf,water−Ti, water)
The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields:
(0.449J/g∘C)(360g)(42.7∘C−Ti,rebar)=(4.184J/g∘C)(425g)(42.7∘C−24.0∘C)
Ti, rebar= (4.184J/g∘C)(425g)(42.7∘C−24.0∘C)=42.7∘C (0.449J/g∘C)(360g)
Solving this gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C.
Since the reaction is exothermic, we must put the negative sign next to the calculated enthalpy change value. If you forget to do so, you will lose the credit for the problem since the negative and positive sign make a huge difference. If there is an endothermic reaction, you just put a positive sign to the enthalpy change value.
NB heat is measured in Joules while the enthalpy change is measured in kJ/mol. When you calculate the value for the heat, you must convert the value in J to kJ to get the proper result for enthalpy change.