In the previous section, we saw extensive and intensive properties. We also saw the derivation of Mayer's relation. In this section, we will see calorimetry.
• When a substance burns, heat energy is released.
• Some substances release greater heat energy than others.
♦ For example, one gram of hydrogen releases greater heat energy than one gram of kerosene.
• So we will want to know the exact amount of heat that can be obtained from various substances.
• Let us see how such an information can be obtained. It can be written in 12 steps:
1. In fig.6.8 below, the sample is shown in magenta color.
• It is taken in a crucible. The crucible is shown in pink color.
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| Fig.6.8 |
• A crucible is a ceramic or metal container in which substances may be melted or subjected to very high temperatures. Some images can be seen here
2. The crucible with the sample is placed inside a stainless steel vessel called the bomb.
• The bomb is sealed at the top.
• In the top lid, inlet is provided to supply oxygen into the bomb. This oxygen is necessary for the combustion of the sample.
3. The sample can be ignited by using an electric circuit.
• The leads of the circuit are taken out through the lid of the bomb.
4. The bomb is immersed in water taken in an outer vessel.
• In fig.6.8, this vessel is shown in yellow color.
5. A stirrer is provided to ensure that any ‘temperature change occurring’, is distributed evenly through out the water.
6. The temperatures are measured using a thermometer.
7. The outer vessel is placed inside an insulating jacket, covered at the top.
• In the fig.6.8, the insulating jacket is shown in green color.
8. The sample is ignited using the electric circuit. A combustion reaction takes place.
• During the combustion, the sample reacts with oxygen and heat is released.
• This heat is transmitted through the walls of the bomb and reaches the water.
• From the water, the heat reaches the outer yellow vessel also.
• Thus the heat is absorbed by three items:
♦ The bomb, water and the outer vessel.
9. The heat will not flow beyond the outer vessel because, there is only air space between the outer vessel and the insulating jacket.
10. These three items are in contact with each other. So their initial and final temperatures will be the same.
• The temperatures are measured using the thermometer.
• From the thermometer readings, we get the rise in temperature ΔT.
11. Once we know the ΔT, we can calculate the quantity of heat absorbed. It can be done in 4 steps:
(i) Heat absorbed by each of the three items:
♦ Let Cbomb be the heat capacity of the bomb.
✰ Then heat absorbed by the bomb = Cbomb ΔT
♦ Let Cwater be the heat capacity of the water
✰ Then heat absorbed by the water = Cwater ΔT
♦ Let Couter be the heat capacity of the outer vessel
✰ Then heat absorbed by the outer vessel = Couter ΔT
(ii) So total heat absorbed = (Cbomb ΔT + Cwater ΔT + Couter ΔT)
• We see that, ΔT is common. So the result becomes:
Total heat absorbed = (Cbomb + Cwater + Couter)ΔT
(iii) The three quantities inside the brackets, can be combined into a single value.
• That is:
♦ the heat capacity of the bomb, Cbomb
♦ can be expressed in terms of
♦ the heat capacity of water.
• Also:
♦ the heat capacity of the outer vessel, Couter
♦ can be expressed in terms of
♦ the heat capacity of water.
• A demonstration of this conversion can be seen in solved example 11.12 of physics notes. This technique is generally called: water equivalent
• Thus, the three quantities inside the brackets in (ii) becomes a single quantity.
• It is called the heat capacity of the calorimeter
♦ We will denote it as Cc.meter
• So we can write: (Cbomb + Cwater + Couter) = Cc.meter
(iv) Thus we can easily find the total heat absorbed by the three items:
Total heat absorbed = Cc.meter × ΔT
12. This absorbed heat is same as the heat released by the sample.
• Since the reaction takes place in a closed vessel, volume is a constant.
• So we can denote the heat released as: QV
• From this QV, we can calculate the enthalpy change of the reaction.
• The following solved example demonstrates the procedure.
Solved example 6.11
1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure according to the equation C (graphite) + O2 (g) → CO2 (g). During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?
Solution:
1. In this problem, Cc.meter is directly given. We do not have to calculate the water equivalents.
2. We have:
• Heat released in a bomb calorimeter= QV = (Cc.meter × ΔT)
= [20.7 (kJ K-1) × 1 (K)] = 20.7 kJ
3. To find the enthalpy change, let us use the basic equations.
• The initial state is the instant just before ignition. In this state, the carbon is in contact with oxygen, but the reaction has not begun.
• The final state is the instant at which all the graphite is completely burnt up.
• We have: UB = UA - QV + P (VB -VA)
[QV is given a negative sign because, heat is released from the system]
⇒ (UB + PVB) - (UA + PVA) = -Q
⇒ HB - HA = -QV
⇒ (UB - UA) + P(VB - VA) = -Q
4. (VB - VA) = (nB - nA)RT
• nB = number of gaseous moles in the final state
= number of moles of CO2 = 1
• nA = number of gaseous moles in the initial state
= number of moles of O2 = 1
5. So (VB - VA) = 0
• Thus from (3), we get: [(UB - UA) + 0] = -QV = -20.7 kJ
⇒ ΔU = -20.7 kJ
• Also from (3), we get: ΔH = (HB - HA) = -QV = -20.7 kJ
Solved example 6.12
One gram of a standard substance X is known to
release a heat of 26.38 kJ. When 0.579 grams of this substance was
ignited in a bomb calorimeter, the rise in temperature was 2.08 K. What
is the ΔH for the combustion of glucose, if a combustion of 1.732 grams
of glucose in the same calorimeter caused a rise in temperature of 3.64 K.
Solution:
1. In this problem, the Cc.meter is not given
• We have to calculate the Cc.meter using the data related to the standard substance X.
2. One gram of X gives 26.38 kJ
• So 0.579 grams will give (0.579 × 26.38) kJ
3. This much heat was absorbed by the calorimeter
• We have: Heat absorbed by any calorimeter = Cc.meter × ΔT
♦ So we can write: (0.579 × 26.38) kJ = (Cc.meter × 2.08)
♦ So Cc.meter = 7.34 kJ K-1
4. In the second part, we are given the temperature rise (ΔT) when glucose was burnt.
• We can use this ΔT to find the heat absorbed by the calorimeter.
• We have:
Heat absorbed by the calorimeter = Cc.meter × ΔT
= (7.34 × 3.64) = 26.73 kJ
5. This absorbed heat is equal to the heat released from glucose.
♦ So heat released from 1.732 grams of glucose = 26.73 J
♦ So heat released from 1 gram glucose = 26.73⁄1.732 kJ
6. Molar mass of glucose is 180 gram.
♦ So in 180 grams of glucose, there will be one mole of glucose molecules.
♦ So in 1 gram glucose, there will be 1⁄180 mol of glucose.
7. Thus we can write:
1⁄180 mol glucose will give 26.73⁄1.732 kJ heat.
• So one mol will give (26.73⁄1.732 ÷ 1⁄180) = (26.73⁄1.732 × 180) = 2777.89 kJ
8.
In the previous solved example, we have seen that, the heat released in
the bomb calorimeter is equal to the ΔH of the sample. Also, it
has to be given a negative sign.
• Thus we can write: ΔH of glucose = -2777.89 kJ mol-1
9. Note that:
♦ In the previous solved example, the result is expressed in 'per gram'.
♦ In the present solved example, the result is expressed in 'per mol'.
• We have seen that, in the bomb calorimeter, reaction takes place at constant volume.
• Next we will see another arrangement where reaction takes place at constant pressure. It can be explained in 4 steps:
1. Two styrofoam cups are nested together. Such cups have good insulating properties.
• The solution to be studied is taken in the inner cup. This is shown in fig.6.9 below:
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| Fig.6.9 |
2. The inner cup is covered at top with an insulating stopper.
• Two holes are provided in the stopper.
♦ One hole is for a stirrer and the other for a thermometer.
3. It is assumed that, the heat evolved is entirely used to increase the temperature of the solution.
• This is because, the cups absorb only negligible heat
4. Knowing the mass of the solution, heat capacity of the solution, initial and final temperatures, we can calculate ΔH.
• The following solved example demonstrates the procedure:
Solved example 6.13
When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range is 0.9969 g cm-3. What is ΔH?
Solution:
1. Volume of water = 100.0 mL = 0.1 L = 100 cm3
• Density of water = 0.9969 g cm-3
• So mass of water = Volume × density = 100 × 0.9969 = 99.69 g
2. Total mass of the sample = Mass of water + Mass of KOH = (99.69+5.03) = 104.72 g
3. From the data book, specific heat capacity of water = 4.184 J g-1K-1
• Let us assume that, specific heat capacity of the solution is same as that of water.
• So heat absorbed by the solution
= mass of the solution × specific heat capacity of water × temperature change
= 104.72 × 4.184 × (34.7 - 23.0) = 5126.3 J = 5.13 kJ
4. To find the enthalpy change, let us use the basic equations.
• The initial state A is the instant when KOH is added to water.
• The final state B is the instant when KOH is completely dissolved in water
• We have: UB = UA - QP + P (VB -VA)
[QP is given a negative sign because, heat is released from the system]
⇒ (UB + PVB) - (UA + PVA) = -Q
⇒ HB - HA = -QP
5. So the enthalpy change = -QP = -5.13 kJ
• We can write:
The ΔH for 5.03 g of KOH = -5.13 kJ
6. Molar mass of KOH = (38+16+2) = 56 grams
⇒ Number of mol of KOH in 56 grams of KOH = 1
⇒ Number of mol of KOH in 1 gram of KOH = 1⁄56
⇒ Number of mol of KOH in 5.03 g KOH = (5.03 × 1⁄56) = 5.03⁄56
7. We can write:
ΔH for 5.03⁄56 mol KOH = -5.13
• So ΔH for 1 mol KOH = (-5.13 ÷ 5.03⁄56) = (-5.13 × 56⁄5.03) = -57.1 kJ mol-1
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