Dr. Pedro Julio Villegas Aguilar
[email protected]
OBJECTIVES • To become familiar with problems about thermo-chemistry. TEXTBOOK Brown, Lemay & Bursten, Chemistry: The Central Science, 10th Ed. (Chapter 5)
QUESTION 1: What are the most important thermo-chemical equations?
IMPORTANT EQUATIONS:
1st Law of Thermodynamics: ∆E= Ef – Ei
∆E= q + w
Enthalpy: ∆H = Hfinal – Hinitial Enthalpy of reaction: ∆Hrxn ∆ Hprod.–∆Hreact.
Molar heat capacity: Cp=∆H/∆T Cp = c*M Heat: q = mc∆T
Work: w = PDV = DnRT
CALORIMETRY QUESTIONS To find the heat given out (or taken in) in a reaction, use this equation. Be careful with m, which is the mass of water heated (or cooled). When finding DH (i.e. heat given out or taken in by one mole), be careful over units (check if q is in J or kJ), and remember DH is –ve if heat is given out. q = mcDT q = heat given out (or taken in) m = mass of water heated c = specific heat capacity DT = temperature change DH = q / moles
QUESTION 2: • A 2.05 g solid is heated to 74.21°C and immersed in 26.05 g H2O in a constantpressure calorimeter. The initial temperature of the water is 27.20°C and csolid = 0.519 J/g K. The changes in temperature are DTwater = 0.45 K; DTsolid = -46.56 K. a) Does this answer make sense?
QUESTION 2: A 2.05 g solid is heated to 74.21°C and immersed in 26.05 g H2O in a constant-pressure calorimeter. The initial temperature of the water is 27.20°C and csolid = 0.519 J/g K. The changes in temperature are DTwater = 0.45 K; DTsolid = -46.56 K. a) Does this answer make sense?
• Yes. We expect the water to heat up (positive DT), and the solid to cool down (negative DT). Also, the mass of the water is much greater than the mass of the solid, so the solid will have a larger magnitude of DT.
PROBLEM 1: • Calculate the amount of heat needed to increase the temperature of 250g of water from 20ºC to 56ºC.
PROBLEM 1: Calculate the amount of heat needed to increase the temperature of 250g of water from 20ºC to 56ºC.
m = 250g Tf = 56ºC
c = 4.18 J/ºC*g Ti = 20ºC q=?
q = mcDT q = 250g*4.18J/ºC*g *(56ºC - 20ºC)
q = 37 620 J = 37.6 kJ
PROBLEM 2: • Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25º to 60º.
PROBLEM 2: Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25ºC to 60ºC.
q = 204.75 J Ti = 25ºC
m = 15g Tf = 60ºC cCu = ?
q = mcDT 204.75 = 15g*cCu*(60ºC - 25ºC)
cCu = 0.39 J/ºC*g
PROBLEM 3: • 216 J of energy is required to raise the temperature of aluminum from 15ºC to 35ºC. Calculate the mass of aluminum (specific heat capacity of aluminum is 0.90 J/ºC*g).
PROBLEM 3: 216 J of energy is required to raise the temperature of aluminum from 15ºC to 35ºC. Calculate the mass of aluminum (caluminum = 0.90 J/ºC*g).
q = 216 J Ti = 15ºC
cAl = 0.90 J/ºC*g Tf = 35ºC mAl = ?
q = mcDT 216 J = mAl* 0.90 J/ºC*g * (35ºC - 15ºC)
mAl = 12g
PROBLEM 4: • The initial temperature of 150g of ethanol was 22ºC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol? (specific heat capacity of ethanol: 2.44 J/ºC*g).
PROBLEM 4: The initial temperature of 150g of ethanol was 22ºC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol? (specific heat capacity of ethanol: 2.44 J/ºC*g).
q = 3240 J m = 150g cethanol = 2.44 J/ºC*g
Ti = 22oC Tf = ?
q = mcDT 3240 J = 150g*2.44 J/ºC*g *(Tf - 22ºC) 8.85 ºC= Tf - 22ºC
Tf = 30.9ºC
PROBLEM 5: • 0.55 g of propanone was burned in a calorimeter containing 80 g of water. The temperature rose by 47.3ºC. Calculate DHc for propanone given the specific heat capacity of water is 4.18 J*mol-1*K-1.
PROBLEM 5: 0.55 g of propanone was burned in a calorimeter containing 80 g of water. The temperature rose by 47.3ºC. Calculate DHc for propanone given the specific heat capacity of water is 4.18 J*mol-1*K-1.
mpropanone = 0.55g mwater = 80 g
c(H2O) = 4.18 J/mol*K DT= 47.3ºC DHc = ?
q = mcDT = 80*4.18*47.3 = 15 820 J = 15.82 kJ
moles(CH3COCH3 ) = m q DHc =
moles
= 15.82
M
= 0.55
0.0948
58.0
= 0.0948
= 1669 kJmol
1
PROBLEM 6: • 25 cm3 of 2.0 mol*dm-3 nitric acid was reacted with 25 cm3 of 2.0 mol*dm-3 potassium hydroxide is an insulated cup. The temperature rose from 20.2ºC to 33.9ºC. Calculate DH for the reaction given the specific heat capacity of water is 4.18 J*mol-1*K-1.
PROBLEM 6: 25 cm3 of 2.0 mol*dm-3 nitric acid was reacted with 25 cm3 of 2.0 mol*dm-3 potassium hydroxide is an insulated cup. The temperature rose from 20.2ºC to 33.9ºC. Calculate DH for the reaction given the specific heat capacity of water is 4.18 J*mol-1*K-1.
concnitric acid = 2mol/dm3
concKOH = 2mol/dm3
Vnitric acid = 25 cm3 VKOH = 25 cm3 Ti= 20.2ºC
Tf= 33.9ºC
mwater = 50 g
c(H2O) = 4.18 J/mol*K
q = mcDT = 50*4.18*(33.9 – 20.2) = 2 863 J = 2.86 kJ molesKOH = conc*V = 2.0 x 25/1000 = 0.0500 mol molesnitric acid = conc*V = 2.0 x 25/1000 = 0.0500 mol
q DH =
moles
= 2.86
0.05
= 57.3 kJ / mol
PROBLEM 7: • Nitrogen dioxide gas reacts with liquid water to yield liquid nitric acid and nitrogen monoxide gas. When one mole of nitrogen dioxide reacts at constant pressure, 23.84 kJ of heat are evolved. What are the molar DHº and DHº for the reaction?
PROBLEM 7: Nitrogen dioxide gas reacts with liquid water to yield liquid nitric acid and nitrogen monoxide gas. When one mole of nitrogen dioxide reacts at STP, 23.84 kJ of heat are evolved. What are the molar DHº and DHº for the reaction?
First we need a balanced equation: NO2(g) +1/3H2O(l) → 2/3HNO3(l) + 1/3NO (g)
DHº = -23.84 kJ/mol DEº = DHº – DnRT = -23.84kJ/mol – (-2/3)*0.0008314kJ/molK*(298K) = -22.19 kJ/mol
PROBLEM 8: • When 3 500 g of the gas butane C4H10, is burned in a bomb calorimeter at 25ºC, 85.99 kJ of heat are evolved. Calculate the molar DHº and DEº for this reaction.
PROBLEM 8: When 3.5 g of the gas butane C4H10, is burned in a bomb calorimeter at 25ºC, 85.99 kJ of heat are evolved. Calculate the molar DHº and DEº for this reaction.
First we need a balanced equation: C4H10(g) +13/2O2 (g) → 4CO2(g) + 5H2O(l) Mbutane = 3500 g
T = 25ºC heatevolved = 85.99 kJ
heatevolved 85.99kJ DE = = = 1428 kJ mol 3500g mol 58 g / mol DH = DE DnRT = 1428 kJ DH = 1437 kJ
mol
4 15 * 0.008314 kJ * (298K ) mol 2 Kmol
PROBLEM 9: • The temperature of 89.1 g piece of metal rises from 22.0°C to 51.1°C when the metal absorbs 794J of energy. What is the specific heat of the metal?
PROBLEM 9: The temperature of 89.1 g piece of metal rises from 22.0°C to 51.1°C when the metal absorbs 794J of energy. What is the specific heat of the metal?
m = 89.1 g
q = 794 J
DT = Tf – Ti = 51.1-22.0 = 29.1°C
q 794J q = mcDT c = = mDT (89.1g) * (29.1C) c = 0.306 J gC
CONCLUSIONS • We practice different way to solve problems about thermo-chemical calculations associated with first Law of Thermodynamic. • Is very important be careful with the units that we use.
