Portage Learning Chemistry CHEM 103 Unit 3
Study online at https://quizlet.com/_d7x2jb
1. Heat temp change = qtemp change = m x c x t
-(40.5 x 4.184 x (t - 85.7)) = 36.8 x 4.184 x (t - 26.3)
-(169.452 x (t - 85.7)) = 153.9712 x (t - 26.3)
-(169.452t - 14522.0364) = 153.9712t - 4049.44256
-169.452t + 14522.0364 = 153.9712t - 4049.44256
18571.479 = 323.423t
57.4 C = t: 1. Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water at
85.7C is added to a 36.8 gram sample of water at 26.3C in a coffee cup calorimeter.
c (water) = 4.184 J/g C
2. qs”l = mass x Heat of Fusion = m x Hfusion
120 x 0.334 = 40.08 kJ: 2. Show the calculation of the energy involved in melting 120 grams of ice at 0oC if
the Heat of Fusion for water is 0.334 kJ/g.
3. moles = grams/molecular weight
moles (S) = 42.8/32.07 = 1.335 mols S
= ΔHrx x new moles / original moles
q = -792 x ( 1.335/2 )= -526.7 kJ: 3. Sulfur undergoes combustion to yield sulfur trioxide by the following
reaction equation:
2 S + 3 O2 ’2 SO3 ΔH = - 792 kJ
If 42.8 g of S is reacted with excess O2, what will be the amount of heat given off?
4. moles = grams/molecular weight
moles (H2S ) = 26.2/34.086 = 0.7686 mols H2S
ΔHrx = q / (new moles / original moles)
-431.8 / ( 0.7686/2) = -1123.6 kJ: 4. Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur
dioxide and water by the following reaction equation:
2 H2S + 3 O2 ’2 SO2 + 2 H2O
What is the ΔH of the reaction if 26.2 g of H2S reacts with excess O2 to yield 431.8 kJ?
5. 1)
q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 500 g x (53.13 C
- 25.00 C) = - 58,791 J
q calorimeter = heat capacity x Δt = (10.5 kJ/C) x (53.13 C - 25.00 C) = - 295.365
kJ x 1000 J/1 kJ= - 295,365 J
q reaction = - 58,791 J + (-295,365 J) = - 354,156 J = - 354,156 J x 1 kJ / 1000 J = -
1/9
, Portage Learning Chemistry CHEM 103 Unit 3
Study online at https://quizlet.com/_d7x2jb
354.16 kJ
(new) moles C6H6 = 7.05 g / 78 = 0.0904 mole C6H6
ΔH = q / (new moles / original moles)
ΔH = - 354.16 / (0.) = - 7828 kJ / mole
(2) Isolated system (bomb calorimeter)
(3) Exothermic (temperature of water rises due to heat given off by combustion
reaction): 5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the
following reaction:
2 C6H6 (l) + 15 O2 (g) ’12 CO2 (g) + 6 H2O (l)
The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to
rise from 25.00 to 53.13 C. The heat capacity of water = 4.18 J/g/C and the heat capacity of the calorimeter = 10.5 kJ/C.
(1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter +
contents, (3) the type of process.
6. q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 100 g x 5.14 K
= 2148.52 J
q calorimeter = heat capacity x Δt = 150 x 5.14 K = 771 J
q reaction = 771 J + 2148.52 J = + 2919.52 J = + 2919.52 J x 1 kJ / 1000 J = 2.92 kJ
(1) ΔH reaction = 2.92 kJ / 14.5 g NaHCO g NaHCO3 / mol NaHCO3 = + 16.9
kJ / mol
(2) The calorimeter + contents (no lid) = open system, (3) Endothermic process-
: 6. A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter
with no lid by the following reaction
NaHCO3 (s) ’Na+ (aq) + HCO3- (aq)
If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150 J/C) decreases from 25.00oC
2/9
Study online at https://quizlet.com/_d7x2jb
1. Heat temp change = qtemp change = m x c x t
-(40.5 x 4.184 x (t - 85.7)) = 36.8 x 4.184 x (t - 26.3)
-(169.452 x (t - 85.7)) = 153.9712 x (t - 26.3)
-(169.452t - 14522.0364) = 153.9712t - 4049.44256
-169.452t + 14522.0364 = 153.9712t - 4049.44256
18571.479 = 323.423t
57.4 C = t: 1. Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water at
85.7C is added to a 36.8 gram sample of water at 26.3C in a coffee cup calorimeter.
c (water) = 4.184 J/g C
2. qs”l = mass x Heat of Fusion = m x Hfusion
120 x 0.334 = 40.08 kJ: 2. Show the calculation of the energy involved in melting 120 grams of ice at 0oC if
the Heat of Fusion for water is 0.334 kJ/g.
3. moles = grams/molecular weight
moles (S) = 42.8/32.07 = 1.335 mols S
= ΔHrx x new moles / original moles
q = -792 x ( 1.335/2 )= -526.7 kJ: 3. Sulfur undergoes combustion to yield sulfur trioxide by the following
reaction equation:
2 S + 3 O2 ’2 SO3 ΔH = - 792 kJ
If 42.8 g of S is reacted with excess O2, what will be the amount of heat given off?
4. moles = grams/molecular weight
moles (H2S ) = 26.2/34.086 = 0.7686 mols H2S
ΔHrx = q / (new moles / original moles)
-431.8 / ( 0.7686/2) = -1123.6 kJ: 4. Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur
dioxide and water by the following reaction equation:
2 H2S + 3 O2 ’2 SO2 + 2 H2O
What is the ΔH of the reaction if 26.2 g of H2S reacts with excess O2 to yield 431.8 kJ?
5. 1)
q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 500 g x (53.13 C
- 25.00 C) = - 58,791 J
q calorimeter = heat capacity x Δt = (10.5 kJ/C) x (53.13 C - 25.00 C) = - 295.365
kJ x 1000 J/1 kJ= - 295,365 J
q reaction = - 58,791 J + (-295,365 J) = - 354,156 J = - 354,156 J x 1 kJ / 1000 J = -
1/9
, Portage Learning Chemistry CHEM 103 Unit 3
Study online at https://quizlet.com/_d7x2jb
354.16 kJ
(new) moles C6H6 = 7.05 g / 78 = 0.0904 mole C6H6
ΔH = q / (new moles / original moles)
ΔH = - 354.16 / (0.) = - 7828 kJ / mole
(2) Isolated system (bomb calorimeter)
(3) Exothermic (temperature of water rises due to heat given off by combustion
reaction): 5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the
following reaction:
2 C6H6 (l) + 15 O2 (g) ’12 CO2 (g) + 6 H2O (l)
The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to
rise from 25.00 to 53.13 C. The heat capacity of water = 4.18 J/g/C and the heat capacity of the calorimeter = 10.5 kJ/C.
(1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter +
contents, (3) the type of process.
6. q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 100 g x 5.14 K
= 2148.52 J
q calorimeter = heat capacity x Δt = 150 x 5.14 K = 771 J
q reaction = 771 J + 2148.52 J = + 2919.52 J = + 2919.52 J x 1 kJ / 1000 J = 2.92 kJ
(1) ΔH reaction = 2.92 kJ / 14.5 g NaHCO g NaHCO3 / mol NaHCO3 = + 16.9
kJ / mol
(2) The calorimeter + contents (no lid) = open system, (3) Endothermic process-
: 6. A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter
with no lid by the following reaction
NaHCO3 (s) ’Na+ (aq) + HCO3- (aq)
If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150 J/C) decreases from 25.00oC
2/9
