A sample of liquid acetone weighing 0.700 g was burned in a bomb calorimeter for which the heat capacity (including the sample) is 6937 J KŠ. The observed temperature rise was from 25.00 °C to 26.69°C. Calculate AU for the combustion of 1 mole of acetone. Given Macetone = 0.700 g Cyim = 6937! K T; = 25°C = 298.15 K Tf = 26.69°C = 299.84 K Find : change in internal energy (AU)and enthalpy (AH) %3D SOLUTION: AU = q + W CuimdT = Cyim(T; – T;) = (6937)(2s (299.84 K – 298.15 K) qvim (6937 |(1.69 K) = 11723.53 J = 11.7235 kJ K. qvim w = 0 (at constant volume) Molar mass '58.08 g (a.)AU = qv1m Δυ qvim X 11.7235kJ m 0.700 k] AU = 972.7155- mol
A sample of liquid acetone weighing 0.700 g was burned in a bomb calorimeter for which the heat capacity (including the sample) is 6937 J KŠ. The observed temperature rise was from 25.00 °C to 26.69°C. Calculate AU for the combustion of 1 mole of acetone. Given Macetone = 0.700 g Cyim = 6937! K T; = 25°C = 298.15 K Tf = 26.69°C = 299.84 K Find : change in internal energy (AU)and enthalpy (AH) %3D SOLUTION: AU = q + W CuimdT = Cyim(T; – T;) = (6937)(2s (299.84 K – 298.15 K) qvim (6937 |(1.69 K) = 11723.53 J = 11.7235 kJ K. qvim w = 0 (at constant volume) Molar mass '58.08 g (a.)AU = qv1m Δυ qvim X 11.7235kJ m 0.700 k] AU = 972.7155- mol
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![A sample of liquid acetone weighing 0.700 g was burned in a bomb calorimeter for which the heat
capacity (including the sample) is 6937 J KŠ. The observed temperature rise was from 25.00 °C to
26.69°C.
Calculate AU for the combustion of 1 mole of acetone.
Given
Macetone = 0.700 g
Cyim = 6937!
K
T; = 25°C = 298.15 K
Tf = 26.69°C = 299.84 K
Find : change in internal energy (AU)and enthalpy (AH)
%3D
SOLUTION:
AU = q + W
CuimdT = Cyim(T; – T;) = (6937)(2s
(299.84 K – 298.15 K)
qvim
(6937
|(1.69 K) = 11723.53 J = 11.7235 kJ
K.
qvim
w = 0 (at constant volume)
Molar mass
'58.08 g
(a.)AU = qv1m
Δυ
qvim X
11.7235kJ
m
0.700
k]
AU = 972.7155-
mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbfc2c9d1-a588-4eff-949b-be2e7d08da98%2F5a3850da-e0bf-474e-9a9f-dd6e0dabb23a%2Fjy6ewts_processed.png&w=3840&q=75)
Transcribed Image Text:A sample of liquid acetone weighing 0.700 g was burned in a bomb calorimeter for which the heat
capacity (including the sample) is 6937 J KŠ. The observed temperature rise was from 25.00 °C to
26.69°C.
Calculate AU for the combustion of 1 mole of acetone.
Given
Macetone = 0.700 g
Cyim = 6937!
K
T; = 25°C = 298.15 K
Tf = 26.69°C = 299.84 K
Find : change in internal energy (AU)and enthalpy (AH)
%3D
SOLUTION:
AU = q + W
CuimdT = Cyim(T; – T;) = (6937)(2s
(299.84 K – 298.15 K)
qvim
(6937
|(1.69 K) = 11723.53 J = 11.7235 kJ
K.
qvim
w = 0 (at constant volume)
Molar mass
'58.08 g
(a.)AU = qv1m
Δυ
qvim X
11.7235kJ
m
0.700
k]
AU = 972.7155-
mol
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