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Transcript
218
= 10.428 × 103 K
= 10428 K
As both Δ Hº and Δ Sº are negative
therefore, the reaction will be
spontaneous below 10428K.
Determine whether the following reaction
is spontaneous under standard conditions.
2H 2 O(l ) + O 2 (g) ⎯⎯
→ 2H 2 O 2 (l ) ΔHº
= +196 kJ, ΔSº = –126 JK–1
Does it have a cross-over temperature?
Given :
ΔHº = +196 kJ, ΔSº = –126o JK–1
To find :
State the reaction condition = ?
Solution :
As ΔHº is positive and ΔSº is negative
therefore, the reaction will be
nonspontaneous at all temperatures.
No, it has no cross over temperature.
=
∴
*28.
*29.
What is the value of ΔSsurr for the following
reaction at 298 K?
→ C6H12O6(s) +
6CO2(g) + 6H2O(l) ⎯⎯
6O2(g), ΔGº = 2879 kJ mol–1, ΔSº = –210
JK–1 mol–1.
Given :
ΔGº = 2879 kJmol–1,
ΔSº = –210 JK–1 mol–1
= –0.210 kJ K–1mol–1
T
= 298 K
To find :
ΔSsurr = ?
Solution :
ΔGº = ΔHº – T ⋅ ΔSº
2879 kJmol–1
= ΔHº – 298 K × (–0.210kJ K–1 mol–1)
Unique Solutions ®
ΔHº + 62.58 kJ mol–1
ΔHº = (2879 – 62.58) kJ = 2816.42 kJ.
ΔSsurr = –
ΔHº
T
ΔSsurr = –
2816.42
= –9.45 kJ K–1
298
Δ Ssurr = –9.45 kJ K –1
Calculate ΔSsurr when one mole of methanol
(CH3OH) is formed from its elements
under standard conditions if Δ f Hº
(CH3OH) = –238.9 kJmol–1
Given :
The heat evolved in the reaction is 23.89
kJ. The same quantity of heat is absorbed
by the surroundings. Hence, entropy
change for the surroundings is given by
T = 25ºC = 289K
To find :
ΔSsurr = ?
Solution :
*30.
ΔSsurr = –
–238.9 (kJ)
ΔHº
=–
298 (K)
T
= 0.8017 kJ K–1
= 801.7 JK–1
Δ Ssurr = 801.7 JK–1
*31.
GIBBS FREE ENERGY
Calculate K p for the reaction,
C 2 H 4 (g) + H 2 (g) ⎯⎯
→ C 2 H 6 (g) , Δ Gº
= –100 kJ mol–1 at 25ºC.
Given :
ΔGº = –100kJmol–1 = –100000 Jmol–1
T = 25ºC + 273K = 298K
S.Y.J.C. Science - Chemistry - Part I