hem F acts actshe heet et C hem Number 127
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The Gibb’s Free Energy Change and Spontaneity Before studying this Factsheet you need to understand: 1. Enthalpy Enthalpy changes changes of react reaction, ion, ∆H 2. Entropy Entropy (S) and entropy entropy changes of reaction reaction,, ∆S.
Under standard conditions, ∆G is related to the enthalpy change, the temperature, T and the entropy change, ∆S, by:
∆H,
∆G ê = ∆ H ê -
T ∆S A reaction is spontaneous if, and only if, ∆G < o
After studying this Factsheet you should be confident about deciding if a chemical reaction or a physical change is spontaneous or non-spontaneous.
ê
Note : ∆H , ∆S and ∆G refer to those changes for the reacting system and are sometimes written as ∆H sys, ∆S sys and ∆G sys. In particular, this aims to differentiate ∆S sys and ∆S surr where ∆S surr is the entropy change occurring in the surrounds of a reacting system and caused by the reacting system. ê
The word “spontaneous” means “energetically possible” or “feasible without the input of energy”. “Spontaneous” does not mean quick or immediate; in fact no statement can be made about the rate of the reaction since there is no relationship between activation energy, which controls rate, and spontaneity.
1
- T∆S
∆G ê = ∆H ê + (-T∆S ê)
- ve (exothermic)
+ ve
- ve
∆G
2
- ve (exothermic)
e.g. 3
Zn(s)
+
+ ve (endothermic) e.g. CO(g) CO(g)
4
- ve
→
→
+ ve
ê
∆H
and
∆S
= (-ve) + (-ve) ∆G is - ve at all temperatures. Such reactions are always spontaneous. Increasing the temperature makes ∆G even more negative because -T∆S gets more negative.
∆G
= (-ve) + (+ve) ∆G is - ve providing ∆H is more negative than -T∆S is positive. Decreasing the temperature makes -T∆S less positive and this favours ∆G being negative.
ZnO(s) + ve
→
ê
ê
+ 2H2O(g) + ve
- ve
+ 2H2O(g)
+ ve (endothermic)
e.g. H2O(l)
½O2(g)
CO(g) CO(g)
ê
We will consider four possible combinations of (excluding ∆H or ∆S being zero).
∆S ê
→
ê
ê
∆H ê
e.g. CH4(g) + 1½O2(g)
ê
ê ê ê
A spontaneous chemical reaction may be defined in terms of the Standard Gibbs free energy change, ∆G.
ê
ê
∆G = (+ve) + (+ve) ∆G is + ve at all temperatures. These reactions
can never be spontaneous. No change of T or P can cause ∆G to become negative!
CH4(g) + 1½O2(g) - ve
∆G
= (+ve) + (-ve) ∆G is - ve providing -T∆S is more negative than ∆H is positive. Increasing the temperature makes -T∆S more negative and this favours ∆G being negative.
H2O(g) at a temperature of 1000C or greater.
Note 1. The statements statements about changing changing the temperature temperature assume assume that ∆H and
∆S. remain
constant as temperature changes.
Note 2. For a reaction involving gases, increasing the pressure favours the reaction where there is a decrease in the number of moles of gas since the entropy decrease is less when the molecules are more closely packed together. Note 3. Using standard enthalpy changes and standard entropies gives a value of the standard Gibb’s free energy change. If this is negative but the reaction actually fails to be spontaneous it may be because the conditions are non-standard and the wrong values of ∆H and S have been used. Note 4. If ∆G is clearly negative but the reaction is not seen to occur, it may be that the reaction has a high activation energy causing it to be so slow that it occurs at a negligible rate. This is described as “kinetic control” or “kinetic stability”. Note 5. Look at cases 2 and 4 and note the relationship to Le Chatelier’s Principle which predicts, but does not explain, that higher temperatures cause a shift in the endothermic direction in an equilibrium reaction, whereas lower temperatures favour the exothermic direction.
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127. The Gibb’s Free Energy Change and Spontaneity
Chem Factsheet
Deciding the +ve or –ve nature of the enthalpy change for the reaction,
∆H
1. May be provided provided in the the question question as an actual actual value. value. 2. May be indicated indicated in the question question by a written written statemen statement. t. e.g. “the test-tub test-tubee became hot” hot” suggests suggests an exothermic exothermic reaction. reaction. 3. May be calculate calculated d via an enthalpy enthalpy cycle cycle using Hess’ Hess’ss Law. Law. 4. May be estimate estimated d via a calculation calculation using mean mean bond enthalpie enthalpies. s. 5. May be be calculated calculated if the entropy entropy change change for for the surroundings is given since ∆H = -T∆SSurr / 1000 6. May be indicated by by the type of reaction. reaction. e.g. combustions are all exothermic; exothermic; decompositions are endothermic with very few exceptions; neutralisations are exothermic. 7. May be deduced from from combinations combinations of informatio information n about spontaneity spontaneity of the reaction reaction and the nature of∆S. e.g. if the question says “the reaction occurs and the entropy change for the system is negative”, ∆H must be negative and more negative than -T∆S is positive, so ∆G is –ve.
Deciding the +ve or –ve nature of the entropy change of the system,
∆S
1. May be provided provided in the the question question as an actual actual value. value. 2.
May be calculated from the entropies of the reactants and products if they are provided using : ∆S = ΣS(Products) - ΣS(Reactants).
3. May be deduced by considering the equation and looking for such factors as changes in the numbers of moles and changes in state. e.g. 1.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆S is –ve since the moles of gases decreases, 6 to 3.
e.g. 2.
[Cu(H2O)6]2+(aq) + EDTA4-(aq) → [Cu(EDTA)]2-(aq) + 6H2O(l) ∆S is + ve because of an increase in the number of particles in solution, 2 to 7.
e.g. 3.
KCl(s) + (aq) → K+(aq) + Cl-(aq) ∆S is + ve since the ions move randomly in solution and vibrate about a fixed point in the solid state.
Note. Deciding whether whether the entropy change is positive or negative when when a substance dissolves in water is not straight forward forward since some of the water molecules are attracted to and ordered around the ions. The higher the charges on the ions the greater their hydration numbers and hence the lower the sum of the entropies of the products. 4. May be deduced deduced from combinatio combinations ns of information information about about spontaneity spontaneity of the reaction reaction and the nature nature of∆H. e.g. if the question says “a reaction occurs and is endothermic”, ∆S must be + ve so that -T∆S is –ve enough to outweigh the +ve ∆H, giving a –ve ∆G.
Acknowledgements: This Factsheet was researched and written by Bob Adam. Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU. ChemistryFactsheets ChemistryFactsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publi sher. ISSN 135 1-5 136
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127. The Gibb’s Free Energy Change and Spontaneity
Chem Factsheet
Practice Questions 1. Consid Consider er the reactio reaction: n: 2Na(s) 2Na(s) + ½O2(g) Na2O(s) (a) Explain why this reaction is spontaneous at 298K even though the entropy change for the system is negative. (b) Assuming no changes in ∆H and ∆S, explain how the temperature should be changed to favour the decomposition of sodium oxide? 2. When titanium titanium is manufac manufactured tured by the Kroll Kroll process process the two reactions reactions are: are: TiO2(s) + 2Cl2(g) + 2C(s) and TiCl4(g) +
2Mg(l)
→
→
TiCl4(g) + 2CO(g) 2CO(g) - - - - (1)
Ti(s) +
2MgCl2(l)
- - - - (2)
(a) For which reaction will the entropy change of the system be negative? Explain your answer. (b) For which reaction will increasing the temperature make ∆G more negative? Explain your answer and state any assumptions made? 3. Predict and explain the signs of the (a) enthalpy change, (b) entropy of the the system change, (c) Gibb’s Gibb’s free free energy change for the following reaction. If any of the predictions cannot be made explain why. [Cu(H2O)6]2+(aq) + 4Cl-(aq) → [CuCl4]2-(aq) + 6H2O(l) 4. Conside Considerr the follo followin wing g reacti reaction: on: 3KClO4(s) + KCl(s) 4KClO3(s)
∆H ê
= -16.8 kJ mol-1.
The standard molar entropy, S , (in units of JK-1mol-1) for each substance is: KCl(s) KCl(s) 83, KClO3(s) 112, KClO4(s) 134. ê
(a) Why does potassium chlorate(VII) have the highest entropy? (b) Calculate the standard entropy change for the reaction. (c) Calculate the standard free energy change at 298K for the above reaction. (d) In terms of the standard free energy change explain if the above reaction is spontaneous at 298K. (e) Explain why no noticeable reaction occurs at 298 K. (f) (f) Assumi Assuming ng that that ∆H and
∆S
do not vary with temperature calculate the temperature at which the reaction will fail to be feasible.
K 4 5 4 w . r e w s n A . K 4 5 4 = 7 3 - / 0 0 0 1 × 8 . 6 1 - = S / H = T ∴ . S T - H = 0 ) f ( o l e B ∆ ∆ ∆ ∆
) . h g i h o o t s i t i s a y g r e n e n o i t a v i t c a e h t = s e i g r e n e e v a h l l i w s e l c i t r a p o N ( . o r e z / w o l e b l l i w n o i t c a e r e h t f o e t a r e h t o s ) g n i r r u c c o e b l l i w s n o i s i l l o c o n o s ( s d i l o s n e e w t e b s i n o i t c a e r e h T ) e ( . 0 < s i ê G c n i s s u o e n a t n o p s s i n o i t c a e r e h T ) d ( ∆ e s y S ∆ ∴ ê ∆ ∆ ∆ . 1 - l o m J k 8 . 5 - = ) 0 0 0 1 / 7 3 ( 8 9 2 - 8 . 6 1 - = ê G S T - ê H = ê G ) c ( l o m K J 7 3 - = } 3 8 + ) 4 3 1 ( 3 { – ) 2 1 1 ( 4 = S ) s t n a t c a e R ( S - ) s t c u d o r P ( S = S ) b ( ∆ ∴ Σ Σ ∆ 1 1 . s s a m r a l o m t s e h g i H ) a ( . 4 . s y S ê ∆ ∆ ∆ ∆ S T - ê H = ê G s a d e d i c e d e b t o n n a c ê G f o n g i s e h t o S
. e v i t i s o p s i ê H a h t e v i t a g e n s s e l r o e r o m s i s y S ê S T - f i d e d i c e d e b t o n n a c t I ) c ( ∆ n ∆ 7 o t e g n a h c s e l c i t r a p 5 e s u a c e b . e v i t i s o P ) b (
d e m r o f e r a 4 y l n o , n e k o r b e r a r e p p o c o t s d n o b 6 e s u a c e b . e v i t i s o P ) a ( . 3 ∆ d ∆ : . t n a t s n o c n i a m e r S n a H s n o i t p m u s s A
. e v – e r o m s e m o c e b o s l a G s e v – e r o m s e m o c e b S T - : n o i t a n a l p x E . 1 n o i t c a e R ) b ( ∆ o ∆
. s a g f o s e l o m f o r e b m u n e h t n i e s a e r c e d a s i e r e h T : n o i t a n a l p x E . 2 n o i t c a e R ) a ( . 2
∆ g ∆ n ∆ . 0 < G n i k a m y l l a u t n e v e o s , e v i t i s o p s i H a h t e v i t a g e n e r o m S T - e k a m o T . e r u t a r e p m e t e h t e s a e r c n I ) b (
. 0 < G n i k a m s u h t , e v i t i s o p s i S T - n a h t e v i t a g e n e r o m e b t s u m H c i m r e h t o x e e b t s u m n o i t c a e r e h T ) a ( . 1 ∆ g ∆ ∆ .
r e w s n A ) k r a m e n o s i e n O ( . s
3
