CHEM1901/3 2013-J-8 June 2013 • The atmosphere of Venus contains 96.5 % CO 2 at 95 atm of pressure, leading to an average global surface temperature of 462 °C. The energy density of solar radiation striking Venus is 2625 J m –2 s –1 . The radius of Venus is 6052 km, and the average albedo (the fraction of solar radiation reflected back into space) of its surface is 0.90. Calculate the magnitude of the greenhouse effect on Venus. Marks 5 An albedo of 0.90 means that 90% of the solar energy is reflected and only 10% is available to heat the planet. If Venus has a radius R, the energy input from the Sun is: E in = (2625 × πR 2 × 0.10) J The energy output from the surface area 4πR 2 is: E out = (4πR 2 × 5.67 × 10 -8 × T 4 ) J At equilibrium, the energy input and output are equal and so the temperature can be calculated: 2625 × πR 2 × 0.10 = 4πR 2 × 5.67 × 10 -8 × T 4 T 4 = 0.10 × 2625 / (4 × 5.67 × 10 -8 ) K so T = 184 K As the actual temperature on Venus is 462 °C = 735 K, the magnitude of the greenhouse effect is: greenhouse effect = (735 – 185) K = 551 K Answer: 551 K The main absorption bands of CO 2 lie in the energy range 600 – 750 cm –1 . What range of wavelengths (in nm) corresponds to this energy range? 600 cm -1 corresponds to a wavelength of (1/600) cm = 0.00167 cm. As 1 cm = 0.01 m and 1 nm = 10 -9 m, this corresponds to 16700 nm. 750 cm -1 corresponds to a wavelength of (1/750) cm = 0.00133 cm. As 1 cm = 0.01 m and 1 nm = 10 -9 m, this corresponds to 13300 nm. ANSWER CONTINUES ON THE NEXT PAGE
Transcript
CHEM1901/3 2013-J-8 June 2013
• The atmosphere of Venus contains 96.5 % CO2 at 95 atm of pressure, leading to an average global surface temperature of 462 °C. The energy density of solar radiation striking Venus is 2625 J m–2 s–1. The radius of Venus is 6052 km, and the average albedo (the fraction of solar radiation reflected back into space) of its surface is 0.90. Calculate the magnitude of the greenhouse effect on Venus.
Marks 5
An albedo of 0.90 means that 90% of the solar energy is reflected and only 10% is available to heat the planet. If Venus has a radius R, the energy input from the Sun is: Ein = (2625 × πR2 × 0.10) J The energy output from the surface area 4πR2 is: Eout = (4πR2 × 5.67 × 10-8 × T4) J At equilibrium, the energy input and output are equal and so the temperature can be calculated:
2625 × πR2 × 0.10 = 4πR2 × 5.67 × 10-8 × T4
T4 = 0.10 × 2625 / (4 × 5.67 × 10-8) K so T = 184 K
As the actual temperature on Venus is 462 °C = 735 K, the magnitude of the greenhouse effect is: greenhouse effect = (735 – 185) K = 551 K
Answer: 551 K
The main absorption bands of CO2 lie in the energy range 600 – 750 cm–1. What range of wavelengths (in nm) corresponds to this energy range?
600 cm-1 corresponds to a wavelength of (1/600) cm = 0.00167 cm. As 1 cm = 0.01 m and 1 nm = 10-9 m, this corresponds to 16700 nm.
750 cm-1 corresponds to a wavelength of (1/750) cm = 0.00133 cm. As 1 cm = 0.01 m and 1 nm = 10-9 m, this corresponds to 13300 nm.
ANSWER CONTINUES ON THE NEXT PAGE
CHEM1901/3 2013-J-8 June 2013
Sketch the emission spectrum of Venus on the axes below. Note the wavelength of maximum intensity, and point out any other important features.
The wavelength corresponding to the intensity maximum is given by Wien’s displacement law:
T λmax ≈ 2.898 × 106 K nm
Using T = 735 K, λmax ≈ (2.898 × 106 K nm) / (735 K) = 3940 nm
4000 13,000 17,000 wavelength (nm)
CO2 absorption bands
λmax = 3940 nm
Inte
nsity
y
CHEM1901/3 2013-J-12 June 2013
• Differential scanning calorimetry (DSC) is an experimental technique that measures the temperature of a sample as a function of the heat supplied to it. Negative or positive peaks on a DSC curve therefore indicate endothermic or exothermic processes respectively. The figure below shows a series of DSC curves collected for methane at different pressures. The scales of all the heat flow curves are the same, but they have been offset from zero for clarity. Clearly identify the type of phase change associated with every peak in the DSC curve.
Marks 8
On the 0.1, 1 and 10 bar curves, there are two phase transitions observed so melting and boiling occurs. From the size of the peak is consistent with the single peak in the 100 bar being melting. The size of the peak in the 0.01 bar curve is not consistent with melting or boiling so it must correspond to sublimation.
ANSWER CONTINUES ON THE NEXT PAGE
s → l
s → l
s → l
s → g
l → g
l → g
CHEM1901/3 2013-J-12 June 2013
Use the DSC data shown to sketch a pressure-temperature phase diagram on the graph below (note that pressure is on a log scale). Label all the important regions of the phase diagram.
liquid
supercritical fluid solid
gas
critical point
triple point
CHEM1901 2012-J-8 June 2012
• Explain the following phenomena. (a) When O2 dissolves in water at room temperature, the total entropy of the system decreases.
Marks 6
The decrease in entropy of O2 going from (g) → (aq) outweighs the increase in entropy of pure water going to water with O2 dissolved in it.
(b) When MgSO4 is dissolved in water, there is a very small but measurable decrease in volume.
The hydration spheres around the dissolved ions (especially the small Mg2+ ion) create a denser structure than that of random H2O molecules in pure water.
(c) A crystal (e.g. of NaCl) in its lowest energy configuration (thermodynamic ground state) will always contain defects at finite temperatures.
Defects increase enthalpy, but they also increase entropy. So a small number of
defects will lower the total free energy (except at T = 0 K).
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY
CHEM1901/3 2010-J-10 June 2010
• Iron oxide that has been exposed to the atmosphere for any length of time will generally contain a mixture of magnetite, Fe2O3, and haematite, Fe3O4. This mixture can be converted to pure Fe3O4 by heating it under an excess of flowing hydrogen gas:
If 0.364 kJ of heat is liberated during the conversion of a 7.18 g sample of iron oxide (consisting only of magnetite and haematite) to pure Fe3O4, what was the initial mass percentage of haematite in the sample?
Marks 4
From the chemical equation, 1 mol of H2 liberates 11.0 kJ of heat. Hence, the number of moles of H2 that must be reacting to produce 0.0364 kJ of heat is: number of moles of H2 = (0.0364 / 11.0) mol = 0.00331 mol
From the chemical equation, 1 mol of H2 reacts with 3 mol of Fe2O3 and so: number of moles of Fe2O3 = (3 × 0.00331) mol = 0.00993 mol The molar mass of Fe2O3 is (2 × 55.85 + 3 × 16.00) g mol-1 = 159.7 g mol-1. Hence, this amount of Fe2O3 corresponds to: mass of Fe2O3 = number of moles × molar mass = (0.00993 mol) × (159.7 g mol-1) = 1.59 g The mass percentage of Fe2O3 in the initial 7.18 g sample is therefore:
mass percentage of Fe2O3 = (1.59 / 7.18) × 100% = 22.1 %
The mass percentage of haematite is therefore (100.0 – 22.1)% = 77.9%.
Answer: 77.9%
Fe3O4 can be further reduced to FeO under flowing hydrogen.
Fe3O4(s) + H2(g) → 3FeO(s) + H2O(g)
Based on the following table of thermodynamic data, what is the minimum temperature at which this would be a spontaneous reaction?
At 630 K, ∆G° = ∆H° - T∆S° = (30. × 103) – (630 K) × (24 J K-1 mol-1) = 15000 J mol-1 As ∆G° > 0, the reaction is not spontaneous and FeO is stable.
CHEM1901/3 2009-J-8 June 2009
• A new process has been developed for converting cellulose from corn waste into the biofuel butanol, C4H9OH. A bomb calorimeter with a heat capacity of 3250 J K–1 was used to determine the calorific value by burning 5.0 g of butanol in excess oxygen.
Write a balanced reaction for the combustion of butanol in oxygen.
Marks 5
C4H9OH(l) + 6O2(g) →→→→ 4CO2(g) + 5H2O(g)
Calculate the heat released from this combustion if the temperature of the calorimeter increased from 23.0 to 78.6 °C during the test.
The heat, q, required to change the temperature of the bomb calorimeter by ∆T is given by:
q = C ∆T
The temperature increases from 23.0 to 78.6 °C corresponding to ∆T = 55.6 °C = = 55.6 K. Hence:
q = (3250 J K-1)(55.6 K) = 181000 J = 181 kJ
Answer: 181 kJ
Use this value to determine the calorific value and molar enthalpy of combustion of butanol.
The calorific value is the energy content per gram. As 5.0 g of butanol releases 181 kJ, the calorific value is:
calorific value = ��� ��
�.� = 36 kJ g-1
The molar enthalpy of combustion is the energy released by combusting one mole. The molar mass of C4H9OH is:
molar mass = (4×12.01 (C) + 10×1.008 (H) + 16.00 (O)) g mol-1 = 74.12 g mol-1
5.0 g of butanol thus corresponds to:
number of moles = ���
��� ����
�.�
��.�� ��� = 0.067 mol
As 0.067 mol generates 181 kJ when combusted, the molar enthalpy of combustion is negative and is given by:
∆H = ���� ��
�.��� � = -2.7 × 103 kJ mol-1
CHEM1901/3 2008-J-8 June 2008
• Write a balanced equation for (i) the explosive decomposition, and (ii) the combustion in air, of TNT, C7H5N3O6(s).
Using ∆rH° = Σ m∆fH°(products) - Σ m∆fH°(reactants), the enthalpy change for reaction (i) can be written as ∆rH° = [12∆fH°(CO(g))] – [2∆fH°(C7H5N3O6(s) )] = ([12 × -111] – [2 × 6.9]) kJ mol-1 = --1346 kJ mol-1
ANSWER CONTINUES ON THE NEXT PAGE
CHEM1901/3 2008-J-8 June 2008
Similarly. the enthalpy change for reaction (ii) can be written as ∆rH° = [14∆fH°(CO2(g)) + 5∆fH°(H2O(g))] – [2∆fH°(C7H5N3O6(s) )] = ([14 × -393 + 5 × -242] – [2 × 6.9]) kJ mol-1 = -6726 kJ mol-1
∆fH°(N2(g)), ∆fH°(H2(g)) and ∆fH°(C(s) all represent formation of the elements from their standard states and are all zero. The combustion reaction releases more energy.
Answer: combustion reaction
CHEM1901/3 2008-J-10 June 2008
• A calorimeter containing 300.0 mL of water at 25 °C was calibrated as follows. A 1000.0 W heating coil was run for 10.0 s, after which time the temperature had increased by 7.5 °C. Calculate the heat capacity of the empty calorimeter. The specific heat of water is 4.184 J K–1 g–1.
Marks6
With a power of 1000.0 W = 1000.0 J s-1, the amount of heat generated by the coil is:
q = (1000.0 J s-1) × (10.0 s) = 1.00 × 103 J
As the density of water at 25 °C is 0.997 g mL-1, 300.0 mL of water corresponds to:
mass of water = m = density × volume = (0.997 g mL-1) × (300.0 mL) = 299 g
The heat required to heat this quantity of water by 7.5 °C is:
The remaining heat, (1.00 × 103 – 9380) J = 620 J, is used to heat the calorimeter. This also increases in temperature by 7.5 °C so the heat capacity of the calorimeter is:
ccalorimeter = �
�� = � �
.� = 82 J K-1
Answer: 82 J K-1
A solution containing 0.040 mol Ag+(aq) was mixed with a second solution containing 0.050 mole Br–(aq) in this calorimeter, causing AgBr(s) to precipitate. The temperature increased by 2.4 °C. Given the solubility product constant is Ksp(AgBr) = 5 × 10–13 M2 , calculate the equilibrium concentrations of Ag+(aq) and Br–(aq) present in the final solution of volume 320 mL.
To calculate the equilibrium constants of Ag+(aq) and Br-(aq), a reaction table can be used, with x representing the number of moles which do not precipitate:
Ag+(aq) Br-(aq) AgBr(s)
initial (mol) 0.040 0.050 0
final (mol) x 0.010 + x 0.040 - x
The concentrations in the final solution of volume 320 mL are therefore:
[Ag+(aq)] = � ���
.�� � and [Br -(aq)] = �.�� �� ���
.�� �
Hence, the solubility product is:
ANSWER CONTINUES ON THE NEXT PAGE
CHEM1901/3 2008-J-10 June 2008
Ksp = [Ag+(aq)][Br -(aq)] = � �.��� �.�� �
.�� � = 5 × 10-13
Ksp is very small, x is tiny and so 0.010 – x ~ 0.010. This approximation gives:
� �.��� �.�
.��� = 5 × 10-13
or
[Ag+(aq)] = � �.��� = (5 × 10-13) � �.��
.�� M = 1.6 × 10-11 M
[Br -] = .�.�� M = 0.031 M
[Ag+(aq)]: 1.6 × 10-11 M [Br–(aq)]: 0.031 M
Calculate the enthalpy of solution of AgBr(s).
As the final solution has a volume of 320 mL, its mass is:
mass = density × volume = (0.997 g mL-1) × (320 mL) = 319 g
The heat produced causes the temperature of this mass of solution and the calorimeter to rise by 2.4 °C:
This corresponds to the heat given out in the precipitation of 0.040 mol of AgBr(s). As the precipitation is exothermic, the heat of solution is endothermic:
heat of solution = +�# �
.# ��� = +85 kJ mol-1
Answer: +85 kJ mol-1
CHEM1901/3
• Use the figure below to help answer the following
Write a balanced equation for the smelting of one of these metal oxides with coke in which a major product is CO2. Give the approximatethis reaction is spontaneous and state what happens outside this temperature range.
Between 400 – 950 K, the Ni oxide will be reduced by coke to produce CO
2NiO(s) + C(s) ���� 2NI(l) + CO
Below 400 K, it appears that the Ni reduction will not occur. Above 950 K, the C – CO line is higher than the C produces CO2. (Any CO2 produced would be reduced by C to produce CO.)
Over what temperature range can ZnO be reduced by Fe? What other metal could be used instead to increase the temperature range in which metallic Zn was produced?
The Zn – ZnO line falls below the Fe reduction of ZnO by Fe will thus occur above this temperature. The Zn – ZnO line is below the Ca the entire temperature range of the diagram and so Ca, Al or Mn could be used to reduce ZnO at these temperatures.
THIS QUESTION CONTINUES ON THE NEXT PAGE.
2008-J-11 June 2008
Use the figure below to help answer the following.
Write a balanced equation for the smelting of one of these metal oxides with coke in . Give the approximate temperature range over which
this reaction is spontaneous and state what happens outside this temperature range.
Ni – NiO lines is below the C – CO2 line and henceby coke to produce CO2:
2NI(l) + CO2(g)
Below 400 K, it appears that the Ni – NiO will be above the C – CO2 line so this
CO line is higher than the C – CO2 line so the reduction produced would be reduced by C to produce CO.)
Over what temperature range can ZnO be reduced by Fe? What other metal could be used instead to increase the temperature range in which metallic Zn was produced?
ZnO line falls below the Fe – FeO line at approximately 1450 K and reduction of ZnO by Fe will thus occur above this temperature.
ZnO line is below the Ca – CaO, Al – Al2O3 and Mn – MnO lines over the entire temperature range of the diagram and so Ca, Al or Mn could be used o reduce ZnO at these temperatures.
THIS QUESTION CONTINUES ON THE NEXT PAGE.
June 2008
Write a balanced equation for the smelting of one of these metal oxides with coke in temperature range over which
this reaction is spontaneous and state what happens outside this temperature range.
Marks 4
and hence the
line so this
line so the reduction produced would be reduced by C to produce CO.)
Over what temperature range can ZnO be reduced by Fe? What other metal could be used instead to increase the temperature range in which metallic Zn was produced?
FeO line at approximately 1450 K and
MnO lines over the entire temperature range of the diagram and so Ca, Al or Mn could be used
CHEM1901/3
Estimate the partial pressure of CO that would be expected at equilibrium in the smelting of ZnO by coke at 1500 K.
Metallic copper is produced by smelting chalcopyrite, CuFeSto produce iron oxides and SOsketch the lnKp versus temperature curve for CuClearly label the curve you have drawn.
CuFeS2(s) + 5/2O2(g) →→→→ Cu(s) + FeO(s) + 2SO The lnKp curve for Cu – CuO lies below Fe (in bold / red) is shown below.
2008-J-12 June 2008
Estimate the partial pressure of CO that would be expected at equilibrium in the smelting of ZnO by coke at 1500 K.
CO lnKp(1) ~ 21 ZnO lnKp(2) ~ 14
p(2)
(2)) = lnKp(1) – lnKp(2) = 21 – 14 = 7
(CO) ~ 1000 atm
Metallic copper is produced by smelting chalcopyrite, CuFeS2(s), directly in oxygen to produce iron oxides and SO2. Write a balanced equation for this reaction, and
temperature curve for Cu-CuO on the diagram on page 24. Clearly label the curve you have drawn.
Cu(s) + FeO(s) + 2SO2(g)
CuO lies below Fe – FeO at all temperatures. A (in bold / red) is shown below.
June 2008
Estimate the partial pressure of CO that would be expected at equilibrium in the Marks 5
directly in oxygen . Write a balanced equation for this reaction, and
CuO on the diagram on page 24.
A sketch
CHEM1901/3 2008-J-14 June 2008
• State the Second Law of Thermodynamics, and explain how an exothermic process in a closed system changes the entropy of the surroundings.
4
In any spontaneous process, the entropy of the universe always increases.
A closed system can exchange heat or work with its surroundings. An exothermic reaction releases heat, q, into the surroundings: ∆∆∆∆surrS = q / T > 0
The enthalpy of reaction defines the heat released into or absorbed from the surroundings at constant temperature and pressure.
The standard enthalpy and entropy of solution of poly(oxyethylene) in water are ∆H° = –7.8 kJ mol–1 and ∆S° = –31 J K–1 mol–1. Use these data to predict whether the solubility of poly(oxyethylene) in water increases or decreases when the solution is warmed.
The surroundings receive 7.8 kJ mol-1 from this exothermic reaction. Hence:
∆∆∆∆surrS° = (7.8 × 103 kJ mol-1) / T
As ∆∆∆∆sysS° = –31 J K–1 mol–1, the total change in entropy is:
∆∆∆∆univS°°°° = (–31 + 7800/T) J K–1 mol–1
Spontaneous process occur when ∆∆∆∆univS°°°° is positive:
• At low T, the second term is large and dominates the first term. This leads to ∆∆∆∆univS°°°° > 0 and dissolution being favoured.
• At high T, the second term is small and the first term dominates. This leas
to ∆∆∆∆univS°°°° < 0 and dissolution being unfavoured.
Hence, the solubility decreases with temperature.
CHEM1901/3 2007-J-7 June 2007
• Write a balanced equation for the combustion of methylhydrazine, CH3NHNH2(g). Marks
To burn one mole of methylhydrazine requires 5/2 moles of liquid O2 to be carried so the combined mass is (46.078) + (5/2 × 32.00) = 126.078. The calorific value is therefore:
calorific value = 1combH 12319.8kJ g
molarmass 126.078−−−−∆∆∆∆ = == == == =
Answer: 9.8 kJ g-1
When methylhydrazine is used as a rocket fuel, the usual oxidant is dinitrogen tetroxide rather than liquid oxygen. Why?
N2O4 has a boiling point of 21 °°°°C, so there are no problems associated with handling cryogenic liquids. N2O4 and methylhydrazine are hypergolic - the reaction occurs spontaneously on contact.
CHEM1901/3 2007-J-8 June 2007
• Estimate the average temperature of Mercury given the solar power density at its surface of 9150 J m–2 s–1, and assuming an average albedo of 6% and zero Greenhouse effect.
Marks 2
With an albedo of 6%, 94% is not reflected. Hence,
• Use the figure below to help answer the following.
Write a balanced equation for the smelting of NiO by coke. In what temperature range will this process be spontaneous?
Marks 4
NiO(s) + C(s) Ni(l) + CO(g)
The reaction is spontaneous at temperatures where the C-CO line is above the Ni-NiO line. As marked on the diagram, this occurs at temperatures above ~620 K.
Why are (a) aluminium and (b) tungsten not recovered from their oxides by smelting with coke? What alternative processes are used and why?
Tungsten forms tungsten carbide if smelted with coke. The oxide is usually reduced using another reductant such as H2.
Aluminium is usually recovered electrolytically as smelting only occurs at temperatures well above 1800 K, which are not readily achieved.
CHEM1901/3 2007-J-13 June 2007
• State the Second Law of Thermodynamics, and explain how the enthalpy of reaction is related to the entropy change of the surroundings.
Marks 5
In any spontaneous reaction, the entropy of the universe always increases. The enthalpy of reaction defines the heat released into or absorbed from the surroundings at constant temperature and pressure. The entropy change of the surroundings arises from the heat released.
By definition, ∆∆∆∆surrS = surrqT
As the heat from the reaction is lost to the surroundings, qsurr = –∆∆∆∆rxnH at constant P.
Hence, ∆∆∆∆surrS = −∆−∆−∆−∆ rxn H
T
Formic acid HCOOH, can dimerise in the gas phase according to the reaction
2HCOOH (HCOOH)2
with a standard enthalpy and entropy of dimerisation of ∆H° = –62 kJ mol–1 and ∆S° = –150 J K–1 mol–1 respectively. Predict the temperature-dependence of the dimerisation reaction.
∆universeS = ∆sysS + ∆surrS = ∆sysS rxn HT
∆∆∆∆−−−−
Assuming that ∆S° and ∆H° do not change greatly with temperature, this becomes:
∆universeS = (-150) 3( 62 10 )
T− ×− ×− ×− ×
−−−− = -150 + 3(62 10 )
T××××
J K -1 mol-1
Dimerization is spontanteous if ∆universeS > 0. This occurs when T < 3(62 10 )
(150)××××
.
Hence, dimerization is spontaneous at T < 413 K and the reverse process is spontaneous at T > 413 K. More dimer is formed at low temperature.
Draw a structure that shows the bonding in the dimer.
H C
O
O H
HC
O
OH
H-bond
CHEM1901/3 2006-J-7 June 2006
• Write the equation whose enthalpy change represents the standard enthalpy of formation of hydrazine, N2H4(g).
Marks 4
N2(g) + 2H2(g) ���� N2H4(g)
Write the equation whose enthalpy change represents the enthalpy of combustion of hydrazine, N2H4(g) to produce water vapour.
N2H4(g) + O2(g) ���� N2(g) + 2H2O(g)
Given the following data, calculate the standard enthalpy of formation of N2H4(g).
∆H°f (H2O(g)) = –242 kJ mol–1
∆H°comb (N2H4(g)) = –580 kJ mol–1
Using o o orxn f fH m H (products) n H (reactants)∆ = ∆ − ∆∆ = ∆ − ∆∆ = ∆ − ∆∆ = ∆ − ∆∑ ∑∑ ∑∑ ∑∑ ∑ , the enthalpy
change for the combustion reaction is:
o o ocomb f 2 f 2 4H [2 H (H O(g))] [ H (N H (l))]∆ = ∆ − ∆∆ = ∆ − ∆∆ = ∆ − ∆∆ = ∆ − ∆ as the remaining
compounds are elements in their standard states.
Hence: o 1
f 2 4
o 1f 2 4
[2 242] [ H (N H (l))] 580kJ mol
H (N H (l))] 96kJ mol
−−−−
−−−−
× − − ∆ = −× − − ∆ = −× − − ∆ = −× − − ∆ = −
∆ = +∆ = +∆ = +∆ = +
Answer: +96 kJ mol-1
CHEM1901/3 2006-J-12 June 2006
• State the Second Law of Thermodynamics, and explain how the enthalpy of reaction is related to the entropy change of the surroundings.
Marks 5
The entropy of the universe always increases: ∆univS = ∆sysS + ∆surrS > 0
The surroundings act as a reservoir of heat for the system: heat gained or lost in a reaction is taken or given to the surroundings, respectively. As ∆sysH = -qsurr at constant pressure:
∆surrS = qsurr
T=
-∆ HsysT
Give an example of a chemical reaction or a chemical process that corresponds to each of the following.
∆S > 0, ∆H > 0
As ∆surrS = -∆ Hsys
T < 0, such processes will only occur if the temperature is
sufficiently high that ∆surrS is smaller in magnitude than ∆sysS. Examples include:
• Melting of ice at 300 K: H2O(s) ���� H2O(l) • Boiling of water at 400 K: H2O(l) ���� H2O(g)
• NH4NO3(s) ���� NH4+(aq) + NO3
−−−−(aq)
∆S > 0, ∆H < 0
As ∆surrS = -∆ Hsys
T > 0 and ∆sysS > 0, such processes will occur at all
