Q-1: A thermodynamic state function is _______.
1. A quantity which depends upon temperature only.
2. A quantity which determines pressure-volume work.
3. A quantity which is independent of path.
4. A quantity which determines heat changes.
Ans:
(3) A quantity which is independent of path.
Reason:
Functions like pressure, volume and temperature depends on the state of the system only and not on the path.
Q-2: Which of the following is a correct conditions for adiabatic condition to occur.
1. q = 0
2. w = 0
3. Δp=0
4. ΔT=0
Ans:
1: q = 0
Reason:
For an adiabatic process heat transfer is zero, i.e. q = 0.
Q-3: The value of enthalpy for all elements in standard state is _____.
(1) Zero
(2) < 0
(3) Different for every element
(4) Unity
Ans:
(1) Zero
Q-4: For combustion of methane ΔUΘ is – Y kJmol−1. Then value of ΔHΘ is ____.
(a) > ΔUΘ
(b) = ΔUΘ
(c) = 0
(d) < ΔUΘ
Ans:
(d) < ΔUΘ
Reason:
ΔHΘ=ΔUΘ+ΔngRT ; ΔUΘ = – Y kJmol−1,
ΔHΘ=(–Y)+ΔngRT⇒ΔHΘ<ΔUΘ
Q-5: For, Methane, di-hydrogen and and graphite the enthalpy of combustion at 298K are given -890.3kJ mol−1, -285.8kJmol−1 and -393.5kJmol−1 respectively. Find the enthalpy of formation of Methane gas?
(a) -52.27kJmol−1
(b) 52kJmol−1
(c) +74.8kJmol−1
(d) -74.8kJmol−1
Ans:
(d) -74.8kJmol−1
à CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
ΔH=−890.3kJmol−1
à C(s) + O2(g) → CO2(g)
ΔH=−393.5kJmol−1
à 2H2(g) + O2(g) → 2H2O(g)
ΔH=−285.8kJmol−1
à C(s) + 2H2(g) → CH4(g)
ΔfHCH4 = ΔcHc + 2ΔfHH2 – ΔfHCO2
= [ -393.5 +2(-285.8) – (-890.3)] kJmol−1
= -74.8kJmol−1
Q-6: A reaction, X + Y à U + V + q is having a +ve entropy change. Then the reaction ____.
(a) will be possible at low temperature only
(b) will be possible at high temperature only
(c) will be possible at any temperature
(d) won’t be possible at any temperature
Ans:
(c) will be possible at any temperature
ΔG should be –ve, for spontaneous reaction to occur
ΔG = ΔH – TΔS
As per given in question,
ΔH is –ve ( as heat is evolved)
ΔS is +ve
Therefore, ΔG is negative
So, the reaction will be possible at any temperature.
Q-7: In the process, system absorbs 801 J and work done by the system is 594 J. Find ΔU for the given process.
Ans:
As per Thermodynamics 1st law,
ΔU = q + W(i);
ΔU internal energy = heat
W = work done
W = -594 J (work done by system)
q = +801 J (+ve as heat is absorbed)
Now,
ΔU = 801 + (-594)
ΔU = 207 J
Q-8: The reaction given below was done in bomb calorimeter, and at 298K we get, ΔU = -753.7 kJ mol−1. Find ΔH at 298K.
NH2CN(g) +3/2O2(g) à N2(g) + CO2(g) + H2O(l)
Ans:
ΔH is given by,
ΔH=ΔU+ΔngRT………………(1)
Δng = change in number of moles
ΔU = change in internal energy
Here,
Δng=∑ng(product)–∑ng(reactant)
= (2 – 2.5) moles
Δng = -0.5 moles
Here,
T =298K
ΔU = -753.7 kJmol−1
R = 8.314×10−3kJmol−1K−1
Now, from (1)
ΔH=(−753.7kJmol−1)+(−0.5mol)(298K)(8.314×10−3kJmol−1K−1)
= -753.7 – 1.2
ΔH = -754.9 kJmol−1
Q-9: Calculate the heat (in kJ) required for 50.0 g aluminium to raise the temperature from 45∘Cto65∘C. For aluminium molar haet capacity is 24 Jmol−1K−1
Ans:
Expression of heat(q),
q=mCPΔT;………………….(a)
ΔT = Change in temperature
c = molar heat capacity
m = mass of substance
From (a)
q=(5027mol)(24mol−1K−1)(20K)
q = 888.88 J q
Q-10: Calculate ΔH for transformation of 1 mole of water into ice from10∘Cto(−10)∘C.. ΔfusH=6.03kJmol−1at10∘C.
Cp[H2O(l)]=75.3Jmol−1K−1
Cp[H2O(s)]=36.8Jmol−1K−1
Ans:
ΔHtotal = sum of the changes given below:
(a) Energy change that occurs during transformation of 1 mole of water from 10∘Cto0∘C.
(b) Energy change that occurs during transformation of 1 mole of water at 0∘C to 1 mole of ice at 0∘C.
(c) Energy change that occurs during transformation of 1 mole of ice from 0∘Cto(−10)∘C.
ΔHtotal=Cp[H2OCl]ΔT+ΔHfreezingCp[H2Os]ΔT
= (75.3 Jmol−1K−1)(0 – 10)K + (-6.03*1000 Jmol−1(-10-0)K
= -753 Jmol−1 – 6030Jmol−1 – 368Jmol−1
= -7151 Jmol−1
= -7.151kJmol−1
Thus, the required change in enthalpy for given transformation is -7.151kJmol−1.
Q-11: Enthalpies of formation for CO2(g), CO(g), N2O4(g), N2O(g) are -393kJmol−1,-110kJmol−1, 9.7kJmol−1 and 81kJmol−1 respectively. Then, ΔrH = _____.
N2O4(g) + 3CO(g) à N2O(g) + 3 CO2(g)
Ans:
“ΔrH for any reaction is defined as the fifference between ΔfH value of products and ΔfH value of reactants.”
ΔrH=∑ΔfH(products)–∑ΔfH(reactants)
Now, for
N2O4(g) + 3CO(g) à N2O(g) + 3 CO2(g)
ΔrH=[(ΔfH(N2O)+(3ΔfH(CO2))–(ΔfH(N2O4)+3ΔfH(CO))]
Now, substituting the given values in the above equation, we get:
ΔrH = [{81kJmol−1 + 3(-393) kJmol−1} – {9.7kJmol−1 + 3(-110) kJmol−1}] ΔrH = -777.7 kJmol−1
Q-12 Enthalpy of combustion of C to CO2 is -393.5 kJmol−1. Determine the heat released on the formation of 37.2g of CO2 from dioxygen and carbon.
Ans:
Formation of carbon dioxide from di-oxygen and carbon gas is given as:
C(s) + O2(g) à CO2(g); ΔfH = -393.5 kJmol−1
1 mole CO2 = 44g
Heat released during formation of 44g CO2 = -393.5 kJmol−1
Therefore, heat released during formation of 37.2g of CO2 can be calculated as
= −393.5kJmol−144g×37.2g
= -332.69 kJmol−1
Q-13: N2(g) + 3H2(g) à 2NH3(g) ; ΔrHΘ = -92.4 kJmol−1
Standard Enthalpy for formation of ammonia gas is _____.
Ans:
“Standard enthalpy of formation of a compound is the enthalpy that takes place during the formation of 1 mole of a substance in its standard form, from its constituent elements in their standard state.”
Dividing the chemical equation given in the question by 2, we get
(0.5)N2(g) + (1.5)H2(g) à 2NH3(g)
Therefore, Standard Enthalpy for formation of ammonia gas
= (0.5) ΔrHΘ
= (0.5)(-92.4kJmol−1)
= -46.2kJmol−1
Q-14: Determine Standard Enthalpy of formation for CH3OH(l) from the data given below:
CH3OH(l) + (3/2)O2(g) à CO2(g) + 2H2O(l); ΔrHΘ = -726 kJmol−1
C(g) + O2(g) à CO2(g); ΔcHΘ = -393 kJmol−1
H2(g) + (1/2)O2(g) à H2O(l); ΔfHΘ = -286 kJmol−1
Ans:
C(s) + 2H2O(g) + (1/2)O2(g) à CH3OH(l) …………………………(i)
CH3OH(l) can be obtained as follows,
ΔfHΘ [CH3OH(l)] = ΔcHΘ
2ΔfHΘ – ΔrHΘ
= (-393 kJmol−1) +2(-286kJmol−1) – (-726kJmol−1)
= (-393 – 572 + 726) kJmol−1
= -239kJmol−1
Thus, ΔfHΘ [CH3OH(l)] = -239kJmol−1
Q-15: Calculate ΔH for the following process
CCl4(g) à C(g) + 4Cl(g) and determine the value of bond enthalpy for C-Cl in CCl4(g).
ΔvapHΘ (CCl4) = 30.5 kJmol−1.
ΔfHΘ (CCl4) = -135.5 kJmol−1.
ΔaHΘ (C) = 715 kJmol−1,
ΔaHΘ is a enthalpy of atomisation
ΔaHΘ (Cl2) = 242 kJmol−1.
Ans:
“ The chemical equations implying to the given values of enthalpies” are:
(1) CCl4(l) à CCl4(g) ; ΔvapHΘ = 30.5 kJmol−1
(2) C(s) à C(g) ΔaHΘ = 715 kJmol−1
(3) Cl2(g) à 2Cl(g) ; ΔaHΘ = 242 kJmol−1
(4) C(g) + 4Cl(g) à CCl4(g); ΔfHΘ = -135.5 kJmol−1 ΔH for the process CCl4(g) à C(g) + 4Cl(g) can be measured as:
ΔH=ΔaHΘ(C)+2ΔaHΘ(Cl2)–ΔvapHΘ–ΔfH
= (715kJmol−1) + 2(kJmol−1) – (30.5kJmol−1) – (-135.5kJmol−1)
Therefore, H=1304kJmol−1
The value of bond enthalpy for C-Cl in CCl4(g)
= 13044kJmol−1
= 326 kJmol−1
Q-16: ΔU = 0 for isolated system, then what will be ΔU?
Ans:
ΔU is positive ; ΔU > 0.
As, ΔU = 0 thenΔS will be +ve, as a result reaction will be spontaneous.
Q-17:
Following reaction takes place at 298K,
2X + Y à Z
ΔH = 400 kJmol−1
ΔH = 0.2kJmol−1K−1
Find the temperature at which the reaction become spontaneous considering ΔS and ΔH to be constant over the entire temperature range?
Ans:
Now,
ΔG=ΔH–TΔS
Let, the given reaction is at equilibrium, then ΔT will be:
T = (ΔH–ΔG)1ΔS ΔHΔS; (ΔG = 0 at equilibrium)
= 400kJmol−1/0.2kJmol−1K−1
Therefore, T = 2000K
Thus, for the spontaneous, ΔG must be –ve and T > 2000K.
Q-18: 2Cl(g) à Cl2(g)
In above reaction what can be the sign for ΔS and ΔH?
Ans:
ΔS and ΔH are having negative sign.
The reaction given in the question represents the formation of Cl molecule from Cl atoms. As the formation of bond takes place in the given reaction. So, energy is released. So, ΔH is negative.
Also, 2 moles of Chlorine atoms is having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, ΔS is negative.
Q-19: 2X(g) + Y(g) à 2D(g)
ΔUΘ = -10.5 kJ and ΔSΘ = -44.1JK−1
Determine ΔGΘ for the given reaction, and predict that whether given reaction can occur spontaneously or not.
Ans:
2X(g) + Y(g) à 2D(g)
Δng = 2 – 3
= -1 mole
Putting value of ΔUΘ in expression of ΔH:
ΔHΘ=ΔUΘ+ΔngRT
= (-10.5KJ) – (-1)( 8.314×10−3kJK−1mol−1)(298K)
= -10.5kJ -2.48kJ
ΔHΘ = -12.98kJ
Putting value of ΔSΘ and ΔHΘ in expression of ΔGΘ:
ΔGΘ=ΔHΘ–TΔSΘ
= -12.98kJ –(298K)(-44.1JK−1)
= -12.98kJ +13.14kJ
ΔGΘ = 0.16kJ
As, ΔGΘ is positive, the reaction won’t occur spontaneously.
Q-20: Find the value of ΔGΘ for the reaction, if equilibrium is given 10.given that T = 300K and R = 8.314×10−3kJK−1mol−1.
Ans:
Now,
ΔGΘ = −2.303RTlogeq
= (2.303)( 8.314×10−3kJK−1mol−1)(300K) log10
= -5744.14Jmol−1
-5.744kJmol−1
Q-21: What can be said about the thermodynamic stability of NO(g), given
(1/2)N2(g) + (1/2)O2(g); ΔrHΘ=90kJmol−1
NO(g) + (1/2)O2(g) à NO2(g) ; ΔrHΘ=−74kJmol−1
Ans:
The +ve value of ΔrH represents that during NO(g) formation from O2 and N2, heat is absorbed. The obtained product, NO(g) is having more energy than reactants. Thus, NO(g) is unstable.
The -ve value of ΔrH represents that during NO2(g) formation from O2(g) and NO(g), heat is evolved. The obtained product, NO2(g) gets stabilized with minimum energy.
Thus, unstable NO(g) converts into unstable NO2(g).
Q-22: Determine ΔS in surrounding given that mole of H2O(l) is formed I standard condition. ΔrHΘ=−286kJmol−1.
Ans:
ΔrHΘ=−286kJmol−1 is given so that amount of heat is evolved during the formation of 1 mole of H2O(l).
Thus, the same heat will be absorbed by surrounding. Qsurr = +286kJmol−1.
Now, ΔSsurr = Qsurr/7
= 286kJmol−1298K
Therefore, ΔSsurr=959.73Jmol−1K−1
1. A quantity which depends upon temperature only.
2. A quantity which determines pressure-volume work.
3. A quantity which is independent of path.
4. A quantity which determines heat changes.
Ans:
(3) A quantity which is independent of path.
Reason:
Functions like pressure, volume and temperature depends on the state of the system only and not on the path.
Q-2: Which of the following is a correct conditions for adiabatic condition to occur.
1. q = 0
2. w = 0
3. Δp=0
4. ΔT=0
Ans:
1: q = 0
Reason:
For an adiabatic process heat transfer is zero, i.e. q = 0.
Q-3: The value of enthalpy for all elements in standard state is _____.
(1) Zero
(2) < 0
(3) Different for every element
(4) Unity
Ans:
(1) Zero
Q-4: For combustion of methane ΔUΘ is – Y kJmol−1. Then value of ΔHΘ is ____.
(a) > ΔUΘ
(b) = ΔUΘ
(c) = 0
(d) < ΔUΘ
Ans:
(d) < ΔUΘ
Reason:
ΔHΘ=ΔUΘ+ΔngRT ; ΔUΘ = – Y kJmol−1,
ΔHΘ=(–Y)+ΔngRT⇒ΔHΘ<ΔUΘ
Q-5: For, Methane, di-hydrogen and and graphite the enthalpy of combustion at 298K are given -890.3kJ mol−1, -285.8kJmol−1 and -393.5kJmol−1 respectively. Find the enthalpy of formation of Methane gas?
(a) -52.27kJmol−1
(b) 52kJmol−1
(c) +74.8kJmol−1
(d) -74.8kJmol−1
Ans:
(d) -74.8kJmol−1
à CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
ΔH=−890.3kJmol−1
à C(s) + O2(g) → CO2(g)
ΔH=−393.5kJmol−1
à 2H2(g) + O2(g) → 2H2O(g)
ΔH=−285.8kJmol−1
à C(s) + 2H2(g) → CH4(g)
ΔfHCH4 = ΔcHc + 2ΔfHH2 – ΔfHCO2
= [ -393.5 +2(-285.8) – (-890.3)] kJmol−1
= -74.8kJmol−1
Q-6: A reaction, X + Y à U + V + q is having a +ve entropy change. Then the reaction ____.
(a) will be possible at low temperature only
(b) will be possible at high temperature only
(c) will be possible at any temperature
(d) won’t be possible at any temperature
Ans:
(c) will be possible at any temperature
ΔG should be –ve, for spontaneous reaction to occur
ΔG = ΔH – TΔS
As per given in question,
ΔH is –ve ( as heat is evolved)
ΔS is +ve
Therefore, ΔG is negative
So, the reaction will be possible at any temperature.
Q-7: In the process, system absorbs 801 J and work done by the system is 594 J. Find ΔU for the given process.
Ans:
As per Thermodynamics 1st law,
ΔU = q + W(i);
ΔU internal energy = heat
W = work done
W = -594 J (work done by system)
q = +801 J (+ve as heat is absorbed)
Now,
ΔU = 801 + (-594)
ΔU = 207 J
Q-8: The reaction given below was done in bomb calorimeter, and at 298K we get, ΔU = -753.7 kJ mol−1. Find ΔH at 298K.
NH2CN(g) +3/2O2(g) à N2(g) + CO2(g) + H2O(l)
Ans:
ΔH is given by,
ΔH=ΔU+ΔngRT………………(1)
Δng = change in number of moles
ΔU = change in internal energy
Here,
Δng=∑ng(product)–∑ng(reactant)
= (2 – 2.5) moles
Δng = -0.5 moles
Here,
T =298K
ΔU = -753.7 kJmol−1
R = 8.314×10−3kJmol−1K−1
Now, from (1)
ΔH=(−753.7kJmol−1)+(−0.5mol)(298K)(8.314×10−3kJmol−1K−1)
= -753.7 – 1.2
ΔH = -754.9 kJmol−1
Q-9: Calculate the heat (in kJ) required for 50.0 g aluminium to raise the temperature from 45∘Cto65∘C. For aluminium molar haet capacity is 24 Jmol−1K−1
Ans:
Expression of heat(q),
q=mCPΔT;………………….(a)
ΔT = Change in temperature
c = molar heat capacity
m = mass of substance
From (a)
q=(5027mol)(24mol−1K−1)(20K)
q = 888.88 J q
Q-10: Calculate ΔH for transformation of 1 mole of water into ice from10∘Cto(−10)∘C.. ΔfusH=6.03kJmol−1at10∘C.
Cp[H2O(l)]=75.3Jmol−1K−1
Cp[H2O(s)]=36.8Jmol−1K−1
Ans:
ΔHtotal = sum of the changes given below:
(a) Energy change that occurs during transformation of 1 mole of water from 10∘Cto0∘C.
(b) Energy change that occurs during transformation of 1 mole of water at 0∘C to 1 mole of ice at 0∘C.
(c) Energy change that occurs during transformation of 1 mole of ice from 0∘Cto(−10)∘C.
ΔHtotal=Cp[H2OCl]ΔT+ΔHfreezingCp[H2Os]ΔT
= (75.3 Jmol−1K−1)(0 – 10)K + (-6.03*1000 Jmol−1(-10-0)K
= -753 Jmol−1 – 6030Jmol−1 – 368Jmol−1
= -7151 Jmol−1
= -7.151kJmol−1
Thus, the required change in enthalpy for given transformation is -7.151kJmol−1.
Q-11: Enthalpies of formation for CO2(g), CO(g), N2O4(g), N2O(g) are -393kJmol−1,-110kJmol−1, 9.7kJmol−1 and 81kJmol−1 respectively. Then, ΔrH = _____.
N2O4(g) + 3CO(g) à N2O(g) + 3 CO2(g)
Ans:
“ΔrH for any reaction is defined as the fifference between ΔfH value of products and ΔfH value of reactants.”
ΔrH=∑ΔfH(products)–∑ΔfH(reactants)
Now, for
N2O4(g) + 3CO(g) à N2O(g) + 3 CO2(g)
ΔrH=[(ΔfH(N2O)+(3ΔfH(CO2))–(ΔfH(N2O4)+3ΔfH(CO))]
Now, substituting the given values in the above equation, we get:
ΔrH = [{81kJmol−1 + 3(-393) kJmol−1} – {9.7kJmol−1 + 3(-110) kJmol−1}] ΔrH = -777.7 kJmol−1
Q-12 Enthalpy of combustion of C to CO2 is -393.5 kJmol−1. Determine the heat released on the formation of 37.2g of CO2 from dioxygen and carbon.
Ans:
Formation of carbon dioxide from di-oxygen and carbon gas is given as:
C(s) + O2(g) à CO2(g); ΔfH = -393.5 kJmol−1
1 mole CO2 = 44g
Heat released during formation of 44g CO2 = -393.5 kJmol−1
Therefore, heat released during formation of 37.2g of CO2 can be calculated as
= −393.5kJmol−144g×37.2g
= -332.69 kJmol−1
Q-13: N2(g) + 3H2(g) à 2NH3(g) ; ΔrHΘ = -92.4 kJmol−1
Standard Enthalpy for formation of ammonia gas is _____.
Ans:
“Standard enthalpy of formation of a compound is the enthalpy that takes place during the formation of 1 mole of a substance in its standard form, from its constituent elements in their standard state.”
Dividing the chemical equation given in the question by 2, we get
(0.5)N2(g) + (1.5)H2(g) à 2NH3(g)
Therefore, Standard Enthalpy for formation of ammonia gas
= (0.5) ΔrHΘ
= (0.5)(-92.4kJmol−1)
= -46.2kJmol−1
Q-14: Determine Standard Enthalpy of formation for CH3OH(l) from the data given below:
CH3OH(l) + (3/2)O2(g) à CO2(g) + 2H2O(l); ΔrHΘ = -726 kJmol−1
C(g) + O2(g) à CO2(g); ΔcHΘ = -393 kJmol−1
H2(g) + (1/2)O2(g) à H2O(l); ΔfHΘ = -286 kJmol−1
Ans:
C(s) + 2H2O(g) + (1/2)O2(g) à CH3OH(l) …………………………(i)
CH3OH(l) can be obtained as follows,
ΔfHΘ [CH3OH(l)] = ΔcHΘ
2ΔfHΘ – ΔrHΘ
= (-393 kJmol−1) +2(-286kJmol−1) – (-726kJmol−1)
= (-393 – 572 + 726) kJmol−1
= -239kJmol−1
Thus, ΔfHΘ [CH3OH(l)] = -239kJmol−1
Q-15: Calculate ΔH for the following process
CCl4(g) à C(g) + 4Cl(g) and determine the value of bond enthalpy for C-Cl in CCl4(g).
ΔvapHΘ (CCl4) = 30.5 kJmol−1.
ΔfHΘ (CCl4) = -135.5 kJmol−1.
ΔaHΘ (C) = 715 kJmol−1,
ΔaHΘ is a enthalpy of atomisation
ΔaHΘ (Cl2) = 242 kJmol−1.
Ans:
“ The chemical equations implying to the given values of enthalpies” are:
(1) CCl4(l) à CCl4(g) ; ΔvapHΘ = 30.5 kJmol−1
(2) C(s) à C(g) ΔaHΘ = 715 kJmol−1
(3) Cl2(g) à 2Cl(g) ; ΔaHΘ = 242 kJmol−1
(4) C(g) + 4Cl(g) à CCl4(g); ΔfHΘ = -135.5 kJmol−1 ΔH for the process CCl4(g) à C(g) + 4Cl(g) can be measured as:
ΔH=ΔaHΘ(C)+2ΔaHΘ(Cl2)–ΔvapHΘ–ΔfH
= (715kJmol−1) + 2(kJmol−1) – (30.5kJmol−1) – (-135.5kJmol−1)
Therefore, H=1304kJmol−1
The value of bond enthalpy for C-Cl in CCl4(g)
= 13044kJmol−1
= 326 kJmol−1
Q-16: ΔU = 0 for isolated system, then what will be ΔU?
Ans:
ΔU is positive ; ΔU > 0.
As, ΔU = 0 thenΔS will be +ve, as a result reaction will be spontaneous.
Q-17:
Following reaction takes place at 298K,
2X + Y à Z
ΔH = 400 kJmol−1
ΔH = 0.2kJmol−1K−1
Find the temperature at which the reaction become spontaneous considering ΔS and ΔH to be constant over the entire temperature range?
Ans:
Now,
ΔG=ΔH–TΔS
Let, the given reaction is at equilibrium, then ΔT will be:
T = (ΔH–ΔG)1ΔS ΔHΔS; (ΔG = 0 at equilibrium)
= 400kJmol−1/0.2kJmol−1K−1
Therefore, T = 2000K
Thus, for the spontaneous, ΔG must be –ve and T > 2000K.
Q-18: 2Cl(g) à Cl2(g)
In above reaction what can be the sign for ΔS and ΔH?
Ans:
ΔS and ΔH are having negative sign.
The reaction given in the question represents the formation of Cl molecule from Cl atoms. As the formation of bond takes place in the given reaction. So, energy is released. So, ΔH is negative.
Also, 2 moles of Chlorine atoms is having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, ΔS is negative.
Q-19: 2X(g) + Y(g) à 2D(g)
ΔUΘ = -10.5 kJ and ΔSΘ = -44.1JK−1
Determine ΔGΘ for the given reaction, and predict that whether given reaction can occur spontaneously or not.
Ans:
2X(g) + Y(g) à 2D(g)
Δng = 2 – 3
= -1 mole
Putting value of ΔUΘ in expression of ΔH:
ΔHΘ=ΔUΘ+ΔngRT
= (-10.5KJ) – (-1)( 8.314×10−3kJK−1mol−1)(298K)
= -10.5kJ -2.48kJ
ΔHΘ = -12.98kJ
Putting value of ΔSΘ and ΔHΘ in expression of ΔGΘ:
ΔGΘ=ΔHΘ–TΔSΘ
= -12.98kJ –(298K)(-44.1JK−1)
= -12.98kJ +13.14kJ
ΔGΘ = 0.16kJ
As, ΔGΘ is positive, the reaction won’t occur spontaneously.
Q-20: Find the value of ΔGΘ for the reaction, if equilibrium is given 10.given that T = 300K and R = 8.314×10−3kJK−1mol−1.
Ans:
Now,
ΔGΘ = −2.303RTlogeq
= (2.303)( 8.314×10−3kJK−1mol−1)(300K) log10
= -5744.14Jmol−1
-5.744kJmol−1
Q-21: What can be said about the thermodynamic stability of NO(g), given
(1/2)N2(g) + (1/2)O2(g); ΔrHΘ=90kJmol−1
NO(g) + (1/2)O2(g) à NO2(g) ; ΔrHΘ=−74kJmol−1
Ans:
The +ve value of ΔrH represents that during NO(g) formation from O2 and N2, heat is absorbed. The obtained product, NO(g) is having more energy than reactants. Thus, NO(g) is unstable.
The -ve value of ΔrH represents that during NO2(g) formation from O2(g) and NO(g), heat is evolved. The obtained product, NO2(g) gets stabilized with minimum energy.
Thus, unstable NO(g) converts into unstable NO2(g).
Q-22: Determine ΔS in surrounding given that mole of H2O(l) is formed I standard condition. ΔrHΘ=−286kJmol−1.
Ans:
ΔrHΘ=−286kJmol−1 is given so that amount of heat is evolved during the formation of 1 mole of H2O(l).
Thus, the same heat will be absorbed by surrounding. Qsurr = +286kJmol−1.
Now, ΔSsurr = Qsurr/7
= 286kJmol−1298K
Therefore, ΔSsurr=959.73Jmol−1K−1
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