A₂(g) + B₂(g) ⇌ 2AB(g)

Total no. of moles = 1 – x + 1 – x + 2x = 2
\(K_p={(P_{AB})^2\over (P_{A_2})(P_{B_2})}={({2x\over 2}\times p)^2\over ({(1-x)\over 2}\times p)({1-x\over 2}\times p)}\)
\(K_p={4x^2\over (1-x)^2}\)
Given that K_p = 1; \({4x^2\over (1-x)^2}=1\)
\(\Rightarrow \) 4x² = (1 - x)²
\(\Rightarrow \) 4x² = 1 + x² - 2x
3x² + 2x - 1 = 0
\(X={-2\pm\sqrt{4-4\times 3\times -1}\over 2(3)}\)
\(X={-2\pm\sqrt{4+12}\over 6}\)
\(={-2\pm\sqrt{16}\over 6}\)
\(={-2+4\over 6};{-2-4\over 6}\)
\(={2\over 6};{-6\over 6}\)
X = 0.33 ; -1 (not possible)
\(\therefore\) [A₂]_eq = 1 - x = 1 - 0.33 = 0.67
[B₂]_eq = 1 - x = 1 - 0.33 = 0.67
[AB]_eq = 2X = 2 x 0.33 = 0.66.

Given \(m_{N_2}\) = 28 g \(m_{H_2}\) = 6g
V = 1 L
\((n_{N_2})_{initial}={28\over 28}=1\ mol\)
\((n_{H_2})_{initial}={6\over 2}=3\ mol\)
N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)

\([NH_3]=({17\over 17})=1\ mol=1\ mol\)
Weight of N₂ = (no. of moles of N₂) × molar mass of N₂
= 0.5 x 28 = 14 g
Weight of H₂ = (no. of moles of H₂) × molar mass of H₂
= 1.5 x 2 = 3 g

K_C = 4 x 10^–2 for the reaction,
H₂S(g) ⇌ H₂(g) + 1/2 S₂(g)
\(K_c={[H_2][S_2]^{1/2}\over [H_2S]}\)
\(\Rightarrow 4\times 10^{-2}={[H_2][S_2]^{1/2}\over [H_2S]}\)
For the reaction,
2H₂S(g) ⇌ 2H₂(g) + S₂(g)
\(K_c={[H_2]^2[S_2]\over [H_2S]}=(4\times 10^{-2})^2=16\times 10^{-4}\)
For the reaction,
3H₂S(g) ⇌ 3H₂(g) + 3/2 S₂(g)
\(K_c={[H_2]^3[S_2]^{3/2}\over [H_2S]^3}=(4\times 10^{-2})^3=64\times 10^{-6}\)

First law of thermodynamics:
(i) The total energy of an isolated system remains constant though it may change from one form to another.
(ii) Whenever energy of a particular type disappears equivalent amount of another type must be produced.
(iii) Total energy of a system and surroundings remains constant.
(iv) Energy can neither be created nor destroyed, but may be converted from one form to another.
(v) The change in the internal energy of a closed system is equal to the energy that passes through its boundary as heat or work.
(vi) Heat and work are equivalent ways of changing a system's internal energy.

Characteristics of work:
(i) Work is defined as the force (F) multiplied by the displacement(x).
-w = F.x ......(1)
The - ve sign is introduced to indicate that the work has been done by the system by spending a part of its internal energy.
(ii) Work is a path function.
(iii) Work appears only at the boundary of the system.
(iv) Work appears during the change in the state of the system.
(v) Work brings a permanent effect in the surroundings.
(vi) Units of work: The SI unit of work is the joule (J) or Kilojoule (KJ).
(vii) If work done by the system, the energy of the system decreases, hence by convention work is taken to be negative (- w).
(viii) If work done by the system, the energy of the system increases, hence by convention work is taken to be positive (+ w).

Characteristics of entropy:
(i) Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
(ii) In general, the entropy of gaseous system is greater than liquids and greater than solids. The symbol of entropy is S.
(iii) Entropy is defined as for a reversible change taking place at a constant temperature (T), the change in entropy (\(\Delta\)S) of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).
 \(\Delta S_{sys}=\frac{q_{rev}}{T}\)
(iv) If heat is absorbed, then \(\Delta\)S is positive and there will be increase in entropy. If heat is evolved, \(\Delta\)S is negative and there is a decrease in entropy.
(v)The change in entropy of a process represented by \(\Delta\) S and is given by the equation.
 \(\Delta S_{sys}=S_f-S_i\)
(vi) If S_f > S_i, \(\Delta\)S is positive, the reaction is spontaneous and reversible.
If S_f < S_i, \(\Delta\)S is negative, the reaction is non-spontaneous and irreversible.
(vii) Unit of entropy: SI unit of entropy is J K^-1.

ΔS⁰ = ΣS ⁰ (products) - ΣS ⁰ (reactants)
= [S⁰(CO₂) - S⁰(CO) +  \(\\ \frac { 1 }{ 2 } \)S⁰(O₂)]
= 213.8 - [197.9 + \(\\ \frac { 1 }{ 2 } \)ｘ205]
= - 86.6 Jk^-1 mol^-1
we know, ΔG⁰ = ΔH⁰ - FΔS⁰
= - 282.84 - 298 x (- 86.6 x 10^-3)
= - 282.84 + 25.807
= - 257.033 kJ

ΔH⁰ = ΔH_f⁰ (CO₂) + 2 ΔH_f⁰ (H₂O) - ΔH_f⁰ (CH₄)
= - 393.5 + 2 x (- 285.8) - (-74.8)
= - 890 kJ mol^-1
ΔS⁰ = S⁰(CO₂) + 2S⁰(H₂O) - S⁰(CH4) - 2S⁰(O₂)
= 214 x 2 x 70 -186 - 2 x 205
= - 242 Jk^-1 mol^-1  
ΔG⁰ = ΔH⁰ - TΔS⁰
= - 890 -300 ｘ(-242 ｘ 10^-3)
= - 890 + 72.6 = - 817.4 kJ mol^-1

\({ H }_{ 2(g) }+\frac { 1 }{ 2 } { O }_{ 2 }\longrightarrow { H }_{ 2 }{ O }_{ (1) }\quad \quad { \Delta H }^{ o }=-285.8\quad KJ\quad (1)\)
\({ C }_{ 9graphite) }+{ O }_{ 2 }\longrightarrow { CO }_{ 2 }\quad ​​​​{ \Delta H }^{ o }=-393.5\quad KJ\quad (2)\)
\({ CH }_{ 4(g) }+{ 2O }_{ 2 }\longrightarrow { CO }_{ 2(g) }+{ 2H }_{ 2 }{ O }_{ (1) }\quad { \Delta H }^{ o }=-890\quad KJ\quad (3)\)
equation (1) X₂ + (2) - (3)
\({ C }_{ (graphite) }+{ 2H }_{ 2(g) }\longrightarrow { CH }_{ 4(g) }\quad { \Delta H }_{ f }^{ o }=-74.7\quad KJ\)
(2 x -285.8) + (-393.5) - (-890.4)
= -571.6 - 393.5 + 890.4
= -965.1 + 890.4
= -74.7 KJ

Heat change at constant pressure (at atmospheric pressure) can be measured using a coffee cup calorimeter. A schematic representation of a coffee cup calorimeter is given in Figure. Instead of bomb, a styrofoam cup is used in this calorimeter. It acts as good adiabatic wall and doesn't allow transfer of heat produced during the reaction to its surrounding. This entire heat energy is absorbed by the water inside the cup. This method can be used for the reactions where there is no appreciable change in volume. The change in the temperature of water is measured and used to calculate the amount of heat that has been absorbed or evolved in the reaction using the following expression.
q =m_w C_wΔT

where m_w is the molar mass of water and C_w is the molar heat capacity of water (4184 kJ K^-1 mol^-1)·

Mathematical statement of the first law of thermodynamics is
ΔU = q+w
Case 1: For a cyclic process involving isothermal expansion of an ideal gas
ΔU= 0; ∴q = -w
In other words, during a cyclic process, the amount of heat absorbed by the system is equal to work done by the system.
Case 2: For an isochoric process (no change in volume) there is no work of expansion.
ΔV= 0, w =0, ΔU=q_v
In other words, during isochoric process, the amount of heat supplied to the system is converted to its internal energy.
Case 3: For an adiabatic process there is no change in heat. i.e. q = O. Hence
q =0;ΔU=w
In other words, in an adiabatic process, the decrease in internal energy is exactly equal to the work done by the system on its surroundings.
Case 4: For an isobaric process. There is no change in the pressure. P remains constant. Hence
ΔU=q+w
ΔU = q-PΔV
In other words, in an isobaric process a part of heat absorbed by the system is used for PV expansion work and the remaining is added to the internal energy of the system.

(i) Open System: A System which can exchange both matter and energy with its surroundings is called an open system. Hot water contained in an open beaker is an example for open system.
(ii) Closed System: A system which can exchange only energy but not matter with its surroundings is called a closed system. A gas contained in a cylinder fitted with a piston constitutes a closed system.
(iii) Isolated System: System which can exchange neither matter and nor energy with its surroundings is called an isolated system. Hot water contained in a thermos flask.
(iv) Homogeneous System: A system is called homogeneous if physical states of all its matter are uniform. Example: mixture of gases, completely miscible mixture of liquids etc.
(v) Heterogeneous System: A system is called heterogeneous if physical states of all its matter are not uniform.

= -290830 - 298 (-330)
= -290830 + 98340 = -192490
\(\Delta{G^{o}}\)=-192490 J mol^-1
Since \(\Delta{G^{o}}\) is negative, the reaction is spontaneous. 

\(\Delta n\)=10-12=-2 mol

=-4984ｘ10³J+8.314 JK^-1 mol^-1ｘ298kｘ(-2)mol.
=-4984000 J-4955.144J=-4988955.144J
\(\Delta H\)=-4988.955 kJ

% Efficiency =\(\left[ \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right] \times 100\)
Here
T₁ = 816+273=1089 K;
T₂ = 21+273 = 294 K
% Efficiency= \(\left( \frac { 1089-294 }{ 1089 } \right) \times 100\)
% Efficiency = 73%

C₃H₈+5O₂\(\rightarrow \)3CO₂+4H₂O
\(\triangle { H }_{ C }^{ 0 }=-2220.2KJ\quad mo{ l }^{ -1 }\)....(1)
C+O ₂\(\rightarrow \)3CO₂
\(\triangle { H }_{F}^{0 }=-393.5KJ\quad mo{ l }^{ -1 }\)....(2)
\({ H }_{ 2 }+\frac { 1 }{ 2 } { O }_{ 2 }\rightarrow { H }_{ 2 }O\)
\(\triangle { H }_{F}^{0 }=-285.8KJ\quad mo{ l }^{ -1 }\)...(3)
3C+4H₂\(\rightarrow \)C₃H₈
\(\triangle { H }_{ C }^{ 0 }\)=?
(2) X3 \(\Rightarrow \)3C+3O₂  \(\rightarrow \)3CO₂
\(\triangle { H }_{F}^{0 }= \) -1180.5 KJ ....(4)
(3)X4\(\Rightarrow \) 4H₂ +2O₂ \(\rightarrow \)4H₂O
\(\triangle { H }_{F}^{0 }= \) -1143.2KJ ....(5)
(4)+(5)-(1)\(\Rightarrow \) 3C+3O₂+4H₂+2O₂+3CO₂ +4H2O\(\rightarrow \)3CO₂+4H₂O+C₃H₈+5O₂
\(\triangle { H }_{F}^{0 }= \) -1180.5-1143.2-(-2220.2)KJ
3C+4H₂\(\rightarrow \)C₃H₈
\(\triangle { H }_{F}^{0 }= \) -103.5KJ
Standard heat of formation of propane is
\(\triangle { H }_{ R }^{ 0 }\left( { C }_{ 3 }{ H }_{ 8 } \right) =-103.5KJ\)

N₂O₄(g) ⇌ 2NO₂(g)
T₁ = 298 K        K_P1 = 0.15
T₂ = 100o C = 100 + 273 = 373 K ;
K_P2 = ?
\(\log({K_2\over K_1})={\Delta H^o\over 2.303R}[{T_2-T_1\over T_1T_2}]\)
K₂ > K₁ and T₂ > T₁
\(\log({K_{p_2}\over 0.15})={57.2KJ\ mol^{-1}\over 2.303\times 8.314JK^{-1}\ mol^{-1}}[{373-298\over 373\times 298}]\)
\(\log({K_{P_2}\over 0.15})={57.2\times 10^{+3}\times 75\over 2.303\times 8.314\times 373\times 298}\)
\(\log({K_{p_2}\over 0.15})=2.02\)
\({K_{p_2}\over 0.15}=104.7\)
\(K_{p_2}=104.7\times 0.15\)
\(K_{p_2}=15.705\)

T_b = 351 K
ΔH_vap = 39840 Jmol^-1
ΔS_V = ?
ΔS_v = \(\frac { { \triangle H }_{ vap } }{ { T }_{ b } } \)
ΔS_v = \(\frac { 39840 }{ 351 } \)
ΔS_v = 113.5 JK^-1 mol^-1

\(K_{eq}={[Fe(SCN)]^{2+}\over [Fe^{3+}][SCN^-]}\)
\(={2\times 10^{-4}M\over 8\times 10^{-4}M\times 6\times 10^{-4}M}\)
= 0.0416 x 10⁴
K_eq = 41.6 x 10² M^-1