Given data:
[A] = [B] = [C]= [D] =1 M
K_c = 100
[D]_eq = ?
Solution:
Let x be the no moles of reactants reacted

	A	B	C	D
Initial concentration	1	1	1	1
At equilibrium (as per reaction stoichiometry)	1-x	1-x	1-x	1-x

\(K_c={[C][D]\over [A][B]}\)
\(100={(1+x)(1+x)\over (1-x)(1-x)}\)
\(\sqrt{100}=\sqrt{{(1+x)(1+x)\over (1-x)(1-x)}}\)
\(10={1+x\over 1-x}\)
10(1 - x) = 1 + x
10 - 10x - 1 - x = 0
9 - 11x = 0
11x = 9
\(x={9\over 11}=0.818\)
[D]eq = 1+x = 1 + 0.818 = 1.818M.

(a) Isothermal process: An isothermal process is defined as one in which the temperature of the system remains constant, during the change from its initial to final state. The system exchanges heat with its surroundings and the temperature of the system remains constant.  
For an isothermal process dT = 0
(b) Adiabatic process: An adiabatic process is defined as one in which there is no exchange of heat (q) between the system and surrounding during the process. For an adiabatic process q = 0
(c) Isobaric process: An isobaric process is defined as one in which the pressure of the system remains constant during its change from the initial to final state. For an isobaric process dP = 0 .
(d) Isochoric process: An isochoric process IS defined as the one in which the volume of system remains constant during its change from initial to final state. For an isochoric process, dV= 0.

V = 3L
\([HI]_{initial}={0.3\ mol\over 3L}=0.1\ M\)
[HI]_eq = 0.05 M
2HI(g) ⇌ H₂(g) + I₂(g)

	HI(g)	H2(g)	I2(g)
Initial Concentration	0.1	-	-
Reacted	0.05	-	-
Equilibrium concentration	0.05	0.025	0.025

\(K_c={[H_2][I_2]\over [HI]^2}\)
\(={0.025\times 0.025\over 0.05\times 0.05}\)
K_c = 0.25
K_P = K_C (RT)^Δng
Δn_g = 2 – 2 = 0
K_P = 0.25 (RT)^o
K_P = 0.25

for the reaction,
SrCO₃ (S) ⇌ SrO(S) + CO₂(S)
Δn_g = 1 – 0 = 1
\(\therefore\) K_P = K_C (RT)
2.2 x 10^–24 = K_C (0.0821) (1002)
\(K_c={2.2\times 10^{-4}\over 0.0821\times 1002}\)
KC = 2.674 x 10^-6

\( \therefore\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=0.4 \mathrm{~mol} \mathrm{dm}^{-3} ; \quad\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.6 \mathrm{~mol} \mathrm{dm}^{-3} ; \quad\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.6 \mathrm{~mol}^{-3} \)
\(\mathrm{K}_{c}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{0.6 \times 0.6}{0.4}\)
K_c = 0.9 mol dm^-3

\(K_c={[AB]^2\over [A_2][B_2]}\)
A - green
B - blue

Given that 'V' is constant (closed system)
At equilibrium
\(K_c={({4\over V})^2\over ({2\over V})({2\over V})}={16\over 4}=4\)
K_P = K_C (RT)^Δn
K_P = 4(RT)^o = 4
At Stage 'x'
\(Q={({6\over V})^2\over ({2\over V})({1\over V})}={36\over 2}=18\)
Q > K_C ie., reverse reaction is favoured
At Stage 'y'
\(Q={({3\over V})^2\over ({3\over V})({3\over V})}={9\over 3\times 3}=1\)
K_C > Q ie., forward reaction is favoured.

T₁ = 25 + 273 = 298 K
T₂ = 37 + 273 = 310 K
ΔH = 32.4 KJmol^-1 = 32400 Jmol^-1
R = 8.314 JK^-1 mol^-1
K_P1 = 0.0260
K_p2 = ?
\(\log {K_2\over K_1}={\Delta H^o\over 2.303R}[{T_2-T_1\over T_2T_1}]\)
\(\log {K_2\over K_1}={32400\over 2.303\times 8.314}({310-298\over 310\times 298})\)
\(={32400\times 10\over 2.303\times 8.314\times 310\times 298}\)
= 0.2198
\({K_2\over K_1}=\) antilog 0.2198 = 1.6588
K₂ = 1.6588 x 0.026 = 0.0431

Given data:
[NH₃] = 1.8 x 10^-2 M
[N₂] = 1.2 x 10^-2 M
[H₂] = 3 x 10^-2 M
K_c = ?
Solution:
N_2(g) + 3H_2(g) \(\rightleftharpoons \) 2NH_3(g)
\(K_c={[NH_3]^2\over[N_2][H_2]^3}={1.8\times 10^{-2}\times 1.8\times 10^{-2}\over 1.2\times 10^{-2}\times 3\times 10^{-2}\times 3\times 10^{-2}\times 3\times 10^{-2}}\)
= 1 x 10³l²mol^-2

H_2(g) + I_2(g) \(\rightleftharpoons \) 2HI_(g)
At equilibrium, [HI] = 0.4 mol L^-1 Kc= ?

	H2	I2	HI
Initial number of moles	1	1	-
Number of moles at equilibrium	1 - x	1 - x	2x = 0.4     x = 0.2
	0.8	0.8	0.4

\(\therefore K_c={[HI]^2\over [H_2[I_2]]}={0.4\times 0.4\over 0.8\times 0.8}=0.25\)

N₂O₂(g) \(\rightleftharpoons \) 2NO₂(g)
Kc = 0.21 at 373 K. The concentrations N₂O₄ and NO₂ are found to be 0.125 mol dm^-3 and 0.5 mol dm^-3 respectively at a given time. From the above information we can predict the direction of reaction as follows.
\(Q={[NO_2]^2\over [N_2O_4]}={0.5\times 0.5\over 0.125}=2\)
The Q value is greater than K_c. Hence, the reaction will proceed in the reverse direction until the Q value reaches 0.21.

H₂(g) + I₂(g) \(\rightleftharpoons \) 2HI(g)
At a particular instant, the concentration of H₂, I₂and HI are found to be 0.2 mol L^-1, 0.2 mol L^-1 and 0.6 mol L^-1 respectively. From the above information we can predict the direction of reaction as follows.
\(Q={[HI]^2\over[H_2][I_2]}={0.6\times 0.6\over 0.2\times 0.2}=9\)
Since Q < K_c, the reaction will proceed in the forward direction.

We know \(\triangle_r{G}^{\ominus}\) = -2.303 RO log K_p
and R = 8.314 JK^-1 mol^-1
Therefore, \(\triangle_r{G}^{\ominus}\) = -2.303 (8.314 J K^-1 mol^-1) x (log 2.47 x 10^-29)
= 163000 J mol^-1 = 163 mol^-1

(i) It is clear that two heat capacities are not equal and C_p is greater than C_v by a factor which is related to the work done.
(ii) At a constant pressure, a part of heat absorbed by the system is used up in increasing the internal energy of the system and the other for doing work by the system.
(iii) At constant volume, the whole of heat absorbed is utilized in increasing the temperature
of the system as there is no work done by the system.
Thus C_P is greater than C_V.
\(C_{P}=\frac{dH}{dT}; C_{V}=\frac{dU}{dT}\)
iv) By definition, H = U +PV for 1 mole of an ideal gas
H= U + RT
By differentiating this equation with respect to temperature T, we get,
\(\frac{dH}{dT}=\frac{dU}{dT}+R\)
\(C_P=C_V+R\)
\(C_P-C_V=R\)
Thus for an ideal gas, C_P is greater than C_V by the gas constant R.

H₂O \(\rightleftharpoons\) H₂O₍₁₎ \(\triangle_{fus}\) H is 6 kJ mol-1
(a) ΔG=ΔH-TΔS
When the reaction is carried out at 0°K
or ΔS=0
ΔG=ΔH
(b) \(H_2O_{s} \rightleftharpoons H_2O{_{(I)}}\)
\(\Delta_{fus}H= 6\) kJ mol^-1
= 6000 kJ mol^-1
\(\Delta_{fus}H= 6\) kJ mol^-1
= 6000 J mol^-1

The equation representing the enthalpy of formation of methane is:
\(C_{(s)}+2H_{2_{(g)}} \rightarrow CH_{4_{(g)}}\)  ΔH = -78.84 kJ
ΔH = 78.84 kJ; Δ^ng = 1 - 2 = -1 mol
R= \(8.314 \times 10 ^{-3}\)kJ-1 K^-1 mol^-1;  T=300 K
According to the relation,
ΔH = ΔU + Δ^ng RT
ΔU = ΔH + Δ^ng RT
=(-78.84 kJ) - (1 mol) x (\(8.314 \times 10 ^{-3}\)kJ K^-1 mol^-1) х 300 K
= -78.84-2.49 = -81.35 kJ.

We know  ΔU=n C_V(T₂-T₁)
ΔH=n C_P(T₂-T₁)
Here \(n=\frac{128}{32}=4\)moles;
T₂ = 100°C = 373 K; T₁ = 0°C = 273 K
ΔU = n C_V (T₂ - T₁)
ΔU = 4 x 21 x (373-273)
ΔU = 8400 J
ΔU = 8.4 kJ
ΔH =n C_p(T₂-T₁)
ΔH = 4 x 29 x (373 - 273)
ΔH = 11600 J
ΔH = 11.6 kJ

The standard enthalpy change for the combustion of ethanol can be calculated from the standard enthalpies of formation of \(C_2 H_5 OH_{I}, \)\(CO_{2{g}}\) and \(H_2 O(_{I})\). The enthalpies of formation are -277, - 393.5 and -285.5 kJ mol^-1 respectively.
\(C_2O_5 OH_{(I)}+3O_{2{g}} \rightarrow 2CO_{2{g}}+3 H_2 O_{(I)}\)
\(\Delta H^{0}_{1} = [(\Delta H^{0}_{f})_{products}-(\Delta H^{0}_{f})_{reactants})]\)
\(\Delta H^{0}_{r}=[2(\Delta H^{0}_{f})_{CO_{2}}+3(\Delta H^{0}_{f})_{H_2O}]-[1(\Delta H^{0}_{f})_{C_2H_5OH}+3(\Delta H^{0}_{f})_{O_2}]\)
\(\triangle {H}_{r}^{0}=\begin{bmatrix} 2\ mol(-393.5)\ kJ\ {mol}^{-1} \\ +3\ mol(-285.5)\ kJ\ {mol}^{-1} \end{bmatrix}-\begin{bmatrix} 1\ mol(-227)\ kJ\ {mol}^{-1} \\ +3\ mool(0)\ kJ\ {mol}-1^{} \end{bmatrix}\)
= [-787-856.5]-[-277]
=-1643.5+277
\(\Delta H^{0}_{r}\)= -1366.5 kJ

Given data q = 400 J; V₁= 5L; V₂ = 10L
Δu = q- w (heat is given to the system (+q); work is done by the system(-w)
Δu=q-PdV
=400 J - 1 atm (10 - 5)L
= 400 J - 5 atm L
[∴ 1L atm=101.33 J]
= 400 J - 5 x 10l.33 J
= 400 J - 506.65 J
= - 106.65 J

Given:
S⁰ (urea) = 173.8 J mol^-1 K^-1
S⁰ (H₂O) = 70 J mol^-1 K^-1
S⁰ (CO₂) = 213.5 J mol^-1K^-1
S⁰ (NH₃) = 192.5 J mol^-1K^-1
NH₂ - CO - NH₂ + H₂O ⟶ 2NH₃ + CO₂
ΔS_r⁰ = Σ(S⁰)_products - Σ(S⁰)_reactants
ΔS_r⁰ = [2 S⁰(NH₃) + S⁰(CO₂)] - [S⁰(urea) + S⁰(H₂O)]
ΔS_r⁰ = [2 x 192.5 + 213.5]-[173.8+70]
ΔS_r⁰ = [598.5] - [243.8]
ΔS_r⁰ = 354.7 J mol^-1K^-1.

\(\Delta { H }_{ f }^{ o }({ N }_{ 2 }{ O }_{ 4 })=+9.7\quad J\quad mol^{ -1 }\)
\(\Delta { H }_{ f }^{ O }(CO)=-110\quad kJ\quad mol^{ -1 }\)
\(\Delta { H }_{ f }^{ o }({ N }_{ 2 }O)=+81.0\quad kJ\quad { mol }^{ -1 }\)
\(\Delta { H }_{ f }^{ O }({ CO }_{ 2 })=-393\quad kJ\quad { mol }^{ -1 }\)
\(\Delta H°=\sum { \Delta { H }_{ f }^{ o } } (products)-\sum { \Delta { H }_{ f }^{ o } } (reactants)\)
\(=\left[ \Delta { H }_{ f }^{ o }({ N }_{ 2 }O)+3\Delta { H }_{ f }^{ o }{ CO }_{ 2 } \right] -\left[ \Delta { H }_{ f }^{ o }({ N }_{ 2 }{ O }_{ 4 })-3\Delta { H }_{ f }^{ 0 }(CO) \right] \)
= [81+3(-393)]-[9.7+3(-110)]
=-777.7 KJ

w=- 2.303 nRT log \(\frac { { P }_{ 1 } }{ { P }_{ 2 } } \)
= -2.303 \(\times\) 5 \(\times\)8.314 \(\times\)300 log\(\frac { 10 }{ 2 } \)
= -20.075 \(\times\) 10³ J

(i) ΔG>0: The process is non-spontaneous and non-feasible.
(ii) ΔG<0: The process is spontaneous and feasible.
(iii) ΔG=0: The process is in equilibrium.

(i) Isothermal process: Isothermal process is defined as one in which the temperature of the system remains constant, during the change from its initial to final states.
For an isothermal process dT = 0
(ii) Isobaric process: Isobaric process is defined as one in which the pressure of the system remains constant during its change from the initial to final state.
For an isobaric process dP = 0.
(iii) Isochoric process: Isochoric process is defined as one in which the volume of system remains constant during its change from initial to final state of the process.
For an isochoric processes dV = 0.

EXTENSIVE PROPERTIES	INTENSIVE PROPERTIES
1. The properties that depend on mass or size of the system are called extensive properties	The properties that are independent on the mass or size of the system are known as intensive properties.
2. Eg: volume, mass, energy, internal energy, etc.	Eg: refractive index, surface tension, density, temperature, etc

ΔU=q-w
= 100 - 50 = 50J
Since work is done by the system, it is -ve.

No transfer of heat or work is observed in an isolated system.
∴ΔU=q+w
ΔU=0+0=0.

(i) System: A system is defined as any portion of matter (or universe) under thermodynamic.consideration, which is separated from the rest of the universe by real or imaginary boundaries.
(ii) Surroundings: Everything in the universe that is not the part of system and can interact with system is called as surroundings.
(iii) Boundary: Anything which separates the system from its surroundings is called boundary.

Given K_eq= 10
Gas constant R = 8.314 JK^-1 mol^-1
T=300K
The relationship between Free energy change ΔG and equilibrium constant K is ΔG^o=-RTlnK
Since K, T and R are positive values, ΔG^o will be negative.
When ΔG is -ve, the process is spontaneous and feasible

(i) State function: A state function is a thermodynamic property of a system, which has a specific value for a given state and does not depend on the path (or manner) by which the particular state is reached.
Example: Pressure (P), Volume (V), Temperature(T)
(ii) Path functions: A path function is a thermodynamic property of the system whose value depends on the path by which the system changes from its initial to final states.
Example: Work (w), Heat (q).

The absolute entropy of a substance at 298 K and one atmosphere pressure is called the standard entropy S⁰.