\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\mathrm{K}_{\mathrm{c}}=\frac{4 x^{2}}{(a-x)(b-x)}\)
This expression doesn't involve, V. So, increase or decrease of volume will not affect the equilibrium and hence the yield of the product.

System: A system is defined as any portion of matter under thermodynamic consideration, which is separated from the rest of the universe by real or imaginary boundaries.
e.g., Water taken in a beaker, balloon filled with air, seed, plant, flower and bird.

(i) Hess's law can be applied to calculate the enthalpy of formation of carbon monoxide. It is very difficult to control the oxidation of graphite to give pure CO. However, enthalpy for the oxidation of graphite to CO₂ can be easily measured and enthalpy of oxidation of CO to CO₂ is also measurable.
(ii) The application of Hess's law enables us to estimate the enthalpy of formation of CO.
C + O₂ \(\rightarrow\) CO₂             \(\Delta\)H^o = -393.5 kJ.....(1)
CO + 1/2O₂ \(\rightarrow\) CO₂      \(\Delta\)H^o = -283 kJ..(2)
on inverting equation (2), we get
CO₂ \(\rightarrow\) CO + 1/2 O₂     \(\Delta\)H^o = +283 kJ ....(3)
on adding equations (2) and (3), we get
C + 1/2 O₂ \(\rightarrow\) CO         \(\Delta\)H^o = -393.5 + 283 = -110.5 kJ.

The calorific value is defined as "The amount of heat produced in calories (or joules) when one gram of the substance is completely burnt." The SI unit of calorific value is J kg^-1. It is usually expressed in cal g ^-1.

CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g)
K_C = 4 x 10^–2 mol lit^–2
Volume = 500 ml = 1/2 L
\(\left[\mathrm{CH}_{4}\right]_{\text {in }}=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}} \)
= 2 mol L^-1
\(\left[\mathrm{CS}_{2}\right]_{\text {in }}=\frac{1 \mathrm{~mol}}{1 / 2 \mathrm{~L}}\)
= 2 mol L^-1
\( {\left[\mathrm{H}_{2} \mathrm{~S}\right]_{\text {in }}=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1} \quad\left[\mathrm{H}_{2}\right]=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1}} \)
\(\mathrm{Q}=\frac{\left[\mathrm{CS}_{2}\right]\left[\mathrm{H}_{2}\right]^{4}}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}=\frac{2 \times(4)^{4}}{(2) \times(4)^{2}}=16 \)
Q > K_c
\(\therefore \) The reaction will proceed in the reverse direction to reach the equilibrium.

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

i) \(K_{p}=K_{c}(R T)^{\Delta n_{g}}\)
K_p = Equilibrium constant in term of partial Pressures.
K_c = Equilibrium constant in term of concentration.
R = Gas constant; T = Temperature
\(\Delta \mathrm{n}_{\mathrm{g}}\) = Difference between the sum of number of moles of products and the sum of number of moles of reactants in gas phases.
ii) Synthesis of HI:
\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\Delta n_{g}=0 \therefore K_{p}=K_{c}(R T) \Delta n_{g}\)
\(K_{p}=K_{c}(R T)^{\circ} \)
\(K_{p}=K_{c} \text {. }\)

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.

Molar Heat of Sublimation: Molar Heat of sublimation is defined as "The change in enthalpy when one mole of a solid is directly converted into the gaseous state at its sublimation temperature". For example, the heat of sublimation of iodine is represented as
I_2(s)⟶I_2(g) ΔH_sub = + 62.42 KJ

Heat of combustion is used
(i) to calculate heat of formation
(ii) to find the calorific value of food and fuel

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

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

(i) Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
(ii) We know that \(\Delta H\) = q_p (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, \(\Delta H_r.\)
(iii) In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and  \(\Delta H_r\)will also be negative.
(iv) Similarly, in an endothermic reaction, heat is absorbed, q_p is positive and \(\Delta H_r\) will also be positive.

The required equation is
C(s) + 2S(s) \(\rightarrow\) CS₂(l); \(\Delta\)H= ?
Given: C(s) + O₂(g) \(\rightarrow\) CO₂(g); \(\Delta\)H = -393.3 kJ  .....(1)
S(s) + O₂(g) \(\rightarrow\) SO₂(g); \(\Delta\)H = -293.7 kJ  .....(2)
CS₂(l) + O₂(g) \(\rightarrow\) CO₂(g) + SO₂(g); \(\Delta\) H = -1108.76 kJ .....(3)
First method: Multiply equation (2) by 2
2S(s) + 2O₂(g) \(\rightarrow\) 2SO₂(g)\(\rightarrow\) 2SO₂ (g); \(\Delta\)H = -587.44
Adding equations (10 and (4) and subtracting (3),
C(s) + 2S(s) + 3O₂(g) - CS₂(l) - 3O₂(g) \(\rightarrow\) CO₂(g) + 2SO₂(g)- CO₂(g) - 2SO₂(g)
C(s) + 2S(s) \(\rightarrow\) CS₂(l)
Thus,\(\Delta\)H_f = -393.3 - 587.44 + 1108.76 = 128.02 kJ
i.e., standard heat offormation of CS₂ is 128.02 kJ
Second method: Standard enthalpy of formation of
CO₂ = -393.3 kJ
SO₂ = -293.72 kJ
CS₂ =?
From the first law of thermodynamics, the standard enthalpy of a compound is equal to the standard enthalpy of formation of the compound and the standard enthalpies of elements is equal to zero.
The equation is CS₂(l) + 3SO₂(g) \(\rightarrow\) CO₂(g) + 2SO₂(g); \(\Delta\)H° = -1108.76 kJ
\(\Delta\)H°=\(\sum { H°(products)-\sum { H°(reactants) } } \)
={\(\Delta\)H°_f(CO₂) + 2H°_f(SO₂)} - {\(\Delta\)H°_f(CS₂) + 3H°_f(O₂)}-1108.76 = {-393.3 + 2 \(\times\) (-296.72)}-\(\Delta\)H°_f(CS2₎
\(\Delta\)H°_f(CS₂) =+ 1108.76 - 393.3 - 2 \(\times\)(297.72)
= 1108.76 - 980.74 = 128.02 kJ
\(\therefore\) \(\Delta\)H°_f(CS₂) = 128.02 kJ mol^-1
Bond energies or Bond enthalpies:
The bond dissociation energy of a diatomic molecule is also called bond energy.
When a molecule or a compound contains more than one bond of the same kind, the average value of the dissociation energies of a given bond is taken. This average bond dissociation energy required to break each bond in a compound is called bond energy.
Using bond energy data, heat of reaction is (\(\Delta\) H) can be calculated by using the equation.
\(\Delta H=\sum { B.E\ of\ (reactants)-\sum { B.E\ of\ (products) } } \)
Alternating, of enthalpy change for the reaction and bond energies of all the bonds except one, the bond energy of that bond can be calculated.

ΔH_t = 2090 Jmol^-1
T_t = 13+273 = 286K
ΔS_t = \(\frac { { \triangle H }_{ t } }{ { T }_{ t } } \)
ΔS_t = \(\frac { 2090 }{ 286 } \)
ΔS_t = 7.307 JK^-1mol^-1​​​​​​​