Thermodynamics

Magnitude of work:

• Reversible compression = reversible expansion (maximum)

• Irreversible compression > irreversible expansion
  
  

Using ΔH = ΔU + ΔnRT, with ΔU = -3263.9 kJ/mol, Δn = -1.5, R = 8.314 × 10⁻³ kJ/mol·K, and T = 298 K,

$$\mathrm{\Delta }H=-3263.9-3.717=-3267.6\text{ kJ/mol}$$

Correct answer: (D).

$$\begin{gathered} {\mathrm{\Delta } \mathrm{H}}_{\mathrm{r}}^{\ominus }=2(710 )+2\times 436-611-4\left(414\right) \\ = 1420 + 872 - 611 - 1656 \\ = 25 \mathrm{kJ} {\mathrm{mole}}^{-1} \end{gathered}$$

“ΔG is positive for a spontaneous reaction” is incorrect.
For spontaneity ΔG < 0; a positive ΔG indicates non-spontaneity.

The given reaction is an endothermic process that is non-spontaneous at the freezing point of water but becomes spontaneous at the boiling point of water. This indicates that the process follows Gibbs free energy equation

$\Delta G=\Delta H-T\Delta S$

It is only possible when $\Delta H$ and $\Delta S$ both are positive.

ΔH < 0 ⇒ dissolution of HCl(g) is exothermic.

HCl(g) + 10H₂O(l) → HCl·10${\mathrm{H}}_2\mathrm{O}$
ΔH₁ = −69.01 kJ mol⁻¹

HCl(g) + 40${\mathrm{H}}_2\mathrm{O}$(l) → HCl·40${\mathrm{H}}_2\mathrm{O}$
ΔH₂ = −72.79 kJ mol⁻¹

Heat of dilution
= ΔH₂ − ΔH₁
= −72.79 − (−69.01)
= −3.78 kJ mol⁻¹

⇒ Heat of solution depends on amount of solvent.
⇒ Option (3) is incorrect.

For heat of formation, reactants must be in elemental state ⇒ Option (4) is incorrect.

(a) Using Hess's Law, we sum the given reactions:
1. C (diamond) $\to C$ (graphite) + X
2. C (diamond) $+O_2\to C{O}_2+Y$
3. C (graphite) $+O_2\to C{O}_2+Z$

Reversing (3):

$C{O}_2\to C\left(\text{ graphite }\right)+O_2-Z$

Adding to (2):
C(diamond) → C(graphite) + (Y–Z)
Comparing with (1): X = Y – Z

If $∆\mathrm{H} >0, ∆\mathrm{S}>0$ then process is spontaneous at high temperature
If $∆\mathrm{H} >0, ∆\mathrm{S}<0$ process is always non spontaneous
If $∆\mathrm{H} <0, ∆\mathrm{S}>0$, then process is always spontaneous
If $∆\mathrm{H} <0, ∆\mathrm{S}<0$, then process is spontaneous at low temperature.

• Energy supplied (heat), $Q = 48 \, \text{J}$Q=48J

• Gas: monoatomic, $n = 1 \, \text{mol}$n=1mol

• Temperature change: \Delta T = 2\,^\circ \text{C} = 2 \, \text{K}

We are asked to find work done, W.

For any process:

$$Q = \Delta E + W$$Q=ΔE+W

Where:

• $\Delta E = n C_V \Delta T$ΔE=nCV​ΔT (for a monoatomic ideal gas, $C_V = \frac{3}{2} R$CV​=23​R)

• $W =$W= work done by gas

• $R = 8.314 \, \text{J/mol·K}$R=8.314J/mol\cdotpK

$$C_V = \frac{3}{2} R = \frac{3}{2} \cdot 8.314 \approx 12.471 \, \text{J/mol·K}$$CV​=23​R=23​⋅8.314≈12.471J/mol\cdotpK $$\Delta E = n C_V \Delta T = 1 \cdot 12.471 \cdot 2 \approx 24.942 \, \text{J} \approx 25 \, \text{J}$$ΔE=nCV​ΔT=1⋅12.471⋅2≈24.942J≈25J

$$W = Q - \Delta E = 48 - 24.942 \approx 23 \, \text{J}$$W=Q−ΔE=48−24.942≈23J

In an adiabatic process, where no heat is exchanged, the relationship between temperature (T) and volume (V) for an ideal gas is given by:
T V^{(γ - 1)} = constant, Here, γ (gamma) Adiabatic constant of gas.

T2 =546 K

• Energy supplied (heat), $Q = 48 \, \text{J}$Q=48J

• Gas: monoatomic, $n = 1 \, \text{mol}$n=1mol

• Temperature change: \Delta T = 2\,^\circ \text{C} = 2 \, \text{K}

We are asked to find work done, W.

For any process:

$$Q = \Delta E + W$$Q=ΔE+W

Where:

• $\Delta E = n C_V \Delta T$ΔE=nCV​ΔT (for a monoatomic ideal gas, $C_V = \frac{3}{2} R$CV​=23​R)

• $W =$W= work done by gas

• $R = 8.314 \, \text{J/mol·K}$R=8.314J/mol\cdotpK

$$C_V = \frac{3}{2} R = \frac{3}{2} \cdot 8.314 \approx 12.471 \, \text{J/mol·K}$$CV​=23​R=23​⋅8.314≈12.471J/mol\cdotpK $$\Delta E = n C_V \Delta T = 1 \cdot 12.471 \cdot 2 \approx 24.942 \, \text{J} \approx 25 \, \text{J}$$ΔE=nCV​ΔT=1⋅12.471⋅2≈24.942J≈25J

$$W = Q - \Delta E = 48 - 24.942 \approx 23 \, \text{J}$$W=Q−ΔE=48−24.942≈23J

$$\begin{gathered} \mathrm{The} \mathrm{desired} \mathrm{reaction} \mathrm{is} \mathrm{C}\left(\mathrm{diamond}\right) \to \mathrm{C}\left(\mathrm{graphite}\right) + \mathrm{x} \mathrm{KJ}-----\left(1\right) \\ \mathrm{Now} \mathrm{C}\left(\mathrm{diamond}\right) +{\mathrm{O}}_2\to {\mathrm{CO}}_2+ \mathrm{y} \mathrm{KJ}---------\left(2\right) \\ \mathrm{C}\left(\mathrm{Graphite}\right) +{\mathrm{O}}_2\to {\mathrm{CO}}_2+ \mathrm{z} \mathrm{KJ}---------\left(3\right) \\ \mathrm{By} \mathrm{substraction} \mathrm{equation} \left(3\right) \mathrm{from} \left(2\right) \mathrm{we} \mathrm{get} \mathrm{equation} \left(1\right) \\ \mathrm{C}\left(\mathrm{diamond}\right) \to \mathrm{C}\left(\mathrm{graphite}\right) + \mathrm{x} \mathrm{KJ} \\ \mathrm{Now} \mathrm{x}=\mathrm{y}-\mathrm{z} \mathrm{KJ} \end{gathered}$$.

Given:
Insulated thermally means q = 0

Acc. to 1st law,

$$\begin{gathered} ∆\mathrm{U} = \mathrm{q} + \mathrm{w} \\ ∆\mathrm{U} = 0 + \mathrm{w}= \mathrm{w} \\ \mathrm{w} > 0, ∆\mathrm{U} > 0 \end{gathered}$$

The standard enthalpy of formation $\left({∆}_{\mathrm{f}}{\mathrm{H}}^{\mathrm{\theta }}\right)$ for an element in its most stable form at any specified temperature is defined as zero. Oxygen's most stable form is the diatomic gas, O₂(g), at all temperature where it remains a gas.

$$\begin{gathered} \mathrm{The} \mathrm{desired} \mathrm{reaction} \mathrm{is} \mathrm{C}\left(\mathrm{diamond}\right) \to \mathrm{C}\left(\mathrm{graphite}\right) + \mathrm{x} \mathrm{KJ}-----\left(1\right) \\ \mathrm{Now} \mathrm{C}\left(\mathrm{diamond}\right) +{\mathrm{O}}_2\to {\mathrm{CO}}_2+ \mathrm{y} \mathrm{KJ}---------\left(2\right) \\ \mathrm{C}\left(\mathrm{Graphite}\right) +{\mathrm{O}}_2\to {\mathrm{CO}}_2+ \mathrm{z} \mathrm{KJ}---------\left(3\right) \\ \mathrm{By} \mathrm{substraction} \mathrm{equation} \left(3\right) \mathrm{from} \left(2\right) \mathrm{we} \mathrm{get} \mathrm{equation} \left(1\right) \\ \mathrm{C}\left(\mathrm{diamond}\right) \to \mathrm{C}\left(\mathrm{graphite}\right) + \mathrm{x} \mathrm{KJ} \\ \mathrm{Now} \mathrm{x}=\mathrm{y}-\mathrm{z} \mathrm{KJ} \end{gathered}$$.

(a) Using Hess's Law, we sum the given reactions:
1. C (diamond) $\to C$ (graphite) + X
2. C (diamond) $+O_2\to C{O}_2+Y$
3. C (graphite) $+O_2\to C{O}_2+Z$

Reversing (3):

$C{O}_2\to C\left(\text{ graphite }\right)+O_2-Z$

Adding to (2):
C(diamond) → C(graphite) + (Y–Z)
Comparing with (1): X = Y – Z

If $∆\mathrm{H} >0, ∆\mathrm{S}>0$ then process is spontaneous at high temperature
If $∆\mathrm{H} >0, ∆\mathrm{S}<0$ process is always non spontaneous
If $∆\mathrm{H} <0, ∆\mathrm{S}>0$, then process is always spontaneous
If $∆\mathrm{H} <0, ∆\mathrm{S}<0$, then process is spontaneous at low temperature.

Negative value of electron gain enthalpy indicates the amount of energy released when an electron is added to an isolated gaseous system.

More negative value of electron gain enthalpy means large amount of energy is released and it also signifies greater ionic character.

B has most negative electron gain enthalpy hence it will form a compound with greater ionic character with E followed by A , D and C.

The given reaction is an endothermic process that is non-spontaneous at the freezing point of water but becomes spontaneous at the boiling point of water. This indicates that the process follows Gibbs free energy equation

$\Delta G=\Delta H-T\Delta S$

It is only possible when $\Delta H$ and $\Delta S$ both are positive.

The standard enthalpy of formation $\left({∆}_{\mathrm{f}}{\mathrm{H}}^{\mathrm{\theta }}\right)$ for an element in its most stable form at any specified temperature is defined as zero. Oxygen's most stable form is the diatomic gas, O₂(g), at all temperature where it remains a gas.

The given reactions are:

$$\begin{gathered} S\left(g\right)+\frac{3}{2}{O}_2\left(g\right)\to S{O}_3\left(g\right)+2xkcal \\ S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to S{O}_3\left(g\right)+ykcal \end{gathered}$$

The heat of formation of $S{O}_2\left(g\right)$ can be derived by subtracting the heat of the second reaction from the first:

$\Delta H_{\text{formation }}\left(S{O}_2\right)=y-2x$

Thus, the enthalpy of formation of $S{O}_2$ is $y-2x$.

For reaction to be spontaneous according to 2^nd law:

ΔG < 0

ΔH − TΔS < 0

$\mathrm{T} > \left(\frac{\mathrm{\Delta H}}{\mathrm{\Delta S}}\right)= \mathrm{Te}$

T > Te

Using ΔG° = -RT ln Kc, with Kc = 10¹², R = 8.314 J/mol·K, T = 298 K, we get ΔG° = -68.47 kJ/mol. Correct answer: (C).

Negative value of electron gain enthalpy indicates the amount of energy released when an electron is added to an isolated gaseous system.

More negative value of electron gain enthalpy means large amount of energy is released and it also signifies greater ionic character.

B has most negative electron gain enthalpy hence it will form a compound with greater ionic character with E followed by A , D and C.

$$\text{Moles of octane} = \frac{1.14}{114} = 0.01 \text{ mol}$$

$$\text{Heat evolved} = C \times \Delta T = 5 \times 5 = 25 \text{ kJ}$$

$$\begin{gathered} \text{Magnitude of heat of combustion}=\frac{25}{0.01} \\ =2500{\text{ kJ mol}}^{-1} \end{gathered}$$

Using ΔG° = -RT ln Kc, with Kc = 10¹², R = 8.314 J/mol·K, T = 298 K, we get ΔG° = -68.47 kJ/mol. Correct answer: (C).

For reaction to be spontaneous according to 2^nd law:

ΔG < 0

ΔH − TΔS < 0

$\mathrm{T} > \left(\frac{\mathrm{\Delta H}}{\mathrm{\Delta S}}\right)= \mathrm{Te}$

T > Te

Hess's Law Formula
$\Delta H_f^{\theta }\left(H_{2}O\right)=$ Bond enthalpies of reactants - Bond enthalpies of products

$-242.0=D(H-H)$$+\frac{1}{2}D(O=O)-2D(O-H)$

Substituting Known Values

$$\begin{gathered} -242.0=440+\frac{1}{2}\left(500\right)-2D(O-H) \\ -242.0=440+250-2D(O-H) \\ -242.0=690-2D(O-H) \end{gathered}$$

Solving for D ( O - H )

$$\begin{gathered} 2D(O-H)=690+242 \\ 2D(O-H)=932 \\ D(O-H)=\frac{932}{2}=466kJ/mol \end{gathered}$$

For closed insulated container q =0

$$\begin{gathered} ∆\mathrm{E} =\mathrm{q}+\mathrm{w} \\ ∆\mathrm{E} = \mathrm{W} \end{gathered}$$

Given:
Insulated thermally means q = 0

Acc. to 1st law,

$$\begin{gathered} ∆\mathrm{U} = \mathrm{q} + \mathrm{w} \\ ∆\mathrm{U} = 0 + \mathrm{w}= \mathrm{w} \\ \mathrm{w} > 0, ∆\mathrm{U} > 0 \end{gathered}$$

(a) Using Hess's Law, we sum the given reactions:
1. C (diamond) $\to C$ (graphite) + X
2. C (diamond) $+O_2\to C{O}_2+Y$
3. C (graphite) $+O_2\to C{O}_2+Z$

Reversing (3):

$C{O}_2\to C\left(\text{ graphite }\right)+O_2-Z$

Adding to (2):
C(diamond) → C(graphite) + (Y–Z)
Comparing with (1): X = Y – Z

Argon is a monoatomic ideal gas

At constant pressure:

Q = nCₚΔT

For monoatomic gas:

Cₚ = (5/2)R

Q = n × (5/2)R × ΔT

500 = 0.5 × (5/2) × 8.3 × ΔT

(5/2) × 8.3 = 20.75

500 = 0.5 × 20.75 × ΔT
500 = 10.375 × ΔT

ΔT = 500 / 10.375 ≈ 48 K

T₂ = 298 + 48 = 346 K ≈ 348 K

For monoatomic gas:

ΔU = n × (3/2)R × ΔT

= 0.5 × (3/2 × 8.3) × 48

= 0.5 × 12.45 × 48
≈ 300 J

Final Answer: 348 K and 300 J

If $∆\mathrm{H} >0, ∆\mathrm{S}>0$ then process is spontaneous at high temperature
If $∆\mathrm{H} >0, ∆\mathrm{S}<0$ process is always non spontaneous
If $∆\mathrm{H} <0, ∆\mathrm{S}>0$, then process is always spontaneous
If $∆\mathrm{H} <0, ∆\mathrm{S}<0$, then process is spontaneous at low temperature.

ΔH < 0 ⇒ dissolution of HCl(g) is exothermic.

HCl(g) + 10H₂O(l) → HCl·10${\mathrm{H}}_2\mathrm{O}$
ΔH₁ = −69.01 kJ mol⁻¹

HCl(g) + 40${\mathrm{H}}_2\mathrm{O}$(l) → HCl·40${\mathrm{H}}_2\mathrm{O}$
ΔH₂ = −72.79 kJ mol⁻¹

Heat of dilution
= ΔH₂ − ΔH₁
= −72.79 − (−69.01)
= −3.78 kJ mol⁻¹

⇒ Heat of solution depends on amount of solvent.
⇒ Option (3) is incorrect.

For heat of formation, reactants must be in elemental state ⇒ Option (4) is incorrect.

$- \mathrm{q} = \mathrm{w} =- \mathrm{nRT} \ln \left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$

$=- 8.3\times 300\times \ln \left(\frac{20}{10}\right)$

= − 1.718 kJ

q = 1.718 kJ

w = − 1.718 kJ

ΔU = 0 ( ∵ ΔT = 0 )

Using ΔH = ΔU + ΔnRT, with ΔU = -3263.9 kJ/mol, Δn = -1.5, R = 8.314 × 10⁻³ kJ/mol·K, and T = 298 K,

$$\mathrm{\Delta }H=-3263.9-3.717=-3267.6\text{ kJ/mol}$$

Correct answer: (D).

$- \mathrm{q} = \mathrm{w} =- \mathrm{nRT} \ln \left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$

$=- 8.3\times 300\times \ln \left(\frac{20}{10}\right)$

= − 1.718 kJ

q = 1.718 kJ

w = − 1.718 kJ

ΔU = 0 ( ∵ ΔT = 0 )

$$\begin{gathered} \mathrm{For} \mathrm{monoatomic} \mathrm{gas} \mathrm{\gamma } =5/3 \\ \mathrm{As} \mathrm{we} \mathrm{know} \\ \mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{∆\mathrm{H}}{∆\mathrm{U}} (\mathrm{at} \mathrm{constant} \mathrm{P}, \mathrm{H} =\mathrm{Q} ) \\ \mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{∆\mathrm{Q}}{∆\mathrm{U}}, \mathrm{hence} \frac{∆\mathrm{Q}}{∆\mathrm{U}}=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=5/3 \end{gathered}$$

Magnitude of work:

• Reversible compression = reversible expansion (maximum)

• Irreversible compression > irreversible expansion
  
  

The given reactions are:

$$\begin{gathered} S\left(g\right)+\frac{3}{2}{O}_2\left(g\right)\to S{O}_3\left(g\right)+2xkcal \\ S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to S{O}_3\left(g\right)+ykcal \end{gathered}$$

The heat of formation of $S{O}_2\left(g\right)$ can be derived by subtracting the heat of the second reaction from the first:

$\Delta H_{\text{formation }}\left(S{O}_2\right)=y-2x$

Thus, the enthalpy of formation of $S{O}_2$ is $y-2x$.

Option A is correct. The combustion reaction of benzene is:

C6H6(l) + 15/2 O2(g) → 6CO2(g) + 3H2O(l)

Using Hess’s law:

ΔHcombustion = ΣΔHf(products) − ΣΔHf(reactants)

Substituting the given values:

−3267 = [6(−393.5) + 3(−286)] − [ΔHf(C6H6)]

−3267 = [−2361 − 858] − ΔHf(C6H6)

−3267 = −3219 − ΔHf(C6H6)

ΔHf(C6H6) = −3219 + 3267 = 48 kJ mol⁻¹

Therefore, the heat of formation of benzene is 48 kJ mol⁻¹.

Extensive properties are mass-dependent and intensive properties are independent of the amount of mass.

Mass , Volume, and internal energy are all extensive properties

For Isochoric process $\mathrm{P} \mathrm{\alpha } \mathrm{T} \mathrm{at} \mathrm{constant} \mathrm{V}$

$$\begin{gathered} \mathrm{Work} \mathrm{done} \mathrm{in} \mathrm{isochoric} \mathrm{process} \mathrm{W} =-\mathrm{P}∆\mathrm{V}, ∆\mathrm{V} =0 , \mathrm{hence} \mathrm{W}= 0 \\ \mathrm{Now} \mathrm{internal} \mathrm{energy} ∆\mathrm{U} ={\mathrm{nC}}_{\mathrm{v}}∆\mathrm{T}, \mathrm{Hence} \mathrm{as} \mathrm{temperature} \mathrm{increse} , \mathrm{internal} \mathrm{energy} \mathrm{increses} \end{gathered}$$

Enthalpy of neutralization of SA & SB is always $-57 \mathrm{kJ}/\mathrm{mol}$ because strong monoacid gives one mole of ${\mathrm{H}}^{+}$and strong mono base gives one mole of${\mathrm{OH}}^{-}$ which form one mole of water.

Enthalpy of neutralization of SA & SB is always $-57 \mathrm{kJ}/\mathrm{mol}$ because strong monoacid gives one mole of ${\mathrm{H}}^{+}$and strong mono base gives one mole of${\mathrm{OH}}^{-}$ which form one mole of water.