Thermodynamics

In a cyclic process, net heat absorbed = work done $=$ area enclosed in $p V$ diagram.

Area $=(3-1) \mathrm{atm} \times(4-2) \mathrm{L}$

$=2 \mathrm{~atm} \times 2 \mathrm{~L}=4 \mathrm{~L} \mathrm{~atm}$.

Now, convert to energy units if needed,

$4 \mathrm{Latm} \times 101.3 \mathrm{~J}=\sim 405 \mathrm{~J}$ (optional)

But in $\mathrm{L} \cdot \mathrm{atm}$, answer is 4 L atm

Let, $A_2=x$, then $B_2=4 x, A B=2 x$

$ \begin{aligned} \Delta H & =\left[\frac{1}{2} A_2+\frac{1}{2} B_2 \longrightarrow A B\right] \\ & =\left(\frac{1}{2} x+2 x\right)-2 x=\frac{1}{2} x=-100 \rightarrow x=-200 \end{aligned} $

So, $B_2=4 x=-800 \mathrm{~kJ} / \mathrm{mol}$

In a cyclic process, the change in internal energy ($\Delta U$) is zero. According to the first law of thermodynamics:

$ \Delta U = \Delta q + \Delta W $

Since $\Delta U = 0$, it follows that:

$ \Delta q = -\Delta W $

From the graph, the work done ($\Delta W$) is equal to the area under the curve. In this case, the process involves a circular path, so the work done can be calculated as the area of the circle:

$ \text{Area of circle} = \pi r^2 $

Given that the radius ($r$) of the circle is half the difference between the maximum and minimum values on the graph,

$ r = \frac{25 - 5}{2} = 10 $

Thus, the area is:

$ \pi \times (10)^2 = \pi \times 100 $

Therefore, the heat absorbed ($\Delta q$) is:

$ \Delta q = 100 \pi \, \text{J} $

To determine the temperature $K$ at which $\Delta G = 0$ for a given reaction, we use the relationship between the Gibbs free energy change ($\Delta G$), enthalpy change ($\Delta H$), and entropy change ($\Delta S$), expressed by the Gibbs free energy equation:

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

Given that $\Delta G = 0$, we set the equation as follows:

$ 0 = \Delta H - T \Delta S $

Solving for the temperature $T$, we get:

$ T = \frac{\Delta H}{\Delta S} $

Now, we can plug in the given values:

$ \Delta H = -20.5 \ \mathrm{~kJ} \ \mathrm{~mol}^{-1} = -20500 \ \mathrm{~J} \ \mathrm{~mol}^{-1} $

$ \Delta S = -50.0 \ \mathrm{~JK}^{-1} \ \mathrm{~mol}^{-1} $

Substituting these values into the equation:

$ T = \frac{-20500 \ \mathrm{~J} \ \mathrm{~mol}^{-1}}{-50.0 \ \mathrm{~JK}^{-1} \ \mathrm{~mol}^{-1}} $

$ T = \frac{20500}{50.0} $

$ T = 410 \ \mathrm{K} $

Therefore, the temperature $K$ at which $\Delta G = 0$ is 410 K. The correct answer is:

Option B

410

To determine the relationship between the enthalpy change ($\Delta H$) and the internal energy change ($\Delta E$) for the given reaction, we need to examine the process and consider the relevant thermodynamic equations.

The reaction given is the combustion of carbon monoxide:

$ 2\text{CO} + \text{O}_2 \rightarrow 2\text{CO}_2 $

We know that this reaction occurs in a closed vessel, which implies a constant volume process. The relationship between enthalpy change and internal energy change for a reaction can be described by the equation:

$ \Delta H = \Delta E + \Delta n_gRT $

Where:

• $\Delta H$ is the enthalpy change
• $\Delta E$ is the internal energy change
• $\Delta n_g$ is the change in the number of moles of gases
• $R$ is the universal gas constant
• $T$ is the temperature in Kelvin

To determine $\Delta n_g$, we need to look at the number of moles of gas before and after the reaction:

• Before the reaction: 2 moles of CO and 1 mole of $\text{O}_2$, totaling 3 moles of gas.
• After the reaction: 2 moles of $\text{CO}_2$.

Hence, $\Delta n_g$ can be calculated as:

$ \Delta n_g = n_{\text{final}} - n_{\text{initial}} = 2 - 3 = -1 $

Substituting $\Delta n_g = -1$ into the enthalpy-internal energy relationship:

$ \Delta H = \Delta E + (-1)RT $

$ \Delta H = \Delta E - RT $

Since $RT$ is always a positive quantity, it means that:

$ \Delta H < \Delta E $

Therefore, the correct answer is:

Option B: $\Delta H < \Delta U$

To find the value of the internal energy change $(\Delta U)$ for the given reaction at $27^{\circ} \mathrm{C}$, we can use the relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$, given by the equation:

$\Delta H = \Delta U + \Delta n \cdot R \cdot T$

Where:

• $\Delta H$ is the enthalpy change
• $\Delta U$ is the internal energy change
• $\Delta n$ is the change in the number of moles of gas
• $R$ is the gas constant, which is $8.314~\mathrm{J~mol^{-1}~K^{-1}}$
• $T$ is the temperature in Kelvin

First, we need to determine $\Delta n$, the change in the number of moles of gas during the reaction:

For the reaction:

$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+ 3 \mathrm{H}_2 \mathrm{O}(l)$

Moles of gas on the reactant side = 3 (from $\mathrm{O}_2$)

Moles of gas on the product side = 2 (from $\mathrm{CO}_2$)

So,

$\Delta n = \text{{moles of gas on product side}} - \text{{moles of gas on reactant side}} = 2 - 3 = -1$

Now, we can convert the temperature to Kelvin:

$T = 27^{\circ} \mathrm{C} + 273 = 300~\mathrm{K}$

Next, we substitute the values into the equation:

$\Delta H = \Delta U + \Delta n \cdot R \cdot T$

Here,

$\Delta H = -1366.5~\mathrm{kJ~mol^{-1}} = -1366.5 \cdot 10^3~\mathrm{J~mol^{-1}}$

$\Delta n = -1$

$R = 8.314~\mathrm{J~mol^{-1}~K^{-1}}$

$T = 300~\mathrm{K}$

Solve for $\Delta U$:

$-1366.5 \cdot 10^3 = \Delta U + (-1) \cdot 8.314 \cdot 300$

Simplify the equation:

$-1366.5 \cdot 10^3 = \Delta U - 2494.2$

So,

$\Delta U = -1366.5 \cdot 10^3 + 2494.2$

$\Delta U \approx -1366.5 \cdot 10^3 + 2.494 \cdot 10^3$

$\Delta U \approx -1364.0 \cdot 10^3~\mathrm{J~mol^{-1}}$

Therefore, the internal energy change $\Delta U$ is:

$\Delta U \approx -1364.0~\mathrm{kJ~mol^{-1}}$

Hence, the correct option is:

Option C: $-1364.0~\mathrm{kJ~mol^{-1}}$

To determine the temperature at which the reaction is at equilibrium, we need to consider the Gibbs free energy change ($\Delta G$) for the reaction. At equilibrium, $\Delta G = 0$.

The Gibbs free energy change is given by:

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

Where:

$\Delta H$ = change in enthalpy

$T$ = temperature in Kelvin

$\Delta S$ = change in entropy

Given that the standard entropies are:

$S^\circ_{X_2} = 60 \ \text{JK}^{-1}\text{mol}^{-1}$

$S^\circ_{Y_2} = 40 \ \text{JK}^{-1}\text{mol}^{-1}$

$S^\circ_{XY_3} = 50 \ \text{JK}^{-1}\text{mol}^{-1}$

The standard entropy change ($\Delta S$) for the reaction is:

$\Delta S = S^\circ_{\text{products}} - S^\circ_{\text{reactants}}$

The reaction is:

$\frac{1}{2} X_2 + \frac{3}{2} Y_2 \to XY_3$

So, the change in entropy can be calculated as:

$\Delta S = S^\circ_{XY_3} - \left( \frac{1}{2} S^\circ_{X_2} + \frac{3}{2} S^\circ_{Y_2} \right)$

Substituting the given values:

$\Delta S = 50 - \left( \frac{1}{2} \cdot 60 + \frac{3}{2} \cdot 40 \right)$

$\Delta S = 50 - \left( 30 + 60 \right)$

$\Delta S = 50 - 90$

$\Delta S = -40 \ \text{JK}^{-1}\text{mol}^{-1}$

Given that $\Delta H = -30 \ \text{kJ} = -30000 \ \text{J}$

To find the temperature at which the reaction is at equilibrium, we set $\Delta G = 0$:

$\Delta G = \Delta H - T \Delta S = 0$

Therefore,

$T = \frac{\Delta H}{\Delta S}$

Substituting the values:

$T = \frac{-30000 \ \text{J}}{-40 \ \text{J}^{-1}\text{K}^{-1}}$

$T = \frac{30000}{40}$

$T = 750 \ \text{K}$

Therefore, the temperature at which the reaction is at equilibrium is 750 K.

Answer: Option C, 750 K

Free expansion, W = 0

Adiabatic process, q = 0

$\Delta$U = q + W = 0, this means that internal energy remains constant. Therefore, $\Delta$T = 0 in ideal gas there is no intermolecular attraction.

Hence, when such a gas expands under adiabatic conditions into a vacuum, no heat is absorbed or evolved, since no external work is done to separate the molecules.

The SI unit of Boltzmann's constant is JK^$-$1

${C_2}{H_5}OH(I) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(I)$

$\Delta U = - 1364.47$ kJ/mol,

$ \Rightarrow \Delta {n_g} = - 1$

T = 25$^\circ$C = 298 K

$\therefore$ $\Delta H = \Delta U + \Delta {n_g}RT$

$\Delta H = - 1364.47 + {{ - 1 \times 8314 \times 298} \over {1000}}$

[Here, value of R in unit of J must be converted into kJ]

$ = - 1364.47 - 2.4776 = - 1366.95$ kJ/mol

(a) $\Delta$G = $\Delta$H $-$ T $\Delta$S

For a system, total entropy change is $\Delta$S_total,

$\Delta$H_total = 0

$\therefore$ $\Delta$G_system = $-$T$\Delta$S_total

$\therefore$ ${{\Delta {G_{system}}} \over {\Delta {S_{total}}}} = - T$

Thus, (a) is true,

(b) For isothermal reversible process, $\Delta$E = 0.

By first law of thermodynamics,

$\Delta$E = q $\times$ W

$\therefore$ W_reversible = $-$q = $-$$\int_{{V_t}}^{{V_f}} {p\,dV} $

W_reversible = $ - nRT\ln {{{V_f}} \over {{V_t}}}$

Thus, (b) is true.

(c) $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ $

$\Delta G^\circ = - nRT\ln \,K = - RT\ln K$ (for n = 1)

$\therefore$ $ - RT\ln K = \Delta H^\circ - T\Delta S^\circ $

$\therefore$ $\ln K = - \left( {{{\Delta H^\circ - T\Delta S^\circ } \over {RT}}} \right)$

Thus, (c) is incorrect.

(d) $\Delta G^\circ = - RT\ln K$

$\therefore$ $\ln K = - {{\Delta G^\circ } \over {RT}}$ $\therefore$ $K = {e^{ - \Delta G^\circ /RT}}$

Thus, (d) is also true.

Statement (a) is correct. Rest of the all statements are incorrect. These are explained as follows:

(a) Cl₂(g) $\to$ 2Cl(g)

As randomness is increasing. Therefore, $\Delta$S is positive for this reaction.

$\therefore$ Statement is correct.

(b) In closed container, $\Delta$V = 0, hence work done is zero. There is no heat exchange. Hence, $\Delta$E = q + W = 0.

$\therefore$ Statement is incorrect.

(c) $\Delta$G will be zero only when equilibrium is reached.

$\therefore$ Statement is incorrect.

(d) $\Delta$G$^\circ$ = $-$RT ln K_eq, not a function of pressure. Thus, this statement is also incorrect.

At top of a mountain,

T₁ = 0 + 273 K, ${p_1} = {{710} \over {760}}$ atm and density = $\rho$₁

At bottom of mountain,

T₂ = 303 K, p₂ = 1 atm and density = $\rho$₂

$\because$ $pV = {w \over M}RT$ or $p = \rho {{RT} \over M}$

$\therefore$ ${{{\rho _1}} \over {{\rho _2}}} = {{{p_1}} \over {{T_1}}} \times {{{T_2}} \over {{p_2}}}$

or, ${{{\rho _1}} \over {{\rho _2}}} = {{710/760 \times 303} \over {273 \times 1}} = {{1.04} \over 1}$

Heat change at constant pressure means enthalpy change ($\Delta$_rH = q_p).

Heat of formation of CO is written as:

$C(s) + {1 \over 2}{O_2}(g) \to CO(g)$; ${\Delta _f}{H^\Theta }$ = ?

Now, we have,

I. $C(g) + {O_2}(g) \to C{O_2}(g)$; ${\Delta _c}H_1^\Theta $ = $-$393.5 kJ/mol

II. $CO(g) + {1 \over 2}{O_2}(g) \to C{O_2}(g)$; ${\Delta _c}H_2^\Theta $ = $-$ 283 kJ/mol

Substracting appropriately (i.e. I-II), we get,

$C(s) + {1 \over 2}{O_2}(g) \to CO(g)$; ${\Delta _f}H_{}^\Theta $

From Hess's law, ${\Delta _f}H_{}^\Theta = {\Delta _c}H_1^\Theta - {\Delta _c}H_2^\Theta $

$ \Rightarrow {\Delta _f}H_{}^\Theta = - 393.5 - ( - 283) = 110.5$ kJ/mol

Now, calculation of the heat of formation at constant volume means that we have to calculate change in internal energy (i.e. $\Delta$_rU)

Using, ${\Delta _r}H = {\Delta _r}U + p\Delta V$ [For a chemical reaction]

$ = {\Delta _r}U + \Delta {n_g}RT$ [As pV = nRT]

[$\Delta$n_g = gaseous moles of products $-$ gaseous moles of reactant]

$ \Rightarrow {\Delta _r}U = {\Delta _r}H - \Delta {n_g}RT$

Now, putting the values : $\Delta {n_g} = 1 - {1 \over 2} = {1 \over 2}$, T = 298 K,

R = 8.314 J/K/mol and ${\Delta _f}{H^\Theta }$ = $-$110.5 kJ/mol

$ \Rightarrow {\Delta _f}{H^\Theta } = - 110.5 - {1 \over 2} \times 8.314 \times 2.98 \times {10^{ - 3}}$

= $-$111.7 kJ/mol