Thermodynamics

$\begin{aligned} & \begin{aligned} \therefore \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{MX}) & =\Delta \mathrm{H}_{\text {sub }}(\mathrm{M})+\text { I.E. }(\mathrm{M})+\frac{1}{2}[\text { B.E. }(\mathrm{X}-\mathrm{X})] \\ & +\mathrm{EG}(\mathrm{X})+\text { L.E. }(\mathrm{MX}) \end{aligned} \\ & -400=(100)+(500)+\frac{1}{2}(\text { B.E. })+(-300)+(-800)\\ & \therefore \text { B.E. }=200 \mathrm{~kJ} \mathrm{~mole}^{-1} \end{aligned} ~ \begin{aligned} \end{aligned}$

$\Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=55 \mathrm{~kJ} / \mathrm{mol}, \quad \mathrm{T}=298 \mathrm{~K}$

$\begin{aligned} & \Delta \mathrm{S}_{\mathrm{rxn}}^{\mathrm{o}}=175 \mathrm{~J} / \mathrm{mol} \\ & \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=\Delta \mathrm{H}_{\mathrm{rxn}}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{~S}_{\mathrm{rxn}}^{\mathrm{o}} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000 \mathrm{~J} / \mathrm{mol}-298 \times 175 \mathrm{~J} / \mathrm{mol} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000-52150 \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=2850 \mathrm{~J} / \mathrm{mol} \end{aligned}$

$ \begin{aligned} & \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \Delta \mathrm{H}_1^{\circ} =-1550 \ldots \text { (i) } \\\\ & \mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta \mathrm{H}_2^{\circ}=-393.5 \ldots \text { (ii) } \\\\ & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \Delta \mathrm{H}_3^{\circ}=-286 .......(iii) \end{aligned} $

$\begin{aligned} & \text { From } 2 \times \mathrm{eq}^{\mathrm{n}}(\mathrm{ii})+3 \times \mathrm{eq}^{\mathrm{n}}(\mathrm{iii})-\mathrm{eq}^{\mathrm{n}}(\mathrm{i}) \\\\ & 2 \mathrm{C}_{\text {(graphite) }}+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}): \Delta \mathrm{H}_{f \mathrm{C}_2 \mathrm{H}_6}^{\circ} \\\\ & \left(\Delta \mathrm{H}_f^{\circ}\right)_{\mathrm{C}_2 \mathrm{H}_6}=2 \times(-393.5)+3 \times(-286)-(-1550) \\\\ & =-95 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$

To find the temperature of vaporization at one atmosphere, we can use the Clausius-Clapeyron equation, which relates the enthalpy of vaporization ($\Delta H_{vap}$) to the change in entropy ($\Delta S_{vap}$) during the phase transition at a particular temperature (T). The relationship can be simplified under the assumption that both the enthalpy of vaporization and the entropy change of vaporization are constant with temperature to the form:

$\Delta H_{vap} = T \cdot \Delta S_{vap}$

This equation states that the enthalpy change of vaporization is equal to the product of the temperature at which the phase change occurs and the entropy change of vaporization. We rearrange this equation to solve for the temperature (T):

$T = \frac{\Delta H_{vap}}{\Delta S_{vap}}$

However, it's crucial to ensure the units are consistent. Given that $\Delta H_{vap}$ is in kJ/mol and $\Delta S_{vap}$ is in J/(mol·K), we need to convert $\Delta H_{vap}$ from kJ/mol to J/mol to match units:

$\Delta H_{vap} = 30 \, \text{kJ/mol} = 30 \times 10^3 \, \text{J/mol}$

Substituting the given values into the equation, we obtain:

$T = \frac{30 \times 10^3 \, \text{J/mol}}{75 \, \text{J/(mol·K)}}$

$T = \frac{30000 \, \text{J/mol}}{75 \, \text{J/(mol·K)}}$

$T = 400 \, \text{K}$

Therefore, the temperature of vaporization at one atmosphere is 400 K.

To solve this problem, we need to understand the concept of neutralization reactions and the heat evolved during these reactions.

When an acid and a base react, they undergo a neutralization reaction to produce water and a salt. The heat released in this process is known as the enthalpy of neutralization.

Consider the neutralization of hydrochloric acid (HCl) and sulfuric acid (H₂SO₄) by sodium hydroxide (NaOH). The balanced chemical equations for these neutralizations are:

$ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} $

$ \mathrm{H_2SO_4} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2SO_4} + 2\mathrm{H_2O} $

For the first reaction, each mole of HCl reacts with one mole of NaOH, releasing a certain amount of heat (let's denote this amount by $ x $ kJ). For the second reaction, each mole of H₂SO₄ reacts with two moles of NaOH. Since we are given equal volumes and molarities of HCl and H₂SO₄, we can infer that one mole of H₂SO₄ will produce twice the heat of one mole of HCl because it produces double the amount of water.

Thus, the heat evolved in the neutralization of H₂SO₄ by NaOH (denoted as $ y $ kJ) will be approximately twice that of HCl. Therefore, $ y = 2x $.

Therefore, the value of $\frac{y}{x}$ is:

$ \frac{y}{x} = \frac{2x}{x} = 2 $

So, the value of $\frac{y}{x}$ is 2.

(I) $\mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4$ Solution $\Delta \mathrm{H}=-70 \mathrm{~kJ}$

(II) $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4$ Solution $\mathrm{\Delta H=12 \mathrm{~kJ}}$

(I) - (II)

$\begin{aligned} & \mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} \\ & \Delta \mathrm{H}=-70-12=-82 \end{aligned}$

To find the change in internal energy for the vaporization of water under the given conditions, we'll use the following relationship between enthalpy change ($\Delta_{\text{vap}} H$) and internal energy change ($\Delta_{\text{vap}} U$):

$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \Delta V$

For vaporization, the change in volume ($\Delta V$) can be approximated by considering the volume of the vapor because the volume of liquid water is relatively small compared to the volume of the vapor.

The ideal gas law gives us:

$P V = n R T$

Since we are dealing with 1 mole of water:

$V = \frac{R T}{P}$

Plugging this into the enthalpy change equation, we get:

$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \left( \frac{R T}{P} \right)$

This simplifies to:

$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + R T$

Rearranging for $\Delta_{\text{vap}} U$:

$\Delta_{\text{vap}} U = \Delta_{\text{vap}} H - R T$

Given:

$\Delta_{\text{vap}} H = 40.79 \, \text{kJ mol}^{-1}$ (or 40790 J/mol)

$R = 8.3 \, \text{JK}^{-1} \text{mol}^{-1}$

$T = 100^\circ \text{C} + 273.15 = 373.15 \, \text{K}$

Now, substitute the values into the equation:

$\Delta_{\text{vap}} U = 40790 \, \text{J mol}^{-1} - (8.3 \, \text{JK}^{-1} \text{mol}^{-1} \times 373.15 \, \text{K})$

$\Delta_{\text{vap}} U = 40790 \, \text{J mol}^{-1} - 3097.145 \, \text{J mol}^{-1}$

$\Delta_{\text{vap}} U = 37692.855 \, \text{J mol}^{-1}$

Converting back to kJ:

$\Delta_{\text{vap}} U = 37.69 \, \text{kJ mol}^{-1}$

Rounding to the nearest integer, the change in internal energy for the vaporization of water under the given conditions is:

38 kJ mol^-1

$\begin{aligned} & \mathrm{V}_1=100 \mathrm{~L} \\ & \mathrm{~V}_2=10 \mathrm{~L} \\ & \mathrm{~W}=-\mathrm{nR} \operatorname{Tl} \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\ & =-1 \times 0.08206 \times 291.15 \times 2.303 \log \frac{10}{100} \\ & =55 \mathrm{~L} \text { atm } \\ & \end{aligned}$

To determine the temperature above which the reaction $2A+B \rightarrow C$ becomes spontaneous, we can use the Gibbs free energy equation:

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

The reaction becomes spontaneous when $\Delta G$ is negative. Therefore, we need to find the temperature at which $\Delta G$ changes from positive to negative. We set $\Delta G$ to zero to find the threshold temperature:

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

Substituting the given values of $\Delta H = 400 \, \text{kJ mol}^{-1} = 400,000 \, \text{J mol}^{-1}$ (since 1 kJ = 1000 J) and $\Delta S = 0.2 \, \text{kJ mol}^{-1} K^{-1} = 200 \, \text{J mol}^{-1} K^{-1}$, we get:

$0 = 400,000 \, \text{J mol}^{-1} - T(200 \, \text{J mol}^{-1} K^{-1})$

Solving for $T$, we have:

$T = \frac{400,000 \, \text{J mol}^{-1}}{200 \, \text{J mol}^{-1} K^{-1}}$

$T = 2000 \, \text{K}$

Therefore, the reaction will become spontaneous above $2000 \, \text{K}$. This means that at temperatures higher than 2000 K, the reaction tends towards product formation without the need for external energy to drive the process.

$-1\left(2 V_1-V_1\right)=n \times \frac{5 R}{2}\left(T_2-T_1\right)$

$\begin{aligned} & -\mathrm{V}_1=\frac{5}{2}\left(n R T_2-5 \mathrm{~V}_1\right) \\ & -\mathrm{V}_1=2.5\left(\mathrm{nRT_{2 } )}-12.5 \mathrm{~V}_1\right. \\ & 11.5 \mathrm{~V}_1=2.5\left(\mathrm{nRT_{2 } )}\right. \\ & 11.5 \times \frac{n R T_1}{P_1}=2.5 \times\left(n R T_2\right) \\ & \mathrm{T}_2=274.16 \mathrm{~k} \\ & \approx 274 \text { (Nearest integer) } \\ & \end{aligned}$

To determine the standard enthalpy of combustion of 2 moles of benzene, we need to use the standard enthalpy of formation values provided and apply Hess's Law. Here is a step-by-step explanation:

Given Data:

1. Standard enthalpy of formation of benzene ($C_6H_6(l)$):

$ \Delta H_f(\text{C}_6\text{H}_6(l)) = 48.5 \, \text{kJ/mol} $

1. Standard enthalpy of formation of carbon dioxide ($CO_2(g)$):

$ \Delta H_f(\text{CO}_2(g)) = -393.5 \, \text{kJ/mol} $

1. Standard enthalpy of formation of water ($H_2O(l)$):

$ \Delta H_f(\text{H}_2O(l)) = -286 \, \text{kJ/mol} $

Combustion Reaction for Benzene:

$ \text{C}_6\text{H}_6(l) + \frac{15}{2} \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 3 \text{H}_2O(l) $

Enthalpy Change Calculation:

Using Hess's Law, the enthalpy change for the reaction can be calculated as follows:

$ \Delta H_{\text{comb}} = \left[ 6 \Delta H_f(\text{CO}_2(g)) + 3 \Delta H_f(\text{H}_2O(l)) \right] - \Delta H_f(\text{C}_6\text{H}_6(l)) $

Substitute the given values:

$ \Delta H_{\text{comb}} = \left[ 6 \times (-393.5) + 3 \times (-286) \right] - 48.5 $

Perform the calculations:

$ \Delta H_{\text{comb}} = \left[ 6 \times (-393.5) \right] + \left[ 3 \times (-286) \right] - 48.5 $

$ \Delta H_{\text{comb}} = \left[ -2361 \right] + \left[ -858 \right] - 48.5 $

$ \Delta H_{\text{comb}} = -3267.5 \, \text{kJ/mol} $

This value is the enthalpy change for the combustion of 1 mole of benzene.

For 2 Moles of Benzene:

$ \Delta H_{\text{comb (2 moles)}} = 2 \times (-3267.5 \, \text{kJ/mol}) $

$ \Delta H_{\text{comb (2 moles)}} = -6535 \, \text{kJ} $

Conclusion:

The standard enthalpy of combustion of 2 moles of benzene is

$ x = 6535 \, \text{kJ} $

Thus, $ x = 6535 $.

$\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{s})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 7 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{I})$

$\begin{aligned} & \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_9 R T \\ & \Delta \mathrm{H}=-321.30-\frac{1}{2} \mathrm{R} \times 300 \end{aligned}$

So, $x=150$

For isothermal process

$\Rightarrow \begin{aligned} & Q=-W \\ & Q=-5 \times 40 \\ & |Q|=+200 \text { Lit atm } \end{aligned}$

$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_4+\mathrm{H}_2 \longrightarrow \mathrm{C}_2 \mathrm{H}_6 \\ & \begin{aligned} \Delta \mathrm{H} & =(615)+(435)-(347)-2(414) \\ & =615+435-347-828 \\ & =-125 \mathrm{~kJ} \end{aligned} \end{aligned}$

To determine the change in standard Gibbs free energy ($\Delta G^{\circ}$) for the reaction at a given temperature when the equilibrium constant ($K$) is known, we can use the following relationship:

$ \Delta G^{\circ} = -RT \ln K $

Here, $R$ is the universal gas constant ($8.314 \ J K^{-1} mol^{-1}$), $T$ is the temperature in Kelvin ($300 \ K$), and $K$ is the equilibrium constant.

Puting the values we get,

$ \Delta G^{\circ} = -8.314 \times 300 \times \ln(10) $

We can convert the natural logarithm of 10 to base-10 logarithm, using the change of base formula, $\ln(10) = 2.3026$. So,

$ \Delta G^{\circ} = -8.314 \times 300 \times 2.3026 $

Calculating this yields:

$ \Delta G^{\circ} = -8.314 \times 300 \times 2.3026 = -5743.914 \ J mol^{-1} $

To convert this to kilojoules per mole, we divide by 1000:

$ \Delta G^{\circ} = -5.743914 \times 10^{3} \ J mol^{-1} \times \frac{1 \ kJ}{10^{3} J} = -5.743914 \ kJ mol^{-1} $

Now, when expressing this in terms of $\times 10^{-1} \ kJ mol^{-1}$, we get:

$ \Delta G^{\circ} = -57.43914 \times 10^{-1} \ kJ mol^{-1} $

Therefore, $\Delta G^{\circ}$ for the reaction is approximately $-57.43914 \times 10^{-1} \ kJ mol^{-1}$ or $-57.4 \times 10^{-1} \ kJ mol^{-1}$ when rounded to three significant figures.

It is isothermal reversible expansion, so work done negative

$\begin{aligned} & \mathrm{W}=-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right) \\ & =-2.303 \times 5 \times 8.314 \times 300 \log \left(\frac{100}{10}\right) \\ & =-28720.713 \mathrm{~J} \\ & \equiv-28721 \mathrm{~J} \end{aligned}$

$\begin{aligned} & \frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29} . \\ & \Delta_{\mathrm{r}} \mathrm{G}^{\Theta}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{P}} \\ & =-8.314 \times 10^{-3} \times 298 \times \ln \left(2.47 \times 10^{-29}\right) \\ & =-8.314 \times 10^{-3} \times 298 \times(-65.87) \\ & =163.19 \mathrm{~kJ} \end{aligned}$

$2 \mathrm{Fe}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}, \Delta \mathrm{H}^{\circ}=-822 \mathrm{~kJ} / \mathrm{mol}$ ........ (1)

$\mathrm{C}_{(\mathrm{s})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}^{\circ}=-110 \mathrm{~kJ} / \mathrm{mol}$ ........ (2)

$3 \mathrm{C}_{(\mathrm{s})}+\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}_3=\text { ? }$

$\begin{aligned} & (3)=3 \times(2)-(1) \\ & \begin{aligned} \Delta \mathrm{H}_3 & =3 \times \Delta \mathrm{H}_2-\Delta \mathrm{H}_1 \\ & =3(-110)+822 \\ & =492 \mathrm{~kJ} / \mathrm{mole} \end{aligned} \end{aligned}$

Work done is given by area enclosed in the P vs V cyclic graph or V vs P cyclic graph.

Sign of work is positive for clockwise cyclic process for V vs P graph.

$\begin{aligned} & W=\frac{1}{2} \times(30-10) \times(30-10)=200 \mathrm{~kPa}-\mathrm{dm}^3 \\ & =200 \times 1000 \mathrm{~Pa}-\mathrm{L}=2 \mathrm{~L}-\mathrm{bar}=200 \mathrm{~J} \end{aligned}$

$\begin{aligned} & \Delta \mathrm{H}_{\text {vap }}^0 \mathrm{CCl}_4=30.5 \mathrm{~kJ} / \mathrm{mol} \\ & \text { Mass of } \mathrm{CCl}_4=284 \mathrm{~gm} \\ & \text { Molar mass of } \mathrm{CCl}_4=154 \mathrm{~g} / \mathrm{mol} \\ & \text { Moles of } \mathrm{CCl}_4=\frac{284}{154}=1.844 \mathrm{~mol} \\ & \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1 \mathrm{~mole}=30.5 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1.844 \mathrm{~mol}=30.5 \times 1.844 \\ & \quad=56.242 \mathrm{~kJ} \end{aligned}$

$\begin{aligned} & \Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\ & =77.2 \times 10^3-400 \times 122=28400 \mathrm{~J} \\ & \Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K} \\ & \Rightarrow 28400=-2.303 \times 8.314 \times 400 \log \mathrm{K} \\ & \Rightarrow \log \mathrm{K}=-3.708=-37.08 \times 10^{-1} \end{aligned}$

The process involves an ideal gas expanding isothermally, meaning temperature ($T$) remains constant. In an isothermal process for an ideal gas, the change in internal energy ($\Delta U$) is zero:

$\Delta U = 0$ because temperature is constant.

According to the first law of thermodynamics, $\Delta U = Q + W$, where $Q$ is the heat transferred to the system and $W$ is the work done by the system. Since $\Delta U = 0$ in an isothermal process, $Q = -W$.

The work done by the gas during the isothermal expansion against a constant external pressure ($P_{ext}$) is given by:

$W = P_{ext} \times \Delta V$

where:

• $P_{ext}$ = constant opposing pressure = $80$ kPa = $80 \times 10^3$ Pa (since $1$ kPa = $10^3$ Pa)
• $\Delta V$ = change in volume = $45$ dm³ - $30$ dm³ = $15$ dm³ = $15 \times 10^{-3}$ m³ (since $1$ dm³ = $10^{-3}$ m³)

$W = -80 \times 10^3 \times 15 \times 10^{-3} = -1200$ J

Therefore, the amount of heat transferred ($Q$) is $1200$ J, taking into consideration the sign convention that work done by the system is negative.

The heat required to melt one mole of a substance is known as its molar enthalpy of fusion, denoted as $\Delta H_\mathrm{fus}$. For sodium chloride, $\Delta H_\mathrm{fus}$ is $30.4\ \mathrm{kJ/mol}$.

The entropy change at the melting point is given by $\Delta S = \frac{\Delta H_\mathrm{fus}}{T_\mathrm{m}}$, where $T_\mathrm{m}$ is the melting point in Kelvin. Substituting the given values, we get:

$28.4\ \mathrm{J/K/mol} = \frac{30.4\ \mathrm{kJ/mol}}{T_\mathrm{m}}$

Solving for $T_\mathrm{m}$, we get:

$T_\mathrm{m} = \frac{30.4\ \mathrm{kJ/mol}}{28.4\ \mathrm{J/K/mol}} = 1070.4\ \mathrm{K}$

Rounding off to the nearest integer, the melting point of sodium chloride is $\boxed{1070\ \mathrm{K}}$.

To find the bond enthalpy of $\mathrm{A}_{2}$, we can use the given information about the reaction and the bond enthalpies' ratio. The reaction is:

$\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}$

The enthalpy change for the reaction, $\Delta H_{f}^{0}$, is given as:

$\Delta H_{f}^{0} = -200 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The bond enthalpies of $\mathrm{A}_{2}$, $\mathrm{B}_{2}$, and $\mathrm{AB}$ are in the ratio of $1: 0.5: 1$.

Let's denote the bond enthalpies of $\mathrm{A}_{2}$, $\mathrm{B}_{2}$, and $\mathrm{AB}$ as $x$, $0.5x$, and $x$, respectively.

The enthalpy change for the reaction can be calculated using the bond enthalpies:

$\Delta H_{f}^{0} = \text{(Sum of bond enthalpies of reactants)} - \text{(Sum of bond enthalpies of products)}$

For the given reaction:

$-200 = (x + 0.5x) - 2x$

Now we can solve for $x$, which represents the bond enthalpy of $\mathrm{A}_{2}$:

$-200 = 1.5x - 2x$

$-200 = -0.5x$

$x = \frac{-200}{-0.5}$

$x = 400$

The bond enthalpy of $\mathrm{A}_{2}$ is $400 ~\mathrm{kJ} ~\mathrm{mol}^{-1}$.

For an isothermal, reversible process, the change in entropy (ΔS) of the system can be calculated using the formula:

$ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) $

where (n) is the number of moles, (R) is the gas constant, and (V_2) and (V_1) are the final and initial volumes, respectively.

Substituting the given values:

$ \Delta S_{\text{system}} = 1 \times 8.314 \, \ln\left(\frac{3}{2}\right) = 3.37 \, \text{J K}^{-1} $

Since the process is reversible and the total entropy change in the universe should be zero for reversible processes, the change in entropy of the surroundings is equal to the negative change of the system's entropy:

$ \Delta S_{\text{surroundings}} = - \Delta S_{\text{system}} = -3.37 \, \text{J K}^{-1} $

But considering the heat transfer from the system to the surroundings, the sign should be positive, which means the entropy of the surroundings also increases:

$ \Delta S_{\text{surroundings}} = 3.37 \, \text{J K}^{-1} $

So, rounded to the nearest integer,

$ \Delta S_{\text{surroundings}} = 3 \, \text{J K}^{-1} $

Intensive properties are independent of the amount of the substance. They don't change with the size or mass of a sample. Examples include temperature, pressure, density, and concentration.

Intensive properties :

1. Molarity : Molarity is a measure of concentration, defined as the number of moles of solute per liter of solution. It doesn't depend on the total volume of the solution but only on the ratio of the amount of solute to the amount of solution.

2. $\mathrm{E}^{\theta}$ cell (standard cell potential) : This is a measure of the potential difference between the anode and cathode in an electrochemical cell under standard conditions. It's independent of the amount of material in the cell.

3. Molar heat capacity : This is the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius (or Kelvin). It's an intrinsic property because it's defined per mole of substance.

4. Molar mass : Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's independent of the quantity of the substance because it's defined per mole.

On the other hand, extensive properties depend on the amount of the substance. They change when the size or mass of a sample changes. For instance, mass and volume are extensive properties.

Extensive properties :

1. Volume: The volume of a substance depends on the amount of that substance. For instance, one liter of water has half the volume of two liters of water.

2. Gibbs free energy change (∆G) : This is the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure. It depends on the number of moles of reactants and products, so it's an extensive property.

3. Mole : The mole is a unit of measurement for amount of substance in the International System of Units (SI). It's a count of a very large number of particles, typically atoms or molecules. Like mass or volume, it's an extensive property because it depends on the quantity of the substance.

First, let's consider the reaction :

$2\mathrm{Al}(s) + \mathrm{Fe}_2\mathrm{O}_3(s) \rightarrow \mathrm{Al}_2\mathrm{O}_3(s) + 2\mathrm{Fe}(s)$

The heat change for this reaction $\Delta H^0$ can be calculated from the heats of formation of the reactants and the products :

$\Delta H^0 = [\Delta H_f^0(\mathrm{Al}_2\mathrm{O}_3) + 2\Delta H_f^0(\mathrm{Fe})] - [2\Delta H_f^0(\mathrm{Al}) + \Delta H_f^0(\mathrm{Fe}_2\mathrm{O}_3)]$

Assuming the elements in their standard states have zero enthalpy of formation, i.e.,

$\Delta H_f^0(\mathrm{Al}) = \Delta H_f^0(\mathrm{Fe}) = 0$, we can simplify this to :

$\Delta H^0 = \Delta H_f^0(\mathrm{Al}_2\mathrm{O}_3) - \Delta H_f^0(\mathrm{Fe}_2\mathrm{O}_3)$

Substitute the given heats of formation into the equation :

$\Delta H^0 = (-1700\ \mathrm{kJ/mol}) - (-840\ \mathrm{kJ/mol}) = -860\ \mathrm{kJ/mol}$

This is the heat of reaction for the above reaction. However, we are asked to find the heat evolved per gram of the mixture.

To find this, we need to determine the molar mass of the reactants in the reaction. The molar mass of $\mathrm{Fe}_2\mathrm{O}_3$ is $2 \times 56 + 3 \times 16 = 160\ \mathrm{g/mol}$ and the molar mass of 2 moles of $\mathrm{Al}$ is $2 \times 27 = 54\ \mathrm{g/mol}$. The total molar mass of the mixture is $160 + 54 = 214\ \mathrm{g/mol}$.

So, the heat evolved per gram of the mixture is $\frac{-860\ \mathrm{kJ/mol}}{214\ \mathrm{g/mol}} = -4.0187\ \mathrm{kJ/g}$

Rounded to the nearest integer, this value is approximately -4 kJ/g. The negative sign indicates that the heat is evolved (exothermic reaction).

An endothermic process is one that absorbs heat from the surroundings. Here are the types of the given reactions:

A. $\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{I}(\mathrm{g})$

This process involves the dissociation of iodine molecules into individual iodine atoms. This requires energy to break the bonds between the iodine atoms, so it is an endothermic process.

B. $\mathrm{HCl}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})$

This is the dissociation of hydrogen chloride into hydrogen and chlorine. Similar to the first reaction, it requires energy to break the bond between the hydrogen and chlorine atoms. Therefore, this is also an endothermic process.

C. $\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

This is the process of evaporation or vaporization, where liquid water is converted to water vapor. Evaporation is an endothermic process because it requires heat to break the hydrogen bonds between water molecules and convert it into a gas.

D. $\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})$

This is a combustion reaction. Combustion reactions are generally exothermic because they release energy in the form of heat and light. So, this is not an endothermic process.

E. Dissolution of ammonium chloride in water

The dissolution of ammonium chloride ($\mathrm{NH}_4\mathrm{Cl}$) in water is known to be endothermic. The ionic bonds in the ammonium chloride and the hydrogen bonds in the water must be broken for the salt to dissolve, which requires heat.

So, in summary, four of these processes (A, B, C, and E) are endothermic.

In the given combustion reaction, the number of moles of gaseous products ($\Delta n_g$) is -2 (since we have 2 moles of CO₂ gas on the product side and 4 moles of gaseous reactants).

The change in internal energy ($\Delta U$) is given as -1406 kJ/mol.

We can calculate the change in enthalpy ($\Delta H$) using the equation $\Delta H = \Delta U + \Delta n_gRT$, where $R$ is the ideal gas constant and $T$ is the temperature.

Substituting the given values:

$ \begin{aligned} \Delta H &= \Delta U + \Delta n_gRT \\\\ &= -1406 \mathrm{kJ/mol} + (-2) \times 8.3 \times 10^{-3} \mathrm{kJ/K/mol} \times 300 \mathrm{K} \\\\ &= -1406 \mathrm{kJ/mol} - 4.98 \mathrm{kJ/mol} \\\\ &\approx -1411 \mathrm{kJ/mol} \end{aligned} $

At equilibrium, $\Delta G = 0$, so $\Delta H = T \Delta S$, which gives $T \Delta S = \Delta H$.

Therefore, the minimum value of $T \Delta S$ needed to reach equilibrium is -1411 kJ

The heat provided by the heater is given by the equation:

$Q = \text{Power} \times \text{Time}$

Substituting the given values:

$Q = 60 \, \text{W} \times 100 \, \text{s} = 6000 \, \text{J}$

The heat capacity (C) is defined as the amount of heat required to raise the temperature of a substance by one degree. It is given by the equation:

$C = Q/\Delta T$

Substituting the given values:

$C = 6000 \, \text{J} / 5 \, \text{°C} = 1200 \, \text{J/K}$

So, the heat capacity of the given gas is approximately 1200 J/K.

The reaction for the formation of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$ is given by:

$2\mathrm{C}(\mathrm{s}) + 6\mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$

The heat of combustion for $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$ is:

$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$

By adding the formation reaction with the combustion reaction times -1, we get:

$2\mathrm{C}(\mathrm{s}) + 3\mathrm{H}_{2}(\mathrm{g}) + \frac{3}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$

This reaction has a heat equal to the heat of formation of ethanol times -1.

Therefore, the heat of formation of ethanol is equal to:

$-1234.7 \mathrm{~kJ/mol} + 2*(-393.5 \mathrm{~kJ/mol}) + 3*(-241.8 \mathrm{~kJ/mol})$

Calculating this gives:

$\Delta H_f(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})) = -277.7 \approx -278 \mathrm{~kJ/mol}$ (rounded to the nearest integer)

For, Spontaneous process $\mathrm{dG}<0$

For, Equilibrium $\mathrm{dG}=0$

For, Nonspontaneous process $\mathrm{dG}>0$

$\therefore $ A Wrong

B Correct

C Correct

D Wrong

Given:

$ \begin{align} \Delta H^0 & = -54.07 \, \text{kJ/mol} = -54070 \, \text{J/mol} \\\\ \Delta S^0 & = 10 \, \text{J/K}\cdot \text{mol} \\\\ T & = 298 \, \text{K} \end{align} $

We find the change in Gibbs free energy $\Delta G^0$:

$ \begin{align} \Delta G^0 & = \Delta H^0 - T \Delta S^0 \\\\ & = -54070 - (10 \times 298) \\\\ & = -54070 - 2980 \\\\ & = -57050 \, \text{J/mol} \end{align} $

Now, we'll use the given expression with the correct constant for $ R $ as 8.314 J/(mol·K):

$ \begin{align} \Delta G^0 & = -2.303 \times 8.314 \times 298 \times \log K \\\\ \log K & = \frac{-57050}{2.303 \times 8.314 \times 298} \\\\ & \approx 10 \end{align} $

So, the answer is $\log K = 10$.

$\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\ell)$

$\begin{aligned} & \text { No. of moles of ethane }=\frac{0.3}{30}=0.01 \\\\ & \text { Heat evolved in Bomb calorimeter }=20 \times 0.5 \\\\ & =10 \mathrm{~kJ} \\\\ & \Delta \mathrm{U}=-\frac{10}{0.01}=-1000 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ & =-1000+(-2.5) \times \frac{8.3 \times 300}{1000} \\\\ & =-1000-6.225 \\\\ & =-1006.225 \\\\ & |\Delta \mathrm{H}| \simeq 1006 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\ & \end{aligned}$

$\frac{\text { (i) }+\text { (iii) }}{2}-$ (ii) gives desired reaction

$ \begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}=\frac{436+78}{2}-(-242) \\\\ & =\frac{436+78}{2}+242=499 \end{aligned} $

$ \mathrm{CCl}_4(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{HCl}(\mathrm{g}) $

Enthalpy of above reaction

$ \begin{aligned} & =\Delta \mathrm{H}_f\left(\mathrm{CO}_2(\mathrm{~g})\right)+4 \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HCl}(\mathrm{g}))-\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{CCl}_4\right)-2 \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}\right) \\\\ & =-394-4 \times 92+105+2 \times 242 \\\\ & =-394-368+105+484 \\\\ & =-173 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $

Hence the magnitude of this will be $173 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

$ \begin{aligned} & \frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{Cl}_{\text {(aq) }}^{-} \quad \Delta \mathrm{H}=? \\\\ & \begin{aligned} \Delta \mathrm{H} & =\frac{1}{2} \Delta_{\mathrm{diss}} \mathrm{H}_{\mathrm{Cl}_2}^{\circ}+\Delta_{\mathrm{eg}} \Delta \mathrm{H}_{\mathrm{Cl}(\mathrm{g})}^{\circ}+\Delta_{\mathrm{hyd}} \mathrm{H}_{\mathrm{Cl}_{(\mathrm{g})}^{-}}^{\circ} \\\\ & =\frac{1}{2} \times 240+(-350)+(-380) \\\\ & =-610 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned} $

$\mathrm{T_1=300~K}$

$\mathrm{w = 3}$ kJ/mole

$\mathrm{w=nC_v\Delta T}$

$3000=1\times20\times(300-\mathrm{T_2})$

$300-\mathrm{T_2}=150$

$\mathrm{T_2=150~K}$

For isothermal process of an ideal gas; $\mathrm{\Delta E=0}$

Let, Volume of C₂H₄ is x litre

Total volume of $\mathrm{CO}_2=2 \mathrm{x}+16.8-\mathrm{x}$

$ \begin{aligned} &\Rightarrow 28=16.8+\mathrm{x} \\\\ &\mathrm{x}=11.2 \mathrm{~L} \\\\ &\mathrm{n}_{\mathrm{CH}_4}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{1 \times 5.6}{0.082 \times 298}=0.229 \text { mole } \\\\ &\mathrm{n}_{\mathrm{C}_2 \mathrm{H}_2}=\frac{11.2}{0.082 \times 298}=0.458 \text { mole } \\\\ &\therefore \text { Heat evolved }=0.229 \times 900+0.458 \times 1400 \\\\ &=206.1+641.2 \\\\ &=847.3 \mathrm{~kJ} \end{aligned} $

$ \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) $

Extra energy used to convert $\mathrm{H}_2 \mathrm{O(l)}$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$

$ \begin{aligned} & =\frac{1800}{2}=900 \mathrm{~kJ} \\\\ & \Rightarrow 900=\mathrm{n}_{\mathrm{H}_2 \mathrm{O}} \times 45 \\\\ & \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{900}{45}=20 \text { mole } \\\\ & \mathrm{W}_{\mathrm{H}_2 \mathrm{O}}=20 \times 18=360 \mathrm{~g} \end{aligned} $

$1 \rightarrow 2 \Rightarrow$ Isobaric process

$2 \rightarrow 3 \Rightarrow$ Isochoric process

$3 \rightarrow 1 \Rightarrow$ Isothermal process

$\mathrm{W}=\mathrm{W}_{1 \rightarrow 2}+\mathrm{W}_{2 \rightarrow 3}+\mathrm{W}_{3 \rightarrow 1}$

$=\left(-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)+0\left[-\mathrm{P}_1 \mathrm{~V}_1 \ln \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\right]\right)$

$=\left[-1 \times(40-20)+0+\left[-1 \times 20 \ln \left(\frac{20}{40}\right)\right]\right]$

$=-20+20 \ln 2$

$=-20+20 \times 2.3 \times 0.3$

$=-6.2$ bar $L$

$|\mathrm{W}|=6.2$ bar $1=620 \mathrm{~J}$

A. $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}>\mathrm{T}_{4}$

B. It is incorrect as particles do not undergo simple harmonic motion.

C. It is correct

D. It is incorrect

E. It is correct

$ \begin{aligned} & \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\\\ & \mathrm{A}: \Delta \mathrm{G}\left(\mathrm{J} \mathrm{mol}^{-1}\right)=-25 \times 10^3+80 \times 300:-\mathrm{ve} \\\\ & \mathrm{B}: \Delta \mathrm{G}\left(\mathrm{J} \mathrm{mol}^{-1}\right)=-22 \times 10^3-40 \times 300:-\mathrm{ve} \\\\ & \mathrm{C}: \Delta \mathrm{G}\left(\mathrm{J} \mathrm{mol}^{-1}\right)=25 \times 10^3+300 \times 50:+\mathrm{ve} \\\\ & \mathrm{D}: \Delta \mathrm{G}\left(\mathrm{J} \mathrm{mol}^{-1}\right)=22 \times 10^3-20 \times 300:+\mathrm{ve} \end{aligned} $

Processes $\mathrm{C}$ and $\mathrm{D}$ are non-spontaneous.