Thermodynamics

Given, temperature $ =300 \mathrm{~K} $, number of moles $ (n)=3 $

Change in volume $ (\Delta V)=-\frac{1}{2}(V) $ where

$ V $ is initial volume and external pressure, $ p_{\text {ext }}=6 \mathrm{~atm} $.

Pressure at 300 K of gas, $ p=3 \mathrm{~atm} $

Work done in isothermal compression is given by

$ \begin{array}{l} W=-p_{\mathrm{ext}} \Delta V \\\\ W=-p_{\mathrm{ext}}\left(-\frac{V}{2}\right) \end{array} $

Using ideal gas equation $ p V=n R T $,

$ V=\frac{n R T}{p} $

Substitute the value of $ V $ in Eq. (ii)

$ W=p_{\mathrm{ext}} \frac{(n R T)}{2 p} $

Substitute all values in Eq. (iv)

$ \begin{array}{l} W=\frac{6 \times 3 \times 0.082 \times 300}{2 \times 3} \times 101.3 \mathrm{~J} \\\\ W=7475.94 \mathrm{~J} \text { or } W=7.476 \mathrm{~kJ} \end{array} $

Given values:

$\Delta_{f} H^{\ominus}$ of $AO(s) = -635 \, \text{kJ/mol}$

$\Delta_{f} H^{\ominus}$ of $BO_{2}(g) = x \, \text{kJ/mol}$

$\Delta_{f} H^{\ominus}$ of $ABO_{3}(s) = -1210 \, \text{kJ/mol}$

$\Delta_{r} H^{\ominus} = 175 \, \text{kJ/mol}$ for the reaction:

$ ABO_{3}(s) \longrightarrow AO(s) + BO_{2}(g) $

The enthalpy change for the reaction, $\Delta_{r} H^{\ominus}$, is calculated using the formula:

$ \Delta_{r} H^{\ominus} = \left[ \Delta_{f} H^{\ominus}(AO) + \Delta_{f} H^{\ominus}(BO_{2}) \right] - \Delta_{f} H^{\ominus}(ABO_{3}) $

Substitute the known values:

$ 175 = [-635 + x] - [-1210] $

Simplifying:

$ 175 = x + 1210 - 635 $

$ 175 = x + 575 $

$ x = 175 - 575 $

$ x = -400 \, \text{kJ/mol} $

Therefore, the value of $x$ is $-400 \, \text{kJ/mol}$.

According to given data

$ \begin{array}{l} \mathrm{C}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \\\\ \Delta H_{C}=-393 \mathrm{~kJ} / \mathrm{mol} \\\\ \mathrm{H}_{2}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \\\\ \Delta H_{C}=-286 \mathrm{~kJ} / \mathrm{mol} \end{array} $

Multiply Eq. (ii) by (2),

$ \begin{array}{r} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \\\\ \Delta H_{C}=-572 \mathrm{~kJ} / \mathrm{mol} . \end{array} $

For combustion of $ \mathrm{CH}_{3} \mathrm{OH}(l) $

$ \begin{array}{r} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\\\ \Delta H_{C}=-726 \mathrm{~kJ} / \mathrm{mol} \\\\ \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\frac{3}{2} \mathrm{O}_{2} \end{array} $

$ \Delta H_{C}=+726 \mathrm{~kJ} / \mathrm{mol} $

Add Eqs. (i), (iii) and (iv), we get

$ \begin{aligned} \mathrm{C}(\mathrm{~s})+2 \mathrm{H}_{2}(g) & +\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH} \\\\ \Delta H_{C} & =-393+(-572)-(-726) \\\\ & =-239 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Given,

Initial volume $\left(V_i\right)=1.0 \mathrm{~L}$

Final volume $\left(V_f\right)=10.0 \mathrm{~L}$

$n$ (number of moles) $=1$

Temperature $=61 \mathrm{~K}$

Now, workdone for isothermal process is given by

$ W=-n R T \times 2.303 \log _{10}\left(\frac{V_f}{V_i}\right) $

Substitute all the values,

$ \begin{aligned} W & =-2.303 \times 1 \times 0.0821 \times 61 \times \log _{10}\left(\frac{10}{1}\right) \\\\ & =-11.52 \mathrm{~L} \mathrm{~atm} \end{aligned} $

Given reaction is

$ \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \longrightarrow 2 \mathrm{NH}_3 $

Two moles of ammonia are formed during the reaction. Thus, the reaction for formation of one mole of ammonia is given

$ \Delta_f H^{\ominus}=-46.2 \mathrm{~kJ} / \mathrm{mol} $

The standard enthalpy of reaction on is

$ \begin{aligned} \Delta_r H^{\ominus} & =2 \times \Delta_f H^{\ominus} \\ & =2 \times-46.2 \\ & =-92.4 \mathrm{~kJ} \end{aligned} $

For each of the given reactions, we will analyze how entropy is affected:

Reaction I:

$ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) $

The change in the number of gas molecules ($\Delta n_g$) is calculated as follows:

$ \Delta n_g = (0 + 1) - 0 = 1 $

Since $\Delta n_g > 0$, this indicates an increase in entropy because the production of gas molecules increases disorder.

Reaction II:

$ \text{Cl}_2(g) \rightarrow 2\text{Cl}(g) $

The change in the number of gas molecules is:

$ \Delta n_g = 2 - 1 = 1 $

Here, $\Delta n_g > 0$, pointing to an increase in entropy because more gas molecules are produced, enhancing disorder.

Reaction III:

$ \text{H}_2 \text{O}(l) \rightarrow \text{H}_2 \text{O}(s) $

This reaction involves water transitioning from a liquid to a solid state. As water solidifies into ice, the molecular arrangement becomes more ordered, resulting in a decrease in entropy.

Therefore, entropy increases in reactions I and II, but decreases in reaction III.

To find $\Delta_r U$ for the given reaction at 1000 K, we start with the thermodynamic relationship:

$ \Delta_r H = \Delta_r U + \Delta n_g R T $

where $\Delta_r H$ is the enthalpy change for the reaction, $\Delta_r U$ is the internal energy change, $R$ is the gas constant $(8.3 \, \mathrm{J} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1})$, and $T$ is the temperature in Kelvin.

For the reaction:

$ A B \mathrm{O}_3(s) \xrightarrow{1000 \, \text{K}} A \mathrm{O}(s) + B \mathrm{O}_2(g) $

$\Delta n_g$ represents the change in the number of moles of gas. Initially, there are 0 moles of gas because the reactant is a solid, while the product is 1 mole of gas ($B \mathrm{O}_2(g)$). Therefore, $\Delta n_g = 1$.

Substitute the values into the thermodynamic equation:

$ \Delta_r U = \Delta_r H - \Delta n_g R T $

$ \Delta_r U = x - (1) \times (8.3 \times 10^3) $

$ \Delta_r U = x - 8.3 \times 10^3 $

Thus, $\Delta_r U$ in $\mathrm{kJ} / \mathrm{mol}$ is:

$ \Delta_r U = (x - 8.3) \, \mathrm{kJ} / \mathrm{mol} $

To find the equilibrium constant $ K $ for the reaction $ A_2(g) \rightleftharpoons B_2(g) $ at 300 K, we are given:

$\Delta_r G^{\Theta} = -11.5 \, \text{kJ/mol} = -11,500 \, \text{J/mol}$

$ T = 300 \, \text{K} $

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

The relationship between $\Delta_r G^{\Theta}$ and the equilibrium constant $ K $ is expressed as:

$ \Delta_r G^{\Theta} = -RT \ln K $

You can also express this in terms of a logarithm base 10:

$ \Delta_r G^{\Theta} = -2.303 \, RT \, \log_{10} K $

Rearranging the equation to solve for $\log_{10} K$, we get:

$ \log_{10} K = \frac{-\Delta_r G^{\Theta}}{2.303 \times R \times T} $

Substituting the given values:

$ \log_{10} K = \frac{-(-11,500)}{2.303 \times 8.314 \times 300} $

Calculating this:

$ = \frac{11,500}{2.303 \times 8.314 \times 300} $

$ = \frac{11,500}{5744.202} $

$ \approx 2.0 $

Thus, solving for $ K $, we have:

$ K = 10^{2.0} \approx 100 $

Therefore, the equilibrium constant at 300 K is approximately 100.

Statement I is accurate, while Statement II is not. Here's the clarification:

In the thermite process described by the reaction:

$ \mathrm{Cr}_2 \mathrm{O}_3 + 2 \mathrm{Al} \longrightarrow \mathrm{Al}_2 \mathrm{O}_3 + 2 \mathrm{Cr} + \mathrm{Q} $

This reaction is both exothermic and thermodynamically feasible, as indicated by the negative Gibbs free energy change ($\Delta G^{\ominus}=-421 \mathrm{~kJ}$). This signifies that the reaction is spontaneous under standard conditions.

However, despite being thermodynamically feasible, the reaction requires a significant amount of activation energy to proceed. At room temperature, the available activation energy is insufficient, which means that the reaction will not occur spontaneously without the application of additional heat or a catalyst to overcome this energy barrier. Thus, the reaction does not naturally occur at room temperature.

To find the enthalpy change for converting 9 grams of $\mathrm{H}_2 \mathrm{O}(l)$ from 10°C to 20°C, we follow these steps:

Calculate the Number of Moles:

The formula for the number of moles is:

$ \text{Number of moles of } \mathrm{H}_2 \mathrm{O} = \frac{\text{Mass}}{\text{Molar mass}} $

Given the mass is 9 g and the molar mass of water is 18 g/mol, we have:

$ \text{Number of moles} = \frac{9}{18} = 0.5 \, \text{mol} $

Determine the Change in Temperature:

The change in temperature is:

$ \Delta T = T_2 - T_1 = 20^\circ\mathrm{C} - 10^\circ\mathrm{C} = 10 \, \text{K} $

Calculate the Enthalpy Change:

The enthalpy change is given by the formula:

$ \Delta H = n \cdot C_p \cdot \Delta T $

Substitute the known values ($n = 0.5$, $C_p = 75 \, \text{J/mol K}$, and $\Delta T = 10 \, \text{K}$):

$ \Delta H = 0.5 \times 75 \times 10 = 375 \, \text{J} $

Therefore, the enthalpy change for this process is 375 Joules.

To calculate the reaction enthalpy change $\Delta_r H$ for the given reaction, we use the following reaction:

$ A(g) + 3B(g) \longrightarrow C(g) + 3D(g) $

The reaction enthalpy change is calculated using the formula:

$ \Delta_r H = \Delta_{\text{formation of C}} + 3 \times \Delta_{\text{formation of D}} - \Delta_{\text{formation of A}} - 3 \times \Delta_{\text{formation of B}} $

Given the enthalpies of formation are:

$\Delta_{\text{formation of A}} = 9.7 \, \text{kJ/mol}$

$\Delta_{\text{formation of B}} = -110 \, \text{kJ/mol}$

$\Delta_{\text{formation of C}} = 81 \, \text{kJ/mol}$

$\Delta_{\text{formation of D}} = -393 \, \text{kJ/mol}$

Substituting these values into the formula gives:

$ \Delta_r H = 81 + (-3 \times 393) - 9.7 - (3 \times -110) $

Calculate each part:

$-3 \times 393 = -1179$

$3 \times -110 = -330$

Now plug these values back into the formula:

$ \Delta_r H = 81 - 1179 - 9.7 + 330 $

Breaking it down:

$81 - 1179 = -1098$

$-1098 - 9.7 = -1107.7$

$-1107.7 + 330 = -777.7$

Thus, the reaction enthalpy change is:

$ \Delta_r H = -777.7 \, \text{kJ/mol} $

To relate the change in enthalpy ($\Delta H$), change in internal energy ($\Delta U$), and change in temperature ($\Delta T$) for 1 mole of an ideal gas, we start with the ideal gas law:

$ pV = nRT $

Assuming pressure $p$ is constant, the change in volume $\Delta V$ can be expressed as:

$ p \Delta V = R \Delta T $

The change in enthalpy $\Delta H$ is related to the change in internal energy $\Delta U$ by:

$ \Delta H = \Delta U + p \Delta V $

Substituting the expression for $p \Delta V$ from the ideal gas law into the enthalpy equation:

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

Rearranging the equation to express $\Delta U$ in terms of $\Delta H$ gives:

$ \Delta U = \Delta H - R \Delta T $

This equation shows the relationship between the change in internal energy, the change in enthalpy, and the change in temperature for an ideal gas.

Let's review the properties:

Extensive properties depend on the size or amount of the system. Examples include enthalpy, volume, and internal energy.

Intensive properties do not depend on the system's size. Examples include density and temperature.

Looking at the list:

Enthalpy: Extensive

Density: Intensive

Volume: Extensive

Internal Energy: Extensive

Temperature: Intensive

Thus, there are 3 extensive properties in the list (enthalpy, volume, and internal energy).

$\text{Answer: Option C (3)}$

All the given statements are correct.

We're dissolving a gas $AB$ in water $(H_2O)$.

The enthalpy change ($\Delta H$) tells us how much heat is absorbed or released during the dissolving process.

Here's what the reactions tell us:

When $AB$ dissolves in 25 moles of water, the heat change is $x \text{ kJ/mol}$.

When $AB$ dissolves in 50 moles of water, the heat change is $y \text{ kJ/mol}$.

Now, let's consider what $y - x$ would represent:

$y - x$ is the difference in enthalpy when we dissolve $AB$ in 50 moles of water compared to when we dissolve it in 25 moles of water. This value tells us about the additional heat change when we add another 25 moles of water to the solution.

Therefore, the correct answer is:

Option A $(y-x)$

Coal gasification is the method of transforming coal and water into sm gas, which is a mixture of CO and $\mathrm{H}_2$

$ \mathrm{C}(\mathrm{~s})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) \xrightarrow{1270 \mathrm{~K}} \mathrm{CO}(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) $

To determine the standard free energy change ($\Delta G^{\circ}$) for the given reaction at $25^{\circ} \text{C}$, we follow these steps:

Identify the Reaction:

$ 3 \mathrm{Ca}(s) + 2 \mathrm{Au}^{3+}(aq, 1 \text{ M}) \rightarrow 3 \mathrm{Ca}^{2+}(aq, 1 \text{ M}) + 2 \mathrm{Au}(s) $

Calculate Electron Exchange:

The electron exchange number, $ n $, is 6.

Apply the Free Energy Formula:

$ \Delta G^{\circ} = -nFE_{\text{Cell}}^{\circ} $

Here, $ F $ is Faraday's constant ($96500 \, \text{C/mol}$).

Determine $ E_{\text{Cell}}^{\circ} $:

$ E_{\text{Cell}}^{\circ} = E_{\text{Cathode}}^{\circ} - E_{\text{Anode}}^{\circ} $

For this reaction:

$ E_{\text{Cell}}^{\circ} = E_{\mathrm{Au}^{3+}/\mathrm{Au}}^{\circ} - E_{\mathrm{Ca}^{2+}/\mathrm{Ca}}^{\circ} $

Using given electrode potentials:

$ E_{\text{Cell}}^{\circ} = +1.50 \, \text{V} - (-2.87 \, \text{V}) = +4.37 \, \text{V} $

Compute $\Delta G^{\circ}$:

$ \Delta G^{\circ} = -6 \times 96500 \times 4.37 = -2530230 \, \text{J} $

Convert $\Delta G^{\circ}$ to kJ:

$ \Delta G^{\circ} = -2530.23 \, \text{kJ} $

Thus, the standard free energy change for the reaction is $-2.53 \times 10^3 \, \text{kJ}$.

Given:

Mass of hydrogen = 3 g

Mass of oxygen = 4 g

Number of Moles Calculation:

Hydrogen:

$ \text{Number of moles of hydrogen} = \frac{3}{2} = 1.5 \, \text{mol} $

Oxygen:

$ \text{Number of moles of oxygen} = \frac{4}{32} = 0.125 \, \text{mol} $

Formula used: $ \frac{\text{Mass}}{\text{Molar mass}} $

Kinetic Energy at a Given Temperature:

The kinetic energy ($ \mathrm{KE} $) of a gas is given by the formula:

$ \mathrm{KE} = \frac{3}{2} nRT $

where:

$ n $ is the number of moles,

$ R $ is the ideal gas constant,

$ T $ is the temperature.

Ratio of Kinetic Energies:

To find the ratio of kinetic energies of hydrogen ($\mathrm{H}_2$) and oxygen ($\mathrm{O}_2$):

$ \frac{(\mathrm{KE})_{\mathrm{H}_2}}{(\mathrm{KE})_{\mathrm{O}_2}} = \frac{n_{\mathrm{H}_2}}{n_{\mathrm{O}_2}} = \frac{1.5}{0.125} = \frac{12}{1} $

Thus, the ratio of the kinetic energies of hydrogen to oxygen is:

$ (\mathrm{KE})_{\mathrm{H}_2} : (\mathrm{KE})_{\mathrm{O}_2} = 12:1 $

Given:

Heat released by the system, $ Q = -900 $ J

Change in volume, $ \Delta V = 10 \text{ L} - 2 \text{ L} = 8 \text{ L} $

Pressure, $ p = 2.0 \text{ atm} $

According to the first law of thermodynamics:

$ \Delta U = Q - W $

where $ W = -p \Delta V $. Thus, the equation modifies to:

$ \Delta U = Q + p \Delta V $

Now substituting the given values:

$ \Delta U = -900 + 2 \times 8 \times 101.3 $

Calculating further:

$ \Delta U = -900 + 2 \times 8 \times 101.3 = -900 + 1620.8 = 720.8 \text{ J} $

Therefore, the change in internal energy, $ \Delta U $, is 720.8 J.

To determine the change in entropy for the transitions of benzene, we need to calculate the entropy changes for both the fusion (solid to liquid) and vaporization (liquid to vapor) processes.

Change in Entropy for Fusion (Solid to Liquid)

The formula for the entropy change due to fusion is:

$ \Delta S_{\text{fusion}} = \frac{\Delta H_{\text{fusion}}}{T_{\text{melting point}}} $

Given:

$ \Delta H_{\text{fusion}} = 10.9 \, \text{kJ/mol} = 10,900 \, \text{J/mol} $

Melting point for benzene = $5.5^\circ \text{C} = 278.5 \, \text{K}$

Calculate:

$ x = \Delta S_{\text{fusion}} = \frac{10,900}{278.5} = 39.11 \, \text{J/K/mol} $

Change in Entropy for Vaporization (Liquid to Vapor)

The formula for the entropy change due to vaporization is:

$ \Delta S_{\text{vaporization}} = \frac{\Delta H_{\text{vaporization}}}{T_{\text{boiling point}}} $

Given:

$ \Delta H_{\text{vaporization}} = 31.0 \, \text{kJ/mol} = 31,000 \, \text{J/mol} $

Boiling point for benzene = $80^\circ \text{C} = 353 \, \text{K}$

Calculate:

$ y = \Delta S_{\text{vaporization}} = \frac{31,000}{353} = 87.81 \, \text{J/K/mol} $

Difference in Entropy Change

Finally, the difference between the entropy changes for vaporization and fusion is:

$ y - x = 87.81 - 39.11 = 48.7 \, \text{J/K/mol} $

Given the reaction:

$ 2 A_2(g) + B_2(g) \longrightarrow 2 A_2B(g) + 600 \, \text{kJ} $

This reaction releases energy, indicating it is exothermic. To find the standard enthalpy of formation $\Delta_f H^{\ominus}$ for $A_2B(g)$, we consider that the total energy released is distributed over the formation of two moles of $A_2B(g)$.

Therefore, the calculation for $\Delta_f H^{\ominus}$ is:

$ \Delta_f H^{\ominus} \text{ for } A_2B(g) = \frac{-600 \, \text{kJ}}{2} $

This results in:

$ \Delta_f H^{\ominus} = -300 \, \text{kJ} \, \text{mol}^{-1} $

Enthalpy of atomisation $\left(\Delta_a H^{\ominus}\right)$ and bond dissociation enthalpy ( $\Delta_{\text {bond }} H^{\ominus}$ ) is not same for polyatomic molecules.

This is because in every dissociation steps, bonds are broken into different species.

Hence, $\mathrm{CH}_4$ is correct answer.

Let's analyze each statement step by step.

For an isothermal process involving an ideal gas, the internal energy change is zero, i.e.,

$\Delta U = 0.$

Since the first law of thermodynamics states that

$\Delta U = q + w,$

it follows that

$q = -w.$

For an irreversible process done against a constant external pressure, the work done by the gas is given by

$w = -p_{\rm ext}\,(V_{\rm final} - V_{\rm initial}).$

Thus, the heat absorbed is

$q = -w = p_{\rm ext}\,(V_{\rm final} - V_{\rm initial}).$

So, Statement I is correct (assuming the process is performed against a constant external pressure).

For an adiabatic process, by definition there is no heat exchange, so

$q = 0.$

Then, the first law reduces to

$\Delta U = q + w = 0 + w = w.$

Expressing the work performed in the adiabatic process as $w_{\rm adiabatic},$ we have

$\Delta U = w_{\rm adiabatic}.$

Therefore, Statement II is also correct.

Since both statements are valid, the correct answer is:

Option A

Both Statement I and Statement II are correct.

From the $p V$ graph, we can observe that it is an isothermal process.

The graph is of isothermal expansion.

Initial state, $p_A$ and $V_C$ as point $B$

Final state, $p_F$ and $V_D$ at point $E$.

So, work done is area under curve of isothermal expansion.

$\therefore$ Workdone $=$ Area of $B \rightarrow C \rightarrow D \rightarrow E$

Given for the reaction $ A \rightarrow P $:

Temperature $ T = 300 \, \text{K} $

Entropy change $ \Delta S_{\text{system}} = 5 \, \text{J/K/mol} $

The heat absorbed by the system can be calculated using the formula:

$ \Delta H_{\text{system}} = T \times \Delta S_{\text{system}} $

Substitute the given values:

$ \Delta H_{\text{system}} = 300 \, \text{K} \times 5 \, \text{J/K/mol} = 1500 \, \text{J/mol} $

Since the answer is needed in kilojoules per mole, convert joules to kilojoules:

$ 1500 \, \text{J/mol} = 1.5 \, \text{kJ/mol} $

Thus, the heat absorbed by the system is $ 1.5 \, \text{kJ/mol} $.

Statement I and statement III are incorrect, while statement II is correct. The corrections for the incorrect statements are as follows:

The total entropy change that accompanies a process is given by:

$ \Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surrounding}} $

The SI units of entropy are $\mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$ or $\mathrm{J} \, \mathrm{K}^{-1}$.

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} $

The standard enthalpy change for the reaction

$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$

can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the same whether it occurs in one step or in many steps.

If we modify the given reactions to reflect the formation of CO, we can add the modified reactions together to get the reaction of interest.

Consider the reactions:

(A) $\mathrm{2CO(g)+O_2(g)\to 2CO_2(g)} \quad \Delta H_1^0=-x~kJ~mol^{-1}$

(B) $\mathrm{C(graphite)+O_2(g)\to CO_2(g)} \quad \Delta H_2^0=-y~kJ~mol^{-1}$

First, divide reaction (A) by 2 to get the reaction for 1 mol of CO:

(A/2) $\mathrm{CO(g)+\frac{1}{2}O_2(g)\to CO_2(g)} \quad \Delta H_1^0=-\frac{x}{2}~kJ~mol^{-1}$

Then, subtract the divided reaction (A) from reaction (B) to get the reaction for the formation of CO:

$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)} $

$\Delta H^0=\Delta H_2^0-\Delta H_1^0=-y-(-\frac{x}{2})= \frac{x}{2} - y~kJ~mol^{-1}$

Therefore, the correct answer is Option A: $\frac{x-2y}{2}$

(A) $\mathrm{\Delta U=q-p\Delta V}$

(B) $\mathrm{\Delta G=\Delta H-T\Delta S}$

(C) $\mathrm{\Delta S=\frac{q_{rev}}{T}}$

(D) $\mathrm{\Delta H=\Delta U+(\Delta nRT)}$

Hence, (B) and (C) relations are correct.