NCERT Solutions for Class 11 Science Chemistry Chapter 6 - Thermodynamics

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Answer:

Thermodynamics doesn't talk about how and at what rate the energy change is being carried out rather it depends upon the state of a system i.e. initial and final state of a system undergoing change.

Hence, the correct answer is option C.

Answer:

In thermodynamics, closed system may be defined as the system which doesn't allow the exchange of material but heat transfer may take place.

Hence, the correct answer is option C.

Answer:

The state of a gas can be described by using state functions like temperature, pressure and volume because their value doesn't depend upon the path followed. It can  also be described using the amount of the gas used.

Hence, the correct  answer is option D.
 

Answer:

The specific heat capacity is the amount of heat required to raise the temperature of one unit mass of a substance by one degree celcius (or kelvin). It means that specific heat capacity is an intensive property and doesn't depend upon mass.

Hence,  the correct answer is option C.

Answer:

The standard enthalpy of combustion is defined as the enthalpy change per mole of a substance when it undergoes combustion provided all the reactants and products are in their standard states. This reaction is an exothermic reaction and therefore, Δ_cH will be negative.
${\mathrm{C}}_4{\mathrm{H}}_{10}\left(\mathrm{g}\right)+\frac{13}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{CO}}_2\left(\mathrm{g}\right)+5{\mathrm{H}}_2\mathrm{O}\left(l\right) {∆}_{\mathrm{c}}H=–2658.0 \mathrm{kJ} {\mathrm{mol}}^{–1}$

Hence, the correct answer is option C.

Answer:

The reaction for the formation of methane may be written as follows:

$\mathrm{C}\left(\mathrm{graphite}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{CH}}_4\left(\mathrm{g}\right)$

The change in the number of gaseous moles may be given as follows:

$$\begin{gathered} ∆n_g=\left[n_g\right(product)-n_g(reactant\left)\right] \\ ∆n_g=1-2 \\ ∆n_g=-1 \end{gathered}$$

We know the relation: ${∆}_f{U}^{\Theta }={∆}_f{H}^{\Theta }-∆n_g\mathrm{R}T$

${∆}_f{U}^{\Theta }={∆}_f{H}^{\Theta }+\mathrm{R}T$

Therefore, $∆{U_f}^{\Theta }>∆{H_f}^{\Theta }$

Hence, the correct answer is option B.


 

Answer:

We know that for an adiabatic process, q = 0 and for free expansion of an ideal gas, w = 0.

From first law of thermodynamics: 

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

Since, there is no change in internal energy, the temperature will remain constant.

Hence, the correct answer is option C.

Answer:

The area under the curve for irreversible compression of an ideal gas is greater than the area under the curve for reversible compression of an ideal gas. Therefore, the irreversible work done is greater than reversible work done.

Hence, the correct answer is option B.

Answer:

Entropy may be defined as the degree of randomness of a system. Freezing is an exothermic process and  the amount of heat releases is absorbed by the surroundings. Therefore, the entropy of surroundings increases whereas the entropy of the system decreases.

$$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{system}}=-\frac{{\mathrm{q}}_{\mathrm{rev}}}{\mathrm{T}} \\ ∆{\mathrm{S}}_{\mathrm{surroundings}}=\frac{q_{rev}}{\mathrm{T}} \end{gathered}$$

Hence, the correct answer is option C.

Answer:

$\mathrm{C} \left(\mathrm{graphite}\right)+{\mathrm{O}}_2 \left(\mathrm{g}\right)\to {\mathrm{CO}}_2 \left(\mathrm{g}\right) ; {∆}_{\mathrm{r}}\mathrm{H}=\mathrm{x} {\mathrm{kJmol}}^{-1} ........................\left(1\right)$
$\mathrm{C} \left(\mathrm{graphite}\right)+\frac{1}{2}{\mathrm{O}}_2 \left(\mathrm{g}\right)\to \mathrm{CO} \left(\mathrm{g}\right);{∆}_{r}H=y \mathrm{kJ} {\mathrm{mol}}^{-1} .................\left(2\right)$
$\mathrm{CO} \left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2 \left(\mathrm{g}\right)\to {\mathrm{CO}}_2 \left(\mathrm{g}\right);{∆}_{r}H=z \mathrm{kJ} {\mathrm{mol}}^{-1} .........................\left(3\right)$

Equation (3) can be obtained by subtracting equation (2) from equation (1)

Therefore, z = x − y or x = y + z

Hence, the correct answer is option C.

Answer:

Same bonds are formed in both the reactions but bonds between the reactant molecules are broken in reaction (b) only. Bond cleavage requires energy and thus energy released will be more in case of (a).

Hence, the correct answer is option C.

Answer:

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is called Standard Molar Enthalpy of Formation. The enthalpy of formation of a compound may be either positive or negative. 

Hence, the correct answer is option C.

Answer:

The process of direct conversion of a substance from solid to vapour state is called sublimation. 

The enthalpy of sublimation may also be shown as:

 $\mathrm{Solid}\underset{\mathrm{Fusion}}{\to }\mathrm{Liquid}\underset{\mathrm{Vapourization}}{\to }\mathrm{Gas}$

Therefore, sublimation is a combination of fusion and vapourization.

Hence, the correct answer is option A.

Answer:

For any reaction to be spontaneous, the change in free energy must be negative i.e. $∆\mathrm{G}=\mathrm{negative}$.

Hence, the correct answer is option B.

Answer:

Thermodynamics deals with the interrelation of various forms of energy and their transformation from one form to another. It also deal with thermal and mechanical equilibrium but doesn't talk about the rate at which  the change has been carried out. It talks about both state and path functions.

Hence, the correct answers are options A and D.
 

Answer:

In an exothermic reaction, system releases heat to the surroundings and the enthalpy of the reaction is negative.

Hence, the correct answers are options A and B.

Answer:

The expansion of gas is an spontaneous process and also the burning of carbon to produce carbon dioxide is also spontaneous.

Hence, the correct answers are options C and D.
 

Answer:

For an isothermal process, the work done is calculated as follows:

$\mathrm{W}=-\mathrm{nRT} \ln \frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}$
For both experiments, n, R and $\ln \frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}$ remains the same. Therefore we have:
$$\begin{gathered} \frac{{\mathrm{W}}_{600}}{{\mathrm{W}}_{300}}=\frac{600}{300}=2 \\ {\mathrm{W}}_{600}=2 {\mathrm{W}}_{300} \end{gathered}$$

For an isothermal reversible process, âˆ†U = 0.

Hence, the correct answers are options C and D.


 

Answer:

We know that the enthalpy of reaction is calculated as follows:

$$\begin{gathered} {∆}_{r}H=\sum _{}\mathrm{Enthalpy} \mathrm{of} \mathrm{formation} \mathrm{of} \mathrm{products}-\sum _{}\mathrm{Enthapy} \mathrm{of} \mathrm{formation} \mathrm{of} \mathrm{reactants} \\ {∆}_{r}H=\sum _{}2{∆}_{f}H\left(\mathrm{ZnO}\right) - \sum _{}2{∆}_{f}H\left(\mathrm{Zn}\right)+{∆}_{f}H\left({\mathrm{O}}_2\right) \end{gathered}$$

Since, for an  exothermic reaction, âˆ†_rH is negative, heat is evolved in the reaction. Therefore, the enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen.

Hence, the correct answers are options A and C.

Answer:

The enthalpy of the reaction is the energy change per mole.

It is given that for 18 g or one mole of water, the energy change is 40.79 kJ mol^–1.

∴ Enthalpy change for 2 moles of water = 2 × 40.79 kJ mol^–1 = 81.58 kJ mol^–1
 
The standard enthalpy of vaporization is the amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar), Therefore, the standard enthalpy of vaporization of water is 40.79 kJ mol^–1.

Answer:

Acetone requires less heat to vaporize due to weak intermolecular dipole-dipole interactions between its molecules as compared to water which contains strong intermolecular hydrogen bonding.
Therefore, the enthalpy of vaporization of water is more than acetone.

Answer:

The standard molar enthalpy of formation is the enthalpy of reaction when 1 mole of a substance is formed from its constituents. In the present case, CaCO₃ has been formed from other compounds and not from its constituent elements. Therefore, standard molar enthalpy of formation and enthalpy of reaction are not the same.

Answer:

The formation of ammonia may be shown as follows:
N₂(g) + 3H₂(g) → 2NH₃(g) ; ∆_fH^⊖ = –91.8 kJ mol^–1

For the reverse reaction, 2NH₃(g) → N₂(g) + 3H₂(g), the value of enthalpy changed is reversed i.e. +91.8 kJ mol^–1.
 

Answer:

According to Hess's law, if a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. 
Therefore, ${∆}_{r}H={∆}_r{H}_1+{∆}_r{H}_2+................$

Answer:

CH₄ contains four (C-H) bonds, therefore, the enthalpy of atomisation for the reaction given here is for four (C-H) bonds.
Thus, the bond energy of one (C-H) bond = $\frac{1665}{4}=416.25 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Answer:

The various steps involved in the calculation of ∆_lattice H^⊖ for NaBr are as follows:
$$\begin{gathered} \mathrm{Na}\left(\mathrm{s}\right)\to \mathrm{Na}\left(\mathrm{g}\right) ; {∆}_{sub}{H}^{\Theta }=108.4 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ \mathrm{Na}\left(\mathrm{g}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{g}\right) +{\mathrm{e}}^{-}\left(\mathrm{g}\right) ; {∆}_i{H}^{\Theta }=496 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ \frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{g}\right)\to \mathrm{Br}\left(\mathrm{g}\right) ; {∆}_{dis}{H}^{\Theta }=\frac{192}{2}=96 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ \mathrm{Br}\left(\mathrm{g}\right)+ {\mathrm{e}}^{-}\left(\mathrm{g}\right)\to {\mathrm{Br}}^{-}\left(\mathrm{g}\right) ; {∆}_{eg}{H}^{\Theta }=-325 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ {\mathrm{Na}}^{+}\left(\mathrm{g}\right) + {\mathrm{Br}}^{-}\left(\mathrm{g}\right) \to {\mathrm{Na}}^{+}{\mathrm{Br}}^{-}\left(\mathrm{s}\right) ; {∆}_f{H}^{\Theta }=-360.1 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

$$\begin{gathered} {∆}_f{H}^{\Theta }={∆}_{sub}{H}^{\Theta }+{∆}_i{H}^{\Theta }+\frac{{∆}_{diss}{H}^{\Theta }}{2}+{∆}_{eg}{H}^{\Theta }+{∆}_{lattice}{H}^{\Theta } \\ {∆}_{lattice}{H}^{\Theta }={∆}_f{H}^{\Theta }-[{∆}_{sub}{H}^{\Theta }+{∆}_i{H}^{\Theta }+\frac{{∆}_{diss}{H}^{\Theta }}{2}+{∆}_{eg}{H}^{\Theta }] \\ {∆}_{lattice}{H}^{\Theta }=-360.1-[108.4+496+96-325] \\ {∆}_{lattice}{H}^{\Theta }=-360.1-108.4-496-96+325 \\ {∆}_{lattice}{H}^{\Theta }=-735.5 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

Answer:

Diffusion of gases increases the degree of randomness i.e. entropy (∆S) is positive.
∆G = âˆ†H − T∆S
Since âˆ†H is 0 and âˆ†S is +ve, âˆ†G will be −ve.
Therefore, the process remains spontaneous.

Answer:

The mathematical relation which relates these three parameters may be written as follows:
$∆\mathrm{S}=\frac{{\mathrm{q}}_{\mathrm{rev}}}{\mathrm{T}}$

Where, $∆\mathrm{S}$ = entropy change
             ${\mathrm{q}}_{\mathrm{rev}}$ = heat of a reversible reaction
              T  =  temperature  

Answer:

Yes, when the system and the surroundings are in thermal equilibrium, their temperatures remain the same.

Answer:

We know the relation between free energy change, ΔG^Θ and equilibrium constant, K_p is as follows:
$∆G^{\Theta }=-\mathrm{R}T \ln K_p$
Here, K_p is given as 0.98 and ln(0.98) will be negative. Therefore, ΔG^Θ will be positive.
Thus, the reaction is non-spontaneous.

 

Answer:

Enthalpy change for a cyclic process is zero. Therefore, the value of ∆H for the cycle as a whole will be zero.

Answer:

In solid water, i.e., ice, the water molecules are closely packed in an ordered arrangement as compared to water molecules in liquid state and result in a decrease in degree of randomness (entropy). 
Therefore, the molar entropy of solid water is less than that of liquid water, i.e., less than 70 J K^–1 mol^–1.

Answer:

Work and heat are path functions as they are path dependent whereas enthalpy, entropy, temperature and free energy are state functions  because their values depend only on the state of the system and not on how it is reached.

Answer:

The molar enthalpy of water is more due to the presence of strong intermolecular hydrogen bonding between its molecules whereas in acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise 1 mol of acetone than it does to vaporize 1 mol of water. 

Answer:

The Gibbs free energy, âˆ†_rG and standard Gibbs free energy, âˆ†_rG^⊖ are related to the equilibrium constant, K as follows:
∆_rG = ∆_rG^⊖ + RT lnK
At equilibrium, âˆ†_rG = 0
∴ 0 = ∆_rG^⊖ + RT lnK
∆_rG^⊖ = −RT lnK
Thus, âˆ†_rG will be zero at equilibrium.

Answer:

We know that for an isolated system:

w = 0 ; q = 0

Therefore, from first law of thermodynamics:

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

Thus, the internal energy change is zero.

Answer:

Heat, q will be independent of path and will be a state function under two conditions:
(i) At constant volume
(ii) At constant pressure

This can be supported as follows:
$$\begin{gathered} ∆U=q+w \\ w=-p∆V \\ ∆U=q-p∆V \\ \mathrm{At} \mathrm{constant} \mathrm{volume}, ∆V=0 \\ \therefore ∆U=q_v \end{gathered}$$

At constant pressure,
$$\begin{gathered} ∆U + p∆V =∆H \\ \therefore ∆H=q_p \end{gathered}$$

Both, ΔU and ΔH are state functions and therefore, heat, q will be a state function.

Answer:

Expansion of a gas in vacuum (p_ex = 0) is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible.
For isothermal expansion, Joule determined experimentally that q = 0.
By first law of thermodynamics,
q = ∆U + (–w)
⇒ 0 = ∆U + 0
∴ Internal energy, âˆ†U = 0

Answer:

Heat capacity at constant pressure (C_p) is the amount of heat required to raise the temperature of a substance by 1 degree celsius (or 1 kelvin).
Specific heat (c) is the quantity of heat required to raise the temperature of one unit mass of a substance by 1 degree celsius (or 1 kelvin).
Mass of 1 mol of water = 18 g
Heat capacity (C_p) is related to specific heat (c) for water as follows:
$C_p=18\times c$
∵ Specific heat (c) of water = 4.18 J g⁻¹ K⁻¹
$$\begin{gathered} \therefore C_p=4.18\times 18 \\ {C}_p=75.24 \mathrm{J} {\mathrm{K}}^{-1} \end{gathered}$$

Answer:

We know that for an  ideal gas:
$C_p-C_v=n\mathrm{R}$
Where, n = number of moles
For 10 moles of an ideal gas:
$C_p-C_v=10 \mathrm{R}=10\times 8.314 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}=83.14 {\mathrm{JK}}^{-1}$

Answer:

Molar enthalpy change of graphite = enthalpy change for 1 g carbon × molar mass of carbon
                                                        = − 20.7 kJ g⁻¹ × 12 g mol⁻¹
                                                        = − 248.4 kJ mol⁻¹
Since, the sign of molar enthalpy change is negative, the reaction is exothermic.

Answer:

The enthalpy change of the given reaction can be calculated as follows:
$$\begin{gathered} {∆}_{r}H=\sum _{}\mathrm{Bond} \mathrm{Energy} \left(\mathrm{reactants}\right)-\sum _{}\mathrm{Bond} \mathrm{Energy} \left(\mathrm{products}\right) \\ {∆}_{r}H={(\mathrm{B}.\mathrm{E})}_{{\mathrm{H}}_2}+{(\mathrm{B}.\mathrm{E})}_{{\mathrm{Br}}_2}-2\times {(\mathrm{B}.\mathrm{E})}_{\mathrm{HBr}} \\ {∆}_{r}H=435+192-2\times 368 \\ {∆}_{r}H=627-736 \\ {∆}_{r}H=-109 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

Answer:

Enthalpy of vapourization of 1 mole of CCl₄ = 30.5 kJ mol^-1
Molar mass of CCl₄ = 154 g mol^-1
Therefore, for 154 g of  CCl₄, heat required for vapourization = 30.5 kJ mol^-1
Thus, for 284 g of CCl₄, heat required for vapourization = $\frac{30.5}{154}\times 284=56.24 \mathrm{kJ}$

Answer:

Standard enthalpy change of formation is defined as the enthalpy change for formation of one mole of a compound from its elements in their most stable states of aggregation.
Therefore, the reaction for formation of one mole of H₂O(l) is as follows:
H₂(g) + $\frac{1}{2}$O₂(g) → H₂O(l)
Thus, standard enthalpy change for the formation of one mole of H₂O(l) = $\frac{1}{2}\times {∆}_r{H}^{\Theta }=-\frac{572}{2}=-286 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Answer:

Considering an ideal gas enclosed in a cylinder fitted with weightless and frictionless piston. The initial volume of the gas is V_iand the pressure is p. If external pressure, p_ext, which is greater than p is applied , the piston is moved inward till the condition (p_ext = p) is achieved.
Let the final volume of the gas is V_f..
The work done is given as follows:
W = −p_ext(V_f − V_i )
W = p_ext(∆V)                  (∵ V_f < V_i)
The work done is equal to the shaded area.

Answer:

A process or change is called reversible if at a certain point it can be reversed by an infinitesimal change. When change in pressure is carried out in infinite steps (reversible condition), the work done can be calculated from P-V curve during compression from initial volume V₁ to final volume V₂.
The work done is represented by the shaded region.

Answer:

The respective graphs for the two cases may be drawn as follows:


The energy is increasing in part (a) and is decreasing in (b). Therefore, enthalpy is a contributing factor for spontaneity in part (b).
 

Answer:

Enthalpy alone can't be the factor defining the spontaneity. We have to consider the entropy also.

$∆\mathrm{G}=\mathrm{H}-\mathrm{T}∆\mathrm{S}$

Thus, it is not possible to decide the spontaneity of a reaction from the given diagram.

Answer:

The given diagram represents that the process is carried out in infinite steps, hence, it is an isothermal reversible expansion of the ideal gas from pressure 2.0 atm to 1.0 atm at 298 K.
We know that for an isothermal reversible expansion of gas, work done can be calculated as follows:
$$\begin{gathered} w=-2.303n\mathrm{R}T \log \left(\frac{P_1}{P_2}\right) \\ w=-2.303\times 1\times 8.314\times 298\times \log \left(\frac{2}{1}\right) \\ w=-1717.46 \mathrm{J} \end{gathered}$$

Answer:

We know that the work done may be calculated as follows:
$$\begin{gathered} w=-p_{ext}(V_f-V_i) \\ w=-2 (50-10) \\ w=-80 \mathrm{L} \mathrm{bar} \\ w=-8000 \mathrm{J} \\ w=-8 \mathrm{kJ} \end{gathered}$$
The negative sign indicates that the work is done by the system.
Work done will be more in case of reversible expansion because the external and internal pressures are almost equal.

Answer:

The correct matching may be done as follows:

(i)$\to$(e) ; (ii)$\to$(d) ; (iii)$\to$(f) ; (iv)$\to$(a) ; (v)$\to$(g),(k),(l) ; (vi)$\to$(b) ; (vii)$\to$(c) ; (viii)$\to$(j) ; (ix)$\to$(h) ; (x)$\to$(i) ; (xi)$\to$(a),(l),(m) ; (xii)$\to$(g),(k)

Answer:

The correct matching may be done as follows:

Answer:

The correct matching may be done as follows:

(i)$\to$(c)

When the enthalpy change is positive, âˆ†_rS^⊖ is negative and ∆_rG^⊖ is positive, the process is non-spontaneous at all temperatures.

(ii)$\to$(a)

When enthalpy change is negative, âˆ†_rS^⊖ is negative and âˆ†_rG^⊖ is positive, the process will be non-spontaneous at very high temperature.

(iii)$\to$(b)

When enthalpy change is negative, ∆_rS^⊖ is positive and âˆ†_rG^⊖ is negative, the process will be spontaneous at all temperatures.

Answer:

The correct matching may be done as follows:

(i) $\to$(b), (d) ; (ii) $\to$(b) ; (iii)$\to$(c) ; (iv)$\to$(a)

Answer:

In combustion, the enthalpy of reactants is greater than that of products. Thus, combustion is exothermic. The enthalpies of all elements in their standard state are zero.

Hence, the correct answer is option B.

Answer:

The spontaneous process is an irreversible process and may be reversed by some external agency. The spontaneous process is somehow associated with an increase in randomness and a decrease in the enthalpy. However, while decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases.

Hence, the correct answer is option B.

Answer:

Crystallization of liquid includes more arranged constituent particles and a decrease of entropy.

Hence, the correct answer is option A.

Answer:

At constant pressure, q_p is heat absorbed by system and –p∆V represent expansion work done by the system. The change in internal energy, ΔU can be expressed as follows:
ΔU = q_p − pΔV
Or, U₂ − U₁ = q_p − p (V₂ − V₁)
Or. q_p = (U₂ + pV₂) − (U₁ + pV₁) …(1)
We can define another thermodynamic function, the enthalpy, H as follows:
H = U +PV
We know that enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure.
∴ âˆ†H = H₂ − H₁ = q_p        ...(2)
Although q is a path dependent function, H is a state function because it depends on U, p and V, all of which are state functions
From  (1) and (2), the reaction can be re-written as follows:
$$\begin{gathered} ∆H=U_2-U_1+P(V_2-V_1) \\ ∆H=∆U+P∆V \end{gathered}$$
Considering the ideal gas equation for states 1 and 2.
PV₁ = n₁RT ; PV₂ = n₂RT
Therefore, we have:
$$\begin{gathered} ∆H=∆U+(n_2-n_1)\mathrm{R}T \\ ∆H=∆U+∆n_g\mathrm{R}T \end{gathered}$$
where, $∆n_g=\mathrm{difference} \mathrm{of} \mathrm{number} \mathrm{of} \mathrm{gaseous} \mathrm{moles} \mathrm{of} \mathrm{products} \mathrm{and} \mathrm{reactants}$
 

Answer:

Extensive properties are those which depend upon the amount of matter present in the system whereas intensive property doesn't.

In this case, mass, internal energy, enthalpy, heat capacity, volume are extensive properties whereas, pressure, molar heat capacity, density, mole fraction, specific heat, temperature, and molarity are intensive properties.

It is to be noted that the ratio of two extensive properties is always intensive. For example- mole fraction, molarity, etc.

Answer:

The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber Cycle.
Let's calculate the lattice enthalpy of NaCl by using the given steps:

• • Sublimation of sodium metal:

; 

• • Ionization of sodium atoms (ionization enthalpy);

; 

• • Dissociation of chlorine (reaction enthalpy is half the bond dissociation enthalpy):

; 

• • Electron gained by chlorine atoms (electron gain enthalpy):

; 

Applying Hess’s Law, we get lattice enthalpy of NaCl(s).

Answer:

$$\begin{gathered} \Delta S_{Total}=\Delta S_{sys}+\Delta S_{surr} \\ \Delta S_{Total}=\Delta S_{sys}+\frac{-\Delta H_{sys}}{T} \\ T\Delta S_{Total}=\Delta S_{sys}-\Delta H_{sys} \\ For spon\tan eous change, \Delta S_{Total}>0 \\ \therefore T\Delta S_{sys}-\Delta H_{sys}>0 \\ \Rightarrow -\left(\Delta H_{sys}-T\Delta S_{sys}\right)>0 \\ But, \Delta H_{sys}-T\Delta S_{sys}=\Delta G_{sys} \\ \therefore -\Delta G>0 \\ \Rightarrow \Delta G_{sys}=\Delta H_{sys}-T\Delta S_{sys}<0 \\ \Delta H_{sys}=\mathrm{Enthalpy} \mathrm{change} \mathrm{of} \mathrm{a} \mathrm{reaction} \\ T\Delta S_{sys}=\mathrm{Energy} \mathrm{which} \mathrm{is} \mathrm{not} \mathrm{available} \mathrm{to} \mathrm{do} \mathrm{useful} \mathrm{work} \\ \Delta G_{sys}=\mathrm{Energy} \mathrm{available} \mathrm{for} \mathrm{doing} \mathrm{useful} \mathrm{work} \end{gathered}$$
So ∆G is the net energy available to do useful work and is thus a measure of the ‘free energy’. For this reason, it is also known as the free energy of the reaction.

The unit of âˆ†G is Joule.

If a reaction has positive enthalpy change and positive entropy change, then, it will be spontaneous at high temperature.

Answer:

(i) Reversible work is represented by the combined areas  and .
(ii) Work against constant pressure, p_f is represented by the area  .
Work (i) > Work (ii)