Thermodynamics

The properties which do not depend upon the quantity or size of matter are called intensive properties. Among the given list molar volume (I), temperature (VI) and density (VII) are intensive properties.

Given,

$ \begin{aligned} & \mathrm{C}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{CO} ; \Delta H_f=-110 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{C}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \Delta H_f^1=-393 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

To find enthalpy of combustion of CO .

i.e. $\mathrm{CO}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \Delta H_{\mathrm{CO}}=$ ?

$ \begin{aligned} \Delta H_{\mathrm{CO}} & =H_f^1-H_f \\ & =-393-(-110) \\ & =-283 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Given the following information about enthalpies of formation at 298 K:

For the formation of $\mathrm{N}_2 \mathrm{O}$:

$ \mathrm{N}_2 + \frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{N}_2 \mathrm{O}; \quad \Delta H_{f(\mathrm{N}_2 \mathrm{O})} = 82 \, \mathrm{kJ/mol} $

For the formation of $\mathrm{NO}$:

$ \frac{1}{2} \mathrm{N}_2 + \frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{NO}; \quad \Delta H_{f(\mathrm{NO})} = 90 \, \mathrm{kJ/mol} $

To find the enthalpy change for the reaction:

$ \mathrm{N}_2 \mathrm{O}(g) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) $

We use the formula for calculating the enthalpy change of the reaction based on the enthalpies of formation:

$ \Delta H = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants}) $

Applying this to the given reaction:

$ \Delta H = \left( 2 \times \Delta H_{f(\mathrm{NO})} \right) - \left( \Delta H_{f(\mathrm{N}_2 \mathrm{O})} + \frac{1}{2} \Delta H_{f(\mathrm{O}_2)} \right) $

Since the formation enthalpy of a pure element in its standard state, like $\mathrm{O}_2$, is zero:

$ \Delta H = 2 \times 90 \, \mathrm{kJ/mol} - 82 \, \mathrm{kJ/mol} $

Calculating gives:

$ \Delta H = 180 \, \mathrm{kJ/mol} - 82 \, \mathrm{kJ/mol} = 98 \, \mathrm{kJ/mol} $

Therefore, the enthalpy change of the reaction is $+98 \, \mathrm{kJ/mol}$.

We can solve this using the formula for the work done by an ideal gas during an isothermal, reversible expansion:

$ W = -nRT \ln\left(\frac{V_2}{V_1}\right) $

where:

• $ W $ is the work done by the gas (given as $-16.5\,\text{kJ}=-16500\,\text{J}$),

• $ n = 2.5\,\text{mol} $,

• $ R = 8.314\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1} $,

• $ V_1 = 2\,\text{dm}^3 $ and $ V_2 = 20\,\text{dm}^3 $.

Follow these steps:

Write down the work formula:

$ W = -nRT\ln\left(\frac{V_2}{V_1}\right) $

Solve for the temperature $ T $:

$ T = -\frac{W}{nR\ln\left(\frac{V_2}{V_1}\right)} $

Substitute the known values:

$ T = -\frac{-16500}{2.5 \times 8.314 \times \ln\left(\frac{20}{2}\right)} $

Calculate the volume ratio and its natural logarithm:

$ \frac{V_2}{V_1} = \frac{20}{2} = 10, \quad \text{so} \quad \ln(10) \approx 2.3026. $

Now compute the temperature:

$ T = \frac{16500}{2.5 \times 8.314 \times 2.3026}. $

First, compute the product in the denominator:

$ 2.5 \times 8.314 = 20.785, $

and then

$ 20.785 \times 2.3026 \approx 47.88. $

Finally, calculate $ T $:

$ T \approx \frac{16500}{47.88} \approx 344.8 \, \text{K}. $

Rounding off to the nearest value gives approximately $ 345 \, \text{K} $.

Thus, the correct option is Option C: 345.

Enthalpy of formation of $\mathrm{D}_2 \mathrm{O}$ will be

$ \begin{aligned} & \mathrm{D}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{D}_2 \mathrm{O} \\ & \Delta H=\Delta H_p \text { (Products) } \\ & =400+\frac{1}{2} \times 498-490 \times 2 \\ & \Delta H_r \text { (Reactants) } \\ & =400+249-980=-331 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

$ \begin{aligned} & \Delta H=\Delta U+\Delta(p V) \\ & \Delta H=\Delta U+\Delta n(R T) \\ & \quad p=1 \mathrm{~atm}, T=0^{\circ} \mathrm{C}=273 \mathrm{~K} \end{aligned} $

$ \begin{aligned}As,& \Delta H=\Delta U \\ & \Delta U=-41 \mathrm{~kJ} / \mathrm{mol}\,\,\,\,\,\,\,\,\,\,\,(given) \end{aligned} $

Therefore, $\Delta H=-41 \mathrm{~kJ} / \mathrm{mol}$

Graphite is the most stable state of carbon and its $\Delta H_f^{\circ}$ is considered as zero.

( ∵ has more van der Waals' force) also $\Delta H_f^{\circ}$ for $\mathrm{C}_{60}>\Delta H_f^{\circ}$ for diamond. $\Delta H_f^{\circ}$ values of graphite, diamond and $\mathrm{C}_{60}$ are $0,1.9$ and $38.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta H_f^{\circ}$ values are in order.

Fullerene $>$ Diamond $>$ Graphite.

In an adiabatic expansion,

$ p V^{\top}=\text { constant } $

where, $p=$ pressure

$V=$ volume

$\gamma=$ ratio of specific heat at constant pressure and constant volume $\left(\frac{C_p}{C_V}\right)$

Differentiating above equation,

$ \begin{aligned} & d p \cdot V^\gamma+\gamma V^{\gamma-1} \cdot p d V=0 \\ & \frac{d p}{p}=\frac{-\gamma d V}{V} \\ & \Rightarrow \quad \frac{d p}{p}=-(1.4) \times 5 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-7 \% \end{aligned} $

∴ Percentage change in pressure is $7 \%$.

Given,

$ \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; $

Also, $\quad 2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_2(g)+\mathrm{O}_2(g)$;

$ \Delta H_1^{\prime}=285 \times 2=+570 \mathrm{~kJ} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \mathrm{N}_2 \mathrm{O}_5(g)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_3(l) \text {; } $

$ \Delta H_2=-76.6 \mathrm{~kJ} $

Also,

$ \begin{gathered} 4 \mathrm{HNO}_3(l) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(g)+2 \mathrm{H}_2 \mathrm{O}(l) ; \\ \Delta H_2^{\prime}=76.6 \times 2=+153.2 \mathrm{~kJ} \quad \mathrm{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{gathered} $

$ \begin{aligned} & \mathrm{N}_2(g)+3 \mathrm{O}_2(g)+\mathrm{H}_2(g) \longrightarrow 2 \mathrm{HNO}_3(l) ; \\ & \Delta H_3=-348.2 \mathrm{~kJ} \end{aligned} $

Also,

$ \begin{aligned} 2 \mathrm{~N}_2(g)+6 \mathrm{O}_2(g)+2 \mathrm{H}_2(g) & \longrightarrow 4 \mathrm{HNO}_3(l) \\ \Delta H_3^{\prime} & =-348.2 \times 2 \\ & =-696.4 \mathrm{~kJ} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

On adding Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & 2 \mathrm{~N}_2(g)+5 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(g) ; \\ & \Delta H=\Delta H_1^{\prime}+\Delta H_2^{\prime}+\Delta H_3^{\prime} \\ & \Delta H=570+153.2-696.4 \\ & \Delta H=723.2-696.4 \\ & \Delta H=26.8 \mathrm{~kJ} . \end{aligned} $

$ \begin{aligned} & \text { } \mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}+2 e^{-} ; \\ & \begin{array}{l} \Delta H_1=\mathrm{IE}_1+\mathrm{IE}_2 \\ 2 \mathrm{~F}+2 e^{-} \longrightarrow 2 \mathrm{~F}^{-} ; \Delta H_2=2 \mathrm{EA}_1 \\ \mathrm{Mg}+2 \mathrm{~F}^{-} \longrightarrow \mathrm{Mg}^{2+}+2 \mathrm{~F}^{-} ; \Delta H=\text { ? } \\ \Delta H=\Delta H_1+\Delta H_2=737+1451+2(-328) \\ =1532 \mathrm{~kJ} / \mathrm{mol} \end{array} \end{aligned} $

$\Delta U=q+w$

$\Delta U$ will be same for path 1,2 and 3 because it depends on initial and final positions and they are same for all of them.

Work done is area under $p \Delta V$ curve. Greater the area more will be the work done. Work is a path function.

Since the area under $p \Delta V$ curve is largest for path 3 and smallest for path 1, therefore, $w_1

The standard enthalpy of formation of all elements in their standard states is zero. Since the natural standard state of carbon is graphite, the enthalpy of formation of graphite is zero, as there is no change involved in its formation.

$2 \mathrm{Na}(s)+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow2 \mathrm{NaOH}(a q)+\mathrm{H}_2(g) $

1 mole of $\mathrm{Na}=23 \mathrm{~g}$ of Na

4 moles of $\mathrm{Na}=92 \mathrm{~g}$ of Na

2 moles of Na reacts with 2 moles of $\mathrm{H}_2 \mathrm{O}$ to form

1 mole of $\mathrm{H}_2$.

4 moles of Na will react with 4 moles of $\mathrm{H}_2 \mathrm{O}$ to form 2 moles of $\mathrm{H}_2$.

Work done by expansion of gas is given by $-p \Delta V$

$ p \Delta V=\Delta n_g R T $

At constant external pressure,

$ \begin{aligned} \Delta V=V_{\mathrm{H}_2}-0 & =V_{\mathrm{H}_2} \\ p \Delta V & =\Delta n_g R T \\ p \Delta V=\Delta n_g R T & =2 \times 8.314 \times 300=4.98 \mathrm{~kJ} \end{aligned} $

Therefore, the work done is 4988.4 J .

Since it is work of expansion so will be with negative sign (-4988.4 J).

(b) At 0 K , the entropy of pure crystalline materials is zero.

$\therefore$ The statement (I) is correct as entropy approach zero, at absolute zero.

Statement (II) entropy for the process, $\mathrm{H}_2 \mathrm{O}(t) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ decreases is incorrect as entropy increases. The degree of randomness when water vaporises to gaseous state increases.

(III) Gibbs' energy is a state function is correct as the Gibbs' free energy depends on initial and final state only.

Hence, statement (I) and (III) are correct.

$2 \mathrm{C}+2 \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{CH}_3 \mathrm{CHO}$

$\Delta H_f=\Sigma \mathrm{BE}_{\text {(reactants) }}-\Sigma \mathrm{BE}_{\text {(products) }}$

$=\left[(2 \times 700)+(2 \times 400)+\left(\frac{1}{2} \times 500\right)\right] -[(4 \times 400)+(250 \times 1)+(700 \times 1)]$

$ \begin{aligned} & =2450-2650 \mathrm{~J} \\ & =-200 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$

From first graph,

If we draw a line parallel to $x$-axis, then at that point pressure will be constant.

So, temperature will vary according to volume i.e, $V_2>V_1$.

So, $T_2$ will be greater than 7, i.e., $T_2>T_1$

Similarly, for second graph,

If we draw a line parallel to $y$-axis. Then, pressure will vary according to volume i.e., $V_2>V_1$, but since pressure and volume are inversely proportional i.e., $p V=k$ hence $p_1>p_2$.

So, the correct option is $T_2>T_1$ and $p_1>p_2$.

Extensive properties are the properties that depends on the amount or mass of the matter present in a sample. Examples of extensive property are volume, enthalpy, heat capacity, internal energy, etc.

Hence, out of the given properties, only four properties, i.e. volume, enthalpy, heat capacity and internal energy are extensive properties.

(A) Work done in an isothermal process can be calculated by using formula, $\begin{equation} W=-n R T \ln \left(\frac{V_f}{V_i}\right) \end{equation}$

(B) For an adiabatic process, $q=0$.

$\begin{aligned} \text{So,} \quad & \Delta U=W \\ & \Delta U=q+W \Rightarrow \Delta U=W \end{aligned}$

(C) Work done for an isobaric process is

$W=-p_{\text {ext }} \times \Delta V$

(D) For an isochoric process, $\Delta V=0$

So, $W=-p \Delta V=0$

Hence, $q=\Delta U$

So, the correct match is

Using equation,

$\Delta G=\Delta G^{\circ}+R T \ln Q$

where, $\Delta G \rightarrow$ Change in free energy

$\Delta G^{\circ} \rightarrow$ Change in standard free-energy

$R \rightarrow$ Gas constant

$T \rightarrow$ Temperature

$Q \rightarrow$ Reaction quotient

At equilibrium,

$\Delta G=0 \text { and } Q=K$

$\begin{aligned} & \text { So, } 0=\Delta G^{\circ}+R T \ln K \\ & \Rightarrow \Delta G^{\circ}=-R T \ln K \end{aligned}$

Entropy order for solid, liquid and gas is :

Gas $>$ Liquid $>$ Solid

(a) $\mathrm{H}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{H}_2(\mathrm{~g})$

Two gaseous atoms combine to form one gaseous molecule. Hence, entropy decreases.

(b) $\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_2 \mathrm{O}(s)$

Entropy decreases when liquid freezes to solid form.

(c) $\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(g)$

Conversion of liquid to gas increases the entropy.

(d) $A(g)+B(g)+C(s) \longrightarrow 2 D(s)$

In the above reaction, the product formed is a solid. Thus, entropy will decreases.

So, option (c) is the correct option.

The given process is isothermal reversible reaction.

State 1 to $2 p_1=15$ bar and $p_2=10$ bar

$\begin{aligned} n & =1 \mathrm{~mol} \\ w_1 & =-2.303 n R T \log \frac{p_1}{p_2} \\ & =-2.303 \times 1 \times 0.083 \times 300 \log \frac{15}{10} \\ & =-10.09 \mathrm{~L} \mathrm{~bar} \quad \text{.... (i)} \end{aligned}$

State 2 to $3 p_1=10$ bar and $p_2=5$ bar

$\begin{aligned} n & =1 \mathrm{~mol} \\ w_2 & =-2.303 n R T \log \frac{p_1}{p_2} \\ & =-2.303 \times 1 \times 0.083 \times 300 \log \frac{10}{5} \\ & =-17.26 \mathrm{~L} \mathrm{bar} \quad \text{... (ii)} \end{aligned}$

State 3 to $1 p_1=5$ bar and $p_2=15$ bar

$\begin{aligned} n & =1 \mathrm{~mol} \\ w_3 & =-2.303 n R T \log \frac{p_1}{p_2} \\ & =-2.303 \times 1 \times 0.083 \times 300 \log \frac{5}{15} \\ & =+27.36 \mathrm{~L} \text { bar } \quad \text{.... (iii)} \end{aligned}$

$\therefore$ Total workdone,

$\begin{aligned} & w=w_1+w_2+w_3 \\ & w=-10.09-17.26+27.36=0 \end{aligned}$