Thermodynamics practice · Mixed difficulty

The latent heat of vaporization of water is 540 cal $\mathrm{g}^{-1}$ at $100^{\circ} \mathrm{C}$. What will be the entropy increase when one mole of water is evaporated at $100^{\circ} \mathrm{C}$ ?

Enthalpy of vaporisation for water is 186.5 kJ $\mathrm{mo}{\mathrm{l}^{ - 1}}$ . The entropy change during vaorisation is …kJ ${\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}$ .

Find the approximate value of $(\Delta \mathrm{H}-\Delta \mathrm{U})$ in $\mathrm{J}$. $\mathrm{mol}^{-1}$, for the formation of $\mathrm{CO}$ from its elements at $298 \mathrm{~K} .\left(\mathrm{R}=8.314 . \mathrm{J}\mathrm{~K}^{-1} \cdot \mathrm{mol}^{-1}\right)$

$\Delta \mathrm{G}$ is equal to

Standard enthalpy (heat) of formation of liquid water at $25^{\circ} \mathrm{C}$ is around

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

For the reaction:

$\begin{aligned} & A(\ell) \rightarrow 2 B(g) \\ & \Delta {U}=2.1 \mathrm{~kcal}, \Delta {S}=20 \mathrm{~cal} / \mathrm{K}, \text { at } {T}=300 \mathrm{~K}. \end{aligned}$

Hence $\Delta {G}$ (in $\mathrm{kcal})$ is $-X$. Find the value of $X$.

Work done in the reversible isothermal process is given by

∆G°  for the reaction X+Y⇌Z is -4.606 kcal. The value of equilibrium constant of the reaction at 227℃ is $\left( {R = 2.0\,\,\mathrm{cal}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}\,{\mathrm{K}^{ - 1}}} \right)$

Statement-I (S-I): Absolute values of internal energy cannot be determined

Statement-II (S-II): It is impossible to determine the exact values of constituent energies of the substances.

Consider the reactions given below. On the basis of these reactions find out which of the algebraic relationship given in options is correct?

1. $\mathrm{C(g})+4 \mathrm{H(g}) \longrightarrow \mathrm{CH_4(g}) ; \Delta_r H=x \mathrm{~kJ~mol}^{-1}$ ,

2. $\mathrm{C}$ (graphite, $\mathrm{s})+2 \mathrm{H_2(g}) \longrightarrow \mathrm{CH_4(g}); \Delta_r H=y \mathrm{~kJ~mol}^{-1}$

Assertion(A): Work and internal energy are not state functions.
Reason(R): The sum of $q+w$ is path function.

For adiabatic expansion of a perfect gas $\frac{d P}{P}$ is :

Calculate the difference between $C_p$ and $C_V$ for 10 moles of an ideal gas.

The enthalpy of neutralization of a substance

For which reaction(s) is the $\Delta \mathrm{H}$ value equal to the $\Delta \mathrm{E}$ value?
(I) $\mathrm{CH}_4(\mathrm{g})+2 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O} (l)$
(II) $\mathrm{C}_2 \mathrm{H}_4(\mathrm{g})+3 \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$

When a certain mass of an ideal gas is adiabatically compressed so that its volume is reduced to $1 / 32$ times, its absolute temperature is increased 4 times. The number of atoms in a molecule (atomicity) of the gas is

Consider the following properties:

"Temperature, Heat, Molar enthalpy, Internal energy, Density"

The number of intensive and extensive properties are respectively

Unit of entropy is

During an adiabatic expansion a gas obeys $V T^3=$ constant . The gas must be :

Calculate the change in entropy $\left(\Delta_{{r}} {S}^{\circ}\right)$ at $298 \mathrm{~K}$ for the reaction in which urea is formed from ${NH}_3$ and ${CO}_2$

$2N{H_3}\left( {\text{g}} \right) + C{O_2}\left( {\text{g}} \right) \to N{H_2} - \mathop C\limits_{\begin{array}{cccccccccccccccccccc} {||} \\ O \end{array}} - N{H_2}\left( {{\text{aq}}} \right) + {H_2}O\left( \ell \right)$

The standard entropy of ${NH}_2{CONH}_2(\mathrm{aq}), {CO}_2(\mathrm{g}), {NH}_3(\mathrm{g})$ and ${H}_2 {O}(\ell)$ are $174.0, 213.7, 195$ and $75 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$ respectively

The heat evolved during the combustion of 112 litre of water gas (mixture of equal volume of ${H_2}$ and CO) is : ${\mathrm{H}_2}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right);\Delta H = - 241.8\mathrm{kJ}$ $\mathrm{CO}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta H = - 283\mathrm{kJ}$

The hydrogen bond between $\mathrm{F}^{-}$and $\mathrm{CHCOOH}$ is very strong and its strength may be analysed by setting up a Born-Haber cycle with the following data. Lattice energy of $\mathrm{KF.CH}_3 \mathrm{COOH}=734;$ enthalpy of vaporization of $\mathrm{CH}_3 \mathrm{COOH}=20;$ enthalpy of solution of $\mathrm{KF}= 35;$ solvation energy of $\mathrm{K}^{+}(\mathrm{g}) =-325;$ solvation energy of $\mathrm{F}^{-}=-389;$ enthalpy of formation of $\mathrm{KF} \cdot \mathrm{CH}_3 \mathrm{COOH}(\mathrm{s})$ from $\mathrm{KF}(\mathrm{s})$ and $\mathrm{CH}_3 \mathrm{COOH}(l) =-25.$ Find the energy of the hydrogen bond between $\mathrm{F}^{-}$ and $\mathrm{CH}_3 \mathrm{COOH}$ in the gas phase (in $\mathrm{kJ} / \mathrm{mol}$).

Enthalpies of formation of $\mathrm{CO}_2(\mathrm{g}), \mathrm{CO}(\mathrm{g}), \mathrm{N}_2 \mathrm{O}(\mathrm{g})$ and $\mathrm{NO}_2(\mathrm{g})$ in $\mathrm{kJ} \mathrm{~mol}^{-1}$ are respectively $-393,-110,81$ and $34$ respectively. Calculate $\Delta_{\mathrm{r}} H$ in $\mathrm{kJ}$ of the following reaction.
$2 \mathrm{NO}_2(\mathrm{g})+2 \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{N}_2 \mathrm{O}(\mathrm{g})+3 \mathrm{CO}_2(\mathrm{g})$

Two moles of an ideal gas $\left[C_{\mathrm{v}, \mathrm{m}}\left(/ \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)=20+0.01 T(/ \mathrm{K})\right]$ is heated at constant pressure from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}.$ The amount of heat absorbed by the gas is

Heat of dissociation of benzene of elements is 5335 kJ/mol. The bond enthalpies of  –C-C-;C=C and –C-H  bonds are 347.3, 615 and 416.2 kJ respectively. Resonance energy of benzene is

A hypothetical reaction, $X\xrightarrow{\,}\,2Y$ proceeds by the following sequence of steps
(i) $\frac{1}{2}X\xrightarrow{\,}\,Z;\,\,\,\,\,\,\,\Delta H = {q_1}$
(ii) $Z\xrightarrow{\,}\,2W;\,\,\,\,\,\,\,\Delta H = {q_2}$
(iii) $W\xrightarrow{\,}\,\frac{1}{2}Y;\,\,\,\,\,\,\,\Delta H = {q_3}$
The value of $\Delta H$ reaction is

The combustion of benzene $(l)$ gives $C{O_2}\left( g \right)$ and ${H_2}O\left( l \right)$. Given that heat of combustion of benzene at constant volume is $-3263.9\,kJ\,mo{l^{ - 1}}$ at $25^\circ C$, heat of combustion (in $kJ\,\,mo{l^{ - 1}}\mathrm{)}$ of benzene at constant pressure will be
$\left( {R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}} \right)$

The heat of combustion of hydrogen gas at constant pressure and at $23^0 \mathrm{C}$ is $280 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The heat of combustion at constant volume and at $23^0 \mathrm{C}$ will be $\left(\mathrm{R}=8.3143 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$

The heat of neutralization of $HCl$ by $NaOH$ is – 55.9 KJ/mol. If the heat of neutralization of $HCN$ by $NaOH$ is – 12.1 KJ/mol. The heat of dissociation of $HCN$ is

Statement I : In the following reaction, $CO(g) + 1/2O(g) \to C{O_2}(g);\Delta H = - 68\,\,k\,cal$ and thus, ${\Delta _f}H(C{O_2}) = - 68\,\,k\,cal\,\,mo{l^{ - 1}}$
Statement II : 1 mol of $C{O_2}$ is formed and enthalpy change is the heat of formation of $C{O_2}$