Thermodynamics practice · Mixed difficulty

Net work done by the system in a cyclic process is equal to:

${C_{\mathrm{diamond}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta H = - 395\mathrm{kJ} \ldots \ldots \mathrm{.}\left( \mathrm{i} \right)$
${C_{\mathrm{graphite}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta H = - 393.5\mathrm{kJ} \ldots \ldots \mathrm{.}\left( {\mathrm{ii}} \right)$ The ∆H, when diamond is formed from graphite, is :

Which equation is correct for adiabatic process

The difference between the heats of reaction at constant pressure and constant volume for the reaction, at a constant temperature $\mathrm{T}$ is

$2 \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+15 \mathrm{O}_2(\mathrm{~g}) \rightarrow 12 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$

Heat liberated by a given amount of an ideal gas undergoing reversible isothermal process is $1200 \, \mathrm{cal}$ at $300 \mathrm{~K}.$ What is the Gibbs free energy change of the gas in this process?

On the basis of the following thermochemical data $\left[\Delta_{\mathrm{f}}{G}^{\circ}\mathrm{H}^{+}(\mathrm{aq})=0\right]$
$\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) ;$ $\Delta \mathrm{H}=57.32 \mathrm{~kJ}$
$\begin{aligned} & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta \mathrm{H}=-286.02 \mathrm{~kJ}\end{aligned}$
The value of enthalpy of formation of $\mathrm{OH}^{-}$ion at $25^{\circ} \mathrm{C}$ is

Which fuel provides the highest calorific value?

Changes in a system from an initial state to the final state were made by a different manner that remains $\Delta H$ remains same but $q$ changes because

A solution of $500 \mathrm{~ml}$ of $2 \mathrm{~M}-{KOH}$ is added to $500 \mathrm{~ml}$ of $2 \mathrm{~M}-{HCl}$ and the mixture is well shaken. The rise in temperature $T_1$ is noted. The experiment is again performed using $250 \mathrm{~ml}$ of ${KOH}$ solution and $500 \mathrm{~ml}$ of ${HCl}$ solution and rise in temperature $T_2$ is again noted. Which of the following is correct?

The temperature of one mole of an ideal gas increases from $298 \mathrm{~K}$ when it absorbs $200 \mathrm{~J}$ of heat at constant volume. The change in the internal energy of the gas is

During complete combustion of one mole of butane, $2658 \mathrm{~kJ}$ of heat is released. The thermochemical reaction for above change is

Statement - I: In an adiabatic expansion of an ideal gas, the entropy of the surrounding
remains unchanged independent whether the expansion is reversible or irreversible.

Statement - II: In an adiabatic process, the surrounding does not exchange heat with the
system and the process may be reversible or irreversible for system but it always remain
reversible for the sorrounding.

For equilibrium $CO(g) + \frac{1}{2}{O_2}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}C{O_2}(g),$ when small amount of caustic soda $(NaOH)$ is put in container then following change will occur

White phosphorus is a tetra atomic solid $\mathrm{P}_4(\mathrm{s})$ at room temperature. Find average $(\mathrm{P}-\mathrm{P})$ bond enthalpy (in $\mathrm{kJ} / \mathrm{mol}$).
Given: $\Delta \mathrm{H}_{\text {sublimation}}$ of $\mathrm{P}_4(\mathrm{s})=59 \mathrm{~kJ} / \mathrm{mol}$
$\Delta \mathrm{H}_{\text {atomization}}$ of $\mathrm{P}_4(\mathrm{s})=1265 \mathrm{~kJ} / \mathrm{mol}$

The calorific value of fat is :

The entropy of the universe:

Assertion (A): The solubility of most salts in water increases with rise in temperature.
Reason (R): For most of the ionic compounds, $\Delta_{\text {sol }} H^{\circ}$ is positive and the dissociation process is endothermic.

Which law states entropy of all pure crystalline
solids is zero at absolute zero?

If values of $\Delta H^{\circ} \,\&\, \Delta S^{\circ}$ for a process/reaction are $77.2 \mathrm{~kJ} \,\&\, 48 \mathrm{~J} / \mathrm{k}$ respectively. Then find value of $\log \frac{1}{k}$. Given: Temp is $300 \mathrm{~K}$

One mole of a liquid ($1 \mathrm{~bar}, 100 \mathrm{~ml}$) is taken in an adiabatic container and the pressure increases steeply to $100 \mathrm{~bar}.$ Then at a constant pressure of $100 \mathrm{~bar},$ volume decrease by $1 \mathrm{~ml}$. Find $\Delta \mathrm{H}$.

A gas is allowed to expand in an insulated container against a constant external pressure of $3.5 \, {~atm}$ from an initial volume of $4.0 \, \mathrm{~L}$ to a final volume of $10.0 \,\mathrm{~L}$. The change in internal energy of the gas will be

For non ideal solution showing positive deviation from Rault's law :

Statement I : Enthalpy changes are positive when $N{a_2}S{O_4}.10\,{H_2}O,\,CuS{O_4}.5{H_2}O$ and salts like $NaCl$, $KCl$ etc., which do not form hydrates are dissolved in water. But enthalpy changes are negative when anhydrous salts capable of forming hydrates are dissolved in water
Statement II : The difference in the behaviour is due to large differences in the molecular weight of hydrated and anhydrous salts. The substances with larger molecular weights usually show positive enthalpy changes on dissolution.

In a flask, colourless ${\mathrm{N}_2}{\mathrm{O}_4}$ (g) is in equilibrium with brown coloured $\mathrm{N}{\mathrm{O}_2}$ (g). At equilibrium when the flask is heated to 100℃, the brown colour deepens and on cooling it becomes coloured. Which statement is incorrect about this observation?

The value of $\Delta \mathrm{H}$ for cooling 2 mole of an ideal monoatomic gas from $225^{\circ} \mathrm{C}$ to $125^{\circ} \mathrm{C}$ to at constant pressure will be $\left[\right.$ given $\left.C_p=\frac{5}{2} R\right]$

Calculate the difference between heat of combustion of carbon monoxide gas at constant pressure and at constant volume at $27^{\circ} \mathrm{C}$ ? $(\mathrm{R}=$ 2 cal K $\mathrm{mol}^{-1}$ )

For spontaneous process:

Statement (S): Heat given out during neutralisation of $\mathrm{NaOH}$ and $\mathrm{HF}$ is $-13.7 \mathrm{~k~cal} / \mathrm{eq}$.

Explanation (E): $\mathrm{F}^{-}$ion is more easily hydrated and thus heat of neutralisation of $\mathrm{HF}$ and $\mathrm{NaOH}$ is more.

Calculate $\Delta H$ for the following reaction at $298 \mathrm{~K},$

$\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{g})$

from the following thermochemical equations.

(i) $3 \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{s})+\mathrm{CO}_2(\mathrm{g});\Delta H=-46.4 \mathrm{~kJ}$

(ii) $\mathrm{FeO} \text{(s)} +\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Fe} \text{(s)} +\mathrm{CO}_2 \text{(g)}; \Delta H=9.0 \mathrm{~kJ}$

(iii) $\mathrm{Fe}_3 \mathrm{O}_4 \text{(s)} +\mathrm{CO}(\mathrm{g}) \rightarrow 3 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}); \Delta H=-41.0 \mathrm{~kJ}$

The correct thermodynamic conditions for the process which is spontaneous only at low temperature is