Answer

A thermodynamic state function is a quantity whose value is independent of a path.

Functions like $p, V, T$ etc. depend only on the state of a system and not on the path.

Hence, alternative (ii) is correct.

Answer

A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, $q=0$.

Therefore, alternative (iii) is correct.

Answer

The enthalpy of all elements in their standard state is zero.

Therefore, alternative (ii) is correct.

Answer

Since $ \Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_gRT \text{ and } \Delta U^{\ominus} = -X\ kJ \ mol^{-1} $

$\Delta H^{\ominus}=(-X)+\Delta n_g R T$.

For combustion of methane:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(l)$

$\Delta n_g = n_p - n_r$

$= 1-(2+1)= -2$

$\therefore \Delta H^{\ominus}= -X - 2RT$

$\Rightarrow \Delta H^{\ominus}<\Delta U^{\ominus}$

Therefore, alternative (iii) is correct.

Answer

According to the question, (i) $\quad CH_4 {(g)}+2 O_2 {(g)} \longrightarrow CO_2 {(g)}+2 H_2 O {(g)}$

$ \Delta H=-890.3\ kJ \ mol^{-1} $

(ii) $C {(s)}+O_2 {(g)} \longrightarrow CO_2 {(g)}$

$ \Delta H=-393.5\ kJ \ mol^{-1} $

(iii) $2 H_2 {(g)}+O_2 {(g)} \longrightarrow 2 H_2 O {(g)}$

$ \Delta H=-285.8\ kJ \ mol^{-1} $

Thus, the desired equation is the one that represents the formation of $CH_4$ (g) i.e.,

$C {(s)}+2 H_2 {(g)} \longrightarrow CH_4 {(g)}$

$\Delta_f H _{CH_4}=\Delta_c H_C+2 \Delta_c H _{H_2}-\Delta_c H _{CH_4}$

$=[-393.5+2(-285.8)-(-890.3)]\ kJ \ mol^{-1}$

$=-74.8\ kJ \ mol^{-1}$

$\therefore$ Enthalpy of formation of $CH _4 {(g)}=- 74.8\ kJ\ mol^{-1}$

Hence, alternative (i) is correct.

Answer

For a reaction to be spontaneous, $\Delta G$ should be negative.

$\Delta G=\Delta H-T \Delta S$

According to the question, for the given reaction,

$\Delta S=$ positive

$\Delta H=$ negative (since heat is evolved)

$\Rightarrow \Delta G=$ negative

Therefore, the reaction is spontaneous at any temperature.

Hence, alternative (iv) is correct.

Answer

According to the first law of thermodynamics,

$\Delta U=q+w \quad…(i)$

Where,

$\Delta U=$ change in internal energy for a process

$q=$ heat

$w=$ work

Given,

$q=+701\ J$ (Since heat is absorbed)

w= $-394\ J$ (Since work is done by the system)

Substituting the values in expression (i), we get

$\Delta U=701\ J+(-394\ J)$

$\Delta U=307\ J$

Hence, the change in internal energy for the given process is $307 J$.

Answer

Enthalpy change for a reaction $(\Delta H)$ is given by the expression,

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

Where,

$\Delta U=$ change in internal energy

$\Delta n_g=$ change in number of gaseous moles

For the given reaction,

$\Delta n_g=\sum n_g$ (products) - $\sum n_g$ (reactants)

=(2 - 1.5) moles

$\Delta n_g=0.5$ moles

And,

$\Delta U=-742.7\ kJ \ mol^{-1}$

$T=298 K$

$R=8.314 \times 10^{-3}\ kJ \ mol^{-1} K^{-1}$

Substituting the values in the expression of $\Delta H$ :

$\Delta H=(-742.7\ kJ \ mol^{-1})+(0.5\ mol)(298\ K)(8.314 \times 10^{-3}\ kJ \ mol^{-1} K^{-1})$

$=-742.7+1.2$

$\Delta H=-741.5\ kJ \ mol^{-1}$

Answer

From the expression of heat $(q)$,

$q=n . C_m . \Delta T$

Where,

$C_m=$ molar heat capacity

$n=$ no. of moles

$\Delta T=$ change in temperature

Substituting the values in the expression of $q$ :

$q=(\dfrac{60}{27} mol)(24\ J\ mol^{-1} K^{-1})(20\ K)$

$q=1066.7\ J$

$q=1.07\ kJ$

Answer

Total enthalpy change involved in the transformation is the sum of the following changes: (a) Energy change involved in the transformation of $1\ mol$ of water at $10^{\circ} C$ to $1 mol$ of water at $0^{\circ} C$.

(b) Energy change involved in the transformation of $1\ mol$ of water at $0^{\circ}$ to $1 mol$ of ice at $0^{\circ} C$.

(c) Energy change involved in the transformation of $1\ mol$ of ice at $0^{\circ} C$ to $1 mol$ of ice at $-10^{\circ} C$.

Total $\Delta H=C_p[H_2 O(l)] \Delta T+\Delta H _{\text{freezing }}+C_p[H_2 O {(s)}] \Delta T$

=$ (75.3\ J\ mol^{-1}K^{-1})(0-10)K + (-6.03 \times 10^3\ J\ mol^{-1} ) + (36.8\ J\ mol^{-1}K^{-1})(-10-0)K $

$=-7151\ J\ mol^{-1}$

Hence, the enthalpy change involved in the transformation is -$7.151\ kJ\ mol^{- 1}$.

Answer

Formation of $CO_2$ from carbon and dioxygen gas can be represented as:

$C {(s)}+O_2 {(g)} \longrightarrow CO_2 {(g)} \quad \Delta_f H=-393.5\ kJ \ mol^{-1}$

$(1$ mole $=44 g)$

Heat released on formation of $44\ g\ CO_2=- 393.5\ kJ\ mol^{-1}$

$\therefore$ Heat released on formation of $35.2\ g\ CO_2$

$=\dfrac{-393.5\ kJ \ mol^{-1}}{44 g} \times 35.2 g$

$=- 314.8\ kJ \ mol^{-1}$

Answer

$\Delta_r H$ for a reaction is defined as the difference between $\Delta_fH$ value of products and $\Delta_f H$ value of reactants.

$\Delta_r H=\sum \Delta_f H$ (products) $-\sum \Delta_f H$ (reactants)

For the given reaction,

$N_2 O _4 {(g)}+3 CO {(g)} \longrightarrow N_2 O {(g)}+3 CO _{2}{(g)}$

$\Delta_r H=[\Delta_f H(N_2 O)+3 \Delta_f H(CO_2)]-[\Delta_f H(N_2 O_4)+3 \Delta_f H(CO)]$

Substituting the values of $\Delta_f H$ for $N_2 O, CO_2, N_2 O_4$ and $CO$ from the question, we get:

$\Delta_r H=[81\ kJ \ mol^{-1}+3(-393)\ kJ \ mol^{-1}]-[9.7\ kJ \ mol^{-1}+3(-110)\ kJ \ mol^{-1}]$

$\Delta_r H=-777.7\ kJ \ mol^{-1}$

Hence, the value of $\Delta_r H$ for the reaction is $-777.7\ kJ \ mol^{-1}$.

Answer

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.

Re-writing the given equation for 1 mole of $NH_3 {(g)}$.

$\dfrac{1}{2} N _2 {(g)}+\dfrac{3}{2} H_2 {(g)} \longrightarrow NH_3 {(g)}$

$\therefore$ Standard enthalpy of formation of $NH_3 {(g)}$

$=1 / 2 \Delta_r H^{\theta}$

$=1 / 2(-92.4\ kJ\ mol^{- 1})$

$=- 46.2\ kJ \ mol^{-1}$

Answer

The reaction that takes place during the formation of $CH_3 OH{(l)}$ can be written as:

$C {(s)}+2 H_2 O {(g)}+\dfrac{1}{2} O_2 {(g)} \longrightarrow CH_3 OH _{(l)}\quad …(1)$

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) $+2 \times$ equation (iii) - equation (i)

$ \Delta _f H^{\ominus} [CH_3OH _{(l)}] = \Delta _cH^{\ominus} + 2\Delta _fH^{\ominus} [H _2O {(l)}] - \Delta _r H^{\ominus} $

$ = (-393\ kJ\ mol^{-1})+2(-286\ kJ \ mol^{-1})- (-726\ kJ \ mol^{-1}) $

$=(-393 -572+726)\ kJ \ mol^{-1}$

$\therefore \Delta_i H^{\ominus}[CH_3 OH {(l)}]=-239\ kJ\ mol^{-1}$

Answer

The chemical equations implying to the given values of enthalpies are:

(i) $CCl_4 {(l)} \longrightarrow CCl_4 {(g)}\ \Delta _{vap} H^{ \ominus }=30.5\ kJ\ mol^{-1 }$

(ii) $C {(s)} \longrightarrow C {(g)}\ \Delta _a H^{ \ominus }=715.0\ kJ \ mol^{-1}$

(iii) $Cl _2 {(g)} \longrightarrow 2 Cl {(g)}\ \Delta _a H^{ \ominus }=242\ kJ\ mol^{{-1}}$

(iv) $C {(s)}+2 Cl_2 {(g)} \longrightarrow CCl_4 {(l)}\ \Delta_f H=- 135.5\ kJ\ mol^{{-1}}$

Enthalpy change for the given process $CCl _4 {(g)} \longrightarrow C {(g)}+4 Cl {(g)}$, can be calculated using the following algebraic calculations as:

Equation (ii) +2 × Equation (iii) -Equation (i) - Equation (iv)

$\Delta_r H=\Delta _a H^{ \ominus }(C)+2 \Delta _a H^{ \ominus }(Cl_2) - \Delta _{\text{vap }} H^{\ominus} - \Delta_f H$

$\Delta_r H=715.0+(2 \times 242 )- 30.5 -(- 135.5)$

$\therefore \Delta H=1304\ kJ \ mol^{-1}$

Bond enthalpy of $C - Cl$ bond in $CCl _4 {(g)}$

$=\dfrac{1304}{4}\ kJ \ mol^{-1}$

$=326\ kJ\ mol^{- 1}$

Answer

$\Delta S=\dfrac{q_{rev}}{T}=\dfrac{\Delta H}{T}=\dfrac{\Delta U + P \Delta V}{T}$

$\Delta S= \dfrac{P \Delta V}{T}$

Since $\Delta U=0, \Delta S$ will be positive and the reaction will be spontaneous.

Answer

From the expression,

$\Delta G=\Delta H- T \Delta S$

Assuming the reaction at equilibrium, $\Delta T$ for the reaction would be:

$T=(\Delta H-\Delta G) \dfrac{1}{\Delta S}$

$=\dfrac{\Delta H}{\Delta S} {(\Delta G=0 \text{ at equilibrium })}$

$=\dfrac{400\ kJ \ mol^{-1}}{0.2\ kJ\ K^{-1} mol^{-1}}$

$T=2000 K$

For the reaction to be spontaneous, $\Delta G$ must be negative. Hence, for the given reaction to be spontaneous, Temperature should be greater than $2000 K$.

Answer

$\Delta H$ and $\Delta S$ are negative

The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, $\Delta H$ is negative.

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, $\Delta S$ is negative for the given reaction.

Answer

For the given reaction,

$2 A {(g)}+B {(g)} \to 2 D {(g)}$

$\Delta n_g=2 - 3$

$=-1$ mole

Substituting the value of $\Delta U^{\ominus}$, in the expression of $\Delta H$ :

$\Delta H^{\ominus}=\Delta U^{\ominus}+\Delta n_g R T$

$=(-10.5\ kJ)+(-1)(8.314 \times 10^{-3}\ kJ\ K^{-1} mol^{-1})(298\ K)$

$=-10.5\ kJ-2.48\ kJ$

$\Delta H^{\ominus}=-12.98\ kJ$

Substituting the values of $\Delta H^{\ominus}$, and $\Delta S^{\ominus}$, in the expression of $\Delta G^{\ominus}$ :

$ \Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\ominus} $

$=-12.98\ kJ-(298\ K)(-44.1\times 10^{-3}\ J K^{-1})$

$=-12.98\ kJ+13.14\ kJ$

$\Delta G^{\ominus}=+0.16\ kJ$

Since $\Delta G^{\ominus}$, for the reaction is positive, the reaction will not occur spontaneously.

Answer

From the expression,

$\Delta G^{\ominus}=-2.303 R T \log K _{e q}$

$\Delta G^{\ominus}$ for the reaction,

$=-(2.303)(8.314\ J\ K^{-1} mol^{-1})(300 K) \log 10$

$=-5744.14\ J\ mol^{-1}$

$=-5.744\ kJ \ mol^{-1}$

Answer

The positive value of $\Delta_r H$ indicates that heat is absorbed during the formation of $NO {(g)}$. This means that $NO {(g)}$ has higher energy than the reactants ($N_2$ and $O_2$ ). Hence, $NO {(g)}$ is unstable.

The negative value of $\Delta_r H$ indicates that heat is evolved during the formation of $NO _2 {(g)}$ from $NO {(g)}$ and $O_2 {(g)}$. The product, $NO_2 {(g)}$ is stabilized with minimum energy.

Hence, unstable $NO {(g)}$ changes to stable $NO_2 {(g)}$.

Answer

It is given that $286\ kJ\ mol^{{-1}}$ of heat is evolved on the formation of $1\ mol$ of $H_2 O (l)$. Thus, an equal amount of heat will be absorbed by the surroundings.

$q _{\text{surr }}=+286\ kJ\ mol^{- 1}$

Entropy change $(\Delta S _{\text{surr }})$ for the surroundings $=\dfrac{q _{\text{surr }}}{T}$ $=\dfrac{286\ kJ \ mol^{-1}}{298\ K}$ $\therefore \Delta S _{\text{surr }}=959.73\ J\ mol^{-1} K^{-1}$