Answer

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

Functions like $p, V$, Tetc. 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 KJmol^{-1} $

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

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

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).e.,

$C _{(s)}+2 H _{2(g)} \longrightarrow CH _{4(g)}$

$\Delta_f H _{CH_4}=\Delta_c H_c+2 \Delta_c H _{H_2}-\Delta_c H _{CO_2}$

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

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

$\therefore$ Enthalpy of formation of $CH _{4(g)}=- 74.8 kJ$ mol

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,

$\Delta U=q+W(i)$

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 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 $(q)$,

$q=m . c . \Delta T$

Where,

$c=$ molar heat capacity

$m=$ mass of substance

$\Delta T=$ change in temperature

Substituting the values in the expression of $q$ :

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

$q=1066.7 J$

$q=1.07 kJ$

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

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

=$ 75.3 J mol^{-1}-6.03 \times 10^3J mol^{-1} -36.8 Jmol^{-1}$

$=-7151 Jmol^{-1}$

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 $44 g CO_2=- 393.5 kJmol^{-1}$

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

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

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

Answer

$\Delta_i H$ for a reaction is defined as the difference between $\Delta_i$ Hvalue of products and $\Delta_i H$ 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 $\Delta_4 H$ 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 $NH _{3(g)}$.

$\frac{1}{2} N _{2(g)}+\frac{3}{2} H _{2(g)} \longrightarrow NH _{3(g)}$

$\therefore$ 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 $CH_3 OH _{(1)}$ can be written as:

$C _{(s)}+2 H_2 O _{(g)}+\frac{1}{2} O _{2(g)} \longrightarrow CH_3 OH _{()(1)}$

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 KJmol^{-1})+2(-286 KJ mol^{-1})- (-726 KJ mol^{-1}) $

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

$\therefore \Delta_i H^{\oplus}[CH_3 OH _{(0)}]=-239 kJmol^{-1}$

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) $C _{(g)}+4 Cl _{(g)} \longrightarrow CCl _{4(g)} \Delta_i H=- 135.5 kJ mol^{{-1}}$

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)

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

$ \Delta _fH^{\ominus} [CH _3OH _{(l)}] = \Delta _cH^{\ominus} + 2\Delta _fH^{\ominus} [H _2O _{(l)}]- \Delta _r H^{\ominus}$

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

Bond enthalpy of $C - Cl$ bond in $CCl _{4(g)}$

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

$=326 kJ mol^{- 1}$

Answer

$\Delta S$ will be positive i.e., greater than zero

Since $\Delta U=0, \Delta$ Swill 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:

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

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

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

$T=2000 K$

For the reaction to be spontaneous, $\Delta$ Gmust be negative. Hence, for the given reaction to be spontaneous, Tshould 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, $\Delta$ His negative.

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, $\Delta$ Sis 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} $

$=-12.98 kJ-(298 K)(-44.1 J K^{-1})$

$=-12.98 kJ+13.14 kJ$

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

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,

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

$=-5744.14 Jmol^{-1}$

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

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 $H_2 O _{(\jmath)}$. 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 $=\frac{q _{\text{surr }}}{7}$ $=\frac{286 kJ mol^{-1}}{298 k}$ $\therefore \Delta S _{\text{surr }}=959.73 J mol^{-1} K^{-1}$