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CHXI06:THERMODYNAMICS

369493 The heat of formation of $C_{2} H_{5} {OH}_{(l)}$ is $- 66 k c a l /$ mol. The heat of combustion of ${CH}_{3} {OCH}_{3 (g)}$ is $- 348 k c a l / m o l .$ $Δ H_{f}$ for $H_{2} O$ and ${CO}_{2}$ are $- 68$ $kcal / mol$ and $- 94 k c a l / m o l$ , respectively. Then, $Δ H$ for the isomerisation reaction ${CH}_{3} {OCH}_{3 (g)}$ is

1

18 k c a l / m o l

2

95 k c a l / m o l

3

22 k c a l / m o l

4

30 k c a l / m o l

Explanation:

$Δ H_{f}^{o} (C_{2} H_{5} {OH}_{(l)}) = - 66 kcal / mol$
$2 C + 3 H_{2} + \frac{1}{2} O_{2} \to C_{2} H_{5} O H; Δ H = - 66 k c a l (1)$
$C H_{3} - O - C H_{3} + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O; Δ H$
$= - 348 k c a l (2)$
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O; Δ H = - 68 k c a l$
$C + O_{2} \to {CO}_{2}; Δ H = - 94 kcal / mol$
$C_{2} H_{5} {OH}_{(1)} \to {CH}_{3} - O - {CH}_{3 (g)}; Δ H_{5} = ?$
$Δ H_{5} = 2 \times (4) + 3 \times (3) - (2) - (1)$
$Δ H = 22 k c a l / m o l$

CHXI06:THERMODYNAMICS

369494 Given,
$N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g);$
$- Δ H_{1}$
$N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g);$
$- Δ H_{2}$
$H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g);$
$Δ H_{3}$
The heat of formation of ${NCl}_{3} (g)$
in terms of $Δ H_{1}, Δ H_{2}$ and $Δ H_{3}$ is

1

Δ H_{f} = - Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

2

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

3

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} + \frac{3}{2} Δ H_{3}

4 None of the above

Explanation:

$I . N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g),$
$- Δ H_{1} (1)$
$I I . N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g),$
$- Δ H_{2} (2)$
$I I I . H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g),$
$Δ H_{3} (3)$
$O n m u l t i p l y i n g E q . (3) b y 3, w e g e t$
$3 H_{2} (g) + 3 C l_{2} (g)$ $⇌$ $6 H C l (g);$
$3 \times Δ H_{3} (4)$
$O n i n v e r t i n g E q . (4), w e g e t$
$6 H C l (g)$ $⇌$ $3 H_{2} (g) + 3 C l_{2} (g)$
$- 3 \times Δ H_{3} (5)$
$O n M u l t i p l y i n g E q . (1) w i t h (2), w e g e t$
$2 N H_{3} + 6 C l_{2}$ $⇌$ $2 N C l_{3} + 6 H C l$
$- 2 \times Δ H_{1} (6)$
$O n a d d i n g E q . (2), (5) a n d (6), w e g e t$
$N_{2} + 3 C l_{2}$ $⇌$ $2 N C l_{3};$
$Δ H_{r} = - Δ H_{2} - 3 \times Δ H_{3} - 2 \times Δ H_{1}$
$∴ Δ H_{f} = \frac{Δ H_{r}}{2} = - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3} - Δ H_{1}$
$= - Δ H_{1} - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

CHXI06:THERMODYNAMICS

369495 In the reaction,
$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$
heat of formation of ${SO}_{2}$ is

1

x - y

2

2 x + y

3

x + y

4

y - 2 x

Explanation:

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J (1)$
$S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J (2)$
$S + O_{2} \to S O_{2} + q k J (3)$
Equation (3) can be obtained by subtracting equation (2) from (1)
$\Rightarrow q = 2 x - ykJ$
Enthalpy of formation $(Δ H_{f}) = - q = - (2 x - y)$
$= y - 2 x$

CHXI06:THERMODYNAMICS

369496 Given
$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = + 117 . 0 k J m o l^{- 1}$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 . 0 k J m o l^{- 1}$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 . 0 k J m o l^{- 1}$
The heat of combustion of
${CS}_{2} + 3 O_{2} \to {CO}_{2} + 2 {SO}_{2}$ is

1

- 807 kJ {mol}^{- 1}

2

- 1104 kJ {mol}^{- 1}

3

+ 1104 kJ {mol}^{- 1}

4

+ 807 kJ {mol}^{- 1}

Explanation:

$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = 117 k J m o l^{- 1} (1)$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 k J m o l^{- 1} (2)$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 k J m o l^{- 1} (3)$
$2 \times$ Eqn. (iii) + Eqn.(ii) - Eqn.(i) gives
$Δ H = 2 (- 297) + (- 393) - 117 = - 1104 k J$

CHXI06:THERMODYNAMICS

369497 A sample of gas changes from $p_{1},$ $V_{1}$ and $T_{1}$ to $p_{2},$ $V_{2}$ and $T_{2}$ by one path and then back to $p_{1},$ $V_{1}$ and $T_{1}$ by another path. How many of the following must be zero for the gas in this cycle?
$Δ T, Δ p, Δ V, q . W$ and $Δ E,$

1 1

2 2

3 3

4 4

Explanation:

supporting img

Original states are restored.
Thus, $Δ p, Δ V, Δ T a n d Δ E = 0.$

CHXI06:THERMODYNAMICS

369493 The heat of formation of $C_{2} H_{5} {OH}_{(l)}$ is $- 66 k c a l /$ mol. The heat of combustion of ${CH}_{3} {OCH}_{3 (g)}$ is $- 348 k c a l / m o l .$ $Δ H_{f}$ for $H_{2} O$ and ${CO}_{2}$ are $- 68$ $kcal / mol$ and $- 94 k c a l / m o l$ , respectively. Then, $Δ H$ for the isomerisation reaction ${CH}_{3} {OCH}_{3 (g)}$ is

1

18 k c a l / m o l

2

95 k c a l / m o l

3

22 k c a l / m o l

4

30 k c a l / m o l

Explanation:

$Δ H_{f}^{o} (C_{2} H_{5} {OH}_{(l)}) = - 66 kcal / mol$
$2 C + 3 H_{2} + \frac{1}{2} O_{2} \to C_{2} H_{5} O H; Δ H = - 66 k c a l (1)$
$C H_{3} - O - C H_{3} + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O; Δ H$
$= - 348 k c a l (2)$
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O; Δ H = - 68 k c a l$
$C + O_{2} \to {CO}_{2}; Δ H = - 94 kcal / mol$
$C_{2} H_{5} {OH}_{(1)} \to {CH}_{3} - O - {CH}_{3 (g)}; Δ H_{5} = ?$
$Δ H_{5} = 2 \times (4) + 3 \times (3) - (2) - (1)$
$Δ H = 22 k c a l / m o l$

CHXI06:THERMODYNAMICS

369494 Given,
$N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g);$
$- Δ H_{1}$
$N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g);$
$- Δ H_{2}$
$H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g);$
$Δ H_{3}$
The heat of formation of ${NCl}_{3} (g)$
in terms of $Δ H_{1}, Δ H_{2}$ and $Δ H_{3}$ is

1

Δ H_{f} = - Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

2

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

3

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} + \frac{3}{2} Δ H_{3}

4 None of the above

Explanation:

$I . N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g),$
$- Δ H_{1} (1)$
$I I . N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g),$
$- Δ H_{2} (2)$
$I I I . H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g),$
$Δ H_{3} (3)$
$O n m u l t i p l y i n g E q . (3) b y 3, w e g e t$
$3 H_{2} (g) + 3 C l_{2} (g)$ $⇌$ $6 H C l (g);$
$3 \times Δ H_{3} (4)$
$O n i n v e r t i n g E q . (4), w e g e t$
$6 H C l (g)$ $⇌$ $3 H_{2} (g) + 3 C l_{2} (g)$
$- 3 \times Δ H_{3} (5)$
$O n M u l t i p l y i n g E q . (1) w i t h (2), w e g e t$
$2 N H_{3} + 6 C l_{2}$ $⇌$ $2 N C l_{3} + 6 H C l$
$- 2 \times Δ H_{1} (6)$
$O n a d d i n g E q . (2), (5) a n d (6), w e g e t$
$N_{2} + 3 C l_{2}$ $⇌$ $2 N C l_{3};$
$Δ H_{r} = - Δ H_{2} - 3 \times Δ H_{3} - 2 \times Δ H_{1}$
$∴ Δ H_{f} = \frac{Δ H_{r}}{2} = - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3} - Δ H_{1}$
$= - Δ H_{1} - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

CHXI06:THERMODYNAMICS

369495 In the reaction,
$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$
heat of formation of ${SO}_{2}$ is

1

x - y

2

2 x + y

3

x + y

4

y - 2 x

Explanation:

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J (1)$
$S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J (2)$
$S + O_{2} \to S O_{2} + q k J (3)$
Equation (3) can be obtained by subtracting equation (2) from (1)
$\Rightarrow q = 2 x - ykJ$
Enthalpy of formation $(Δ H_{f}) = - q = - (2 x - y)$
$= y - 2 x$

CHXI06:THERMODYNAMICS

369496 Given
$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = + 117 . 0 k J m o l^{- 1}$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 . 0 k J m o l^{- 1}$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 . 0 k J m o l^{- 1}$
The heat of combustion of
${CS}_{2} + 3 O_{2} \to {CO}_{2} + 2 {SO}_{2}$ is

1

- 807 kJ {mol}^{- 1}

2

- 1104 kJ {mol}^{- 1}

3

+ 1104 kJ {mol}^{- 1}

4

+ 807 kJ {mol}^{- 1}

Explanation:

$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = 117 k J m o l^{- 1} (1)$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 k J m o l^{- 1} (2)$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 k J m o l^{- 1} (3)$
$2 \times$ Eqn. (iii) + Eqn.(ii) - Eqn.(i) gives
$Δ H = 2 (- 297) + (- 393) - 117 = - 1104 k J$

CHXI06:THERMODYNAMICS

369497 A sample of gas changes from $p_{1},$ $V_{1}$ and $T_{1}$ to $p_{2},$ $V_{2}$ and $T_{2}$ by one path and then back to $p_{1},$ $V_{1}$ and $T_{1}$ by another path. How many of the following must be zero for the gas in this cycle?
$Δ T, Δ p, Δ V, q . W$ and $Δ E,$

1 1

2 2

3 3

4 4

Explanation:

supporting img

Original states are restored.
Thus, $Δ p, Δ V, Δ T a n d Δ E = 0.$

CHXI06:THERMODYNAMICS

369493 The heat of formation of $C_{2} H_{5} {OH}_{(l)}$ is $- 66 k c a l /$ mol. The heat of combustion of ${CH}_{3} {OCH}_{3 (g)}$ is $- 348 k c a l / m o l .$ $Δ H_{f}$ for $H_{2} O$ and ${CO}_{2}$ are $- 68$ $kcal / mol$ and $- 94 k c a l / m o l$ , respectively. Then, $Δ H$ for the isomerisation reaction ${CH}_{3} {OCH}_{3 (g)}$ is

1

18 k c a l / m o l

2

95 k c a l / m o l

3

22 k c a l / m o l

4

30 k c a l / m o l

Explanation:

$Δ H_{f}^{o} (C_{2} H_{5} {OH}_{(l)}) = - 66 kcal / mol$
$2 C + 3 H_{2} + \frac{1}{2} O_{2} \to C_{2} H_{5} O H; Δ H = - 66 k c a l (1)$
$C H_{3} - O - C H_{3} + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O; Δ H$
$= - 348 k c a l (2)$
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O; Δ H = - 68 k c a l$
$C + O_{2} \to {CO}_{2}; Δ H = - 94 kcal / mol$
$C_{2} H_{5} {OH}_{(1)} \to {CH}_{3} - O - {CH}_{3 (g)}; Δ H_{5} = ?$
$Δ H_{5} = 2 \times (4) + 3 \times (3) - (2) - (1)$
$Δ H = 22 k c a l / m o l$

CHXI06:THERMODYNAMICS

369494 Given,
$N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g);$
$- Δ H_{1}$
$N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g);$
$- Δ H_{2}$
$H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g);$
$Δ H_{3}$
The heat of formation of ${NCl}_{3} (g)$
in terms of $Δ H_{1}, Δ H_{2}$ and $Δ H_{3}$ is

1

Δ H_{f} = - Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

2

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

3

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} + \frac{3}{2} Δ H_{3}

4 None of the above

Explanation:

$I . N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g),$
$- Δ H_{1} (1)$
$I I . N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g),$
$- Δ H_{2} (2)$
$I I I . H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g),$
$Δ H_{3} (3)$
$O n m u l t i p l y i n g E q . (3) b y 3, w e g e t$
$3 H_{2} (g) + 3 C l_{2} (g)$ $⇌$ $6 H C l (g);$
$3 \times Δ H_{3} (4)$
$O n i n v e r t i n g E q . (4), w e g e t$
$6 H C l (g)$ $⇌$ $3 H_{2} (g) + 3 C l_{2} (g)$
$- 3 \times Δ H_{3} (5)$
$O n M u l t i p l y i n g E q . (1) w i t h (2), w e g e t$
$2 N H_{3} + 6 C l_{2}$ $⇌$ $2 N C l_{3} + 6 H C l$
$- 2 \times Δ H_{1} (6)$
$O n a d d i n g E q . (2), (5) a n d (6), w e g e t$
$N_{2} + 3 C l_{2}$ $⇌$ $2 N C l_{3};$
$Δ H_{r} = - Δ H_{2} - 3 \times Δ H_{3} - 2 \times Δ H_{1}$
$∴ Δ H_{f} = \frac{Δ H_{r}}{2} = - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3} - Δ H_{1}$
$= - Δ H_{1} - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

CHXI06:THERMODYNAMICS

369495 In the reaction,
$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$
heat of formation of ${SO}_{2}$ is

1

x - y

2

2 x + y

3

x + y

4

y - 2 x

Explanation:

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J (1)$
$S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J (2)$
$S + O_{2} \to S O_{2} + q k J (3)$
Equation (3) can be obtained by subtracting equation (2) from (1)
$\Rightarrow q = 2 x - ykJ$
Enthalpy of formation $(Δ H_{f}) = - q = - (2 x - y)$
$= y - 2 x$

CHXI06:THERMODYNAMICS

369496 Given
$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = + 117 . 0 k J m o l^{- 1}$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 . 0 k J m o l^{- 1}$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 . 0 k J m o l^{- 1}$
The heat of combustion of
${CS}_{2} + 3 O_{2} \to {CO}_{2} + 2 {SO}_{2}$ is

1

- 807 kJ {mol}^{- 1}

2

- 1104 kJ {mol}^{- 1}

3

+ 1104 kJ {mol}^{- 1}

4

+ 807 kJ {mol}^{- 1}

Explanation:

$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = 117 k J m o l^{- 1} (1)$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 k J m o l^{- 1} (2)$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 k J m o l^{- 1} (3)$
$2 \times$ Eqn. (iii) + Eqn.(ii) - Eqn.(i) gives
$Δ H = 2 (- 297) + (- 393) - 117 = - 1104 k J$

CHXI06:THERMODYNAMICS

369497 A sample of gas changes from $p_{1},$ $V_{1}$ and $T_{1}$ to $p_{2},$ $V_{2}$ and $T_{2}$ by one path and then back to $p_{1},$ $V_{1}$ and $T_{1}$ by another path. How many of the following must be zero for the gas in this cycle?
$Δ T, Δ p, Δ V, q . W$ and $Δ E,$

1 1

2 2

3 3

4 4

Explanation:

supporting img

Original states are restored.
Thus, $Δ p, Δ V, Δ T a n d Δ E = 0.$

CHXI06:THERMODYNAMICS

369493 The heat of formation of $C_{2} H_{5} {OH}_{(l)}$ is $- 66 k c a l /$ mol. The heat of combustion of ${CH}_{3} {OCH}_{3 (g)}$ is $- 348 k c a l / m o l .$ $Δ H_{f}$ for $H_{2} O$ and ${CO}_{2}$ are $- 68$ $kcal / mol$ and $- 94 k c a l / m o l$ , respectively. Then, $Δ H$ for the isomerisation reaction ${CH}_{3} {OCH}_{3 (g)}$ is

1

18 k c a l / m o l

2

95 k c a l / m o l

3

22 k c a l / m o l

4

30 k c a l / m o l

Explanation:

$Δ H_{f}^{o} (C_{2} H_{5} {OH}_{(l)}) = - 66 kcal / mol$
$2 C + 3 H_{2} + \frac{1}{2} O_{2} \to C_{2} H_{5} O H; Δ H = - 66 k c a l (1)$
$C H_{3} - O - C H_{3} + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O; Δ H$
$= - 348 k c a l (2)$
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O; Δ H = - 68 k c a l$
$C + O_{2} \to {CO}_{2}; Δ H = - 94 kcal / mol$
$C_{2} H_{5} {OH}_{(1)} \to {CH}_{3} - O - {CH}_{3 (g)}; Δ H_{5} = ?$
$Δ H_{5} = 2 \times (4) + 3 \times (3) - (2) - (1)$
$Δ H = 22 k c a l / m o l$

CHXI06:THERMODYNAMICS

369494 Given,
$N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g);$
$- Δ H_{1}$
$N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g);$
$- Δ H_{2}$
$H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g);$
$Δ H_{3}$
The heat of formation of ${NCl}_{3} (g)$
in terms of $Δ H_{1}, Δ H_{2}$ and $Δ H_{3}$ is

1

Δ H_{f} = - Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

2

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

3

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} + \frac{3}{2} Δ H_{3}

4 None of the above

Explanation:

$I . N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g),$
$- Δ H_{1} (1)$
$I I . N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g),$
$- Δ H_{2} (2)$
$I I I . H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g),$
$Δ H_{3} (3)$
$O n m u l t i p l y i n g E q . (3) b y 3, w e g e t$
$3 H_{2} (g) + 3 C l_{2} (g)$ $⇌$ $6 H C l (g);$
$3 \times Δ H_{3} (4)$
$O n i n v e r t i n g E q . (4), w e g e t$
$6 H C l (g)$ $⇌$ $3 H_{2} (g) + 3 C l_{2} (g)$
$- 3 \times Δ H_{3} (5)$
$O n M u l t i p l y i n g E q . (1) w i t h (2), w e g e t$
$2 N H_{3} + 6 C l_{2}$ $⇌$ $2 N C l_{3} + 6 H C l$
$- 2 \times Δ H_{1} (6)$
$O n a d d i n g E q . (2), (5) a n d (6), w e g e t$
$N_{2} + 3 C l_{2}$ $⇌$ $2 N C l_{3};$
$Δ H_{r} = - Δ H_{2} - 3 \times Δ H_{3} - 2 \times Δ H_{1}$
$∴ Δ H_{f} = \frac{Δ H_{r}}{2} = - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3} - Δ H_{1}$
$= - Δ H_{1} - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

CHXI06:THERMODYNAMICS

369495 In the reaction,
$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$
heat of formation of ${SO}_{2}$ is

1

x - y

2

2 x + y

3

x + y

4

y - 2 x

Explanation:

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J (1)$
$S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J (2)$
$S + O_{2} \to S O_{2} + q k J (3)$
Equation (3) can be obtained by subtracting equation (2) from (1)
$\Rightarrow q = 2 x - ykJ$
Enthalpy of formation $(Δ H_{f}) = - q = - (2 x - y)$
$= y - 2 x$

CHXI06:THERMODYNAMICS

369496 Given
$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = + 117 . 0 k J m o l^{- 1}$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 . 0 k J m o l^{- 1}$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 . 0 k J m o l^{- 1}$
The heat of combustion of
${CS}_{2} + 3 O_{2} \to {CO}_{2} + 2 {SO}_{2}$ is

1

- 807 kJ {mol}^{- 1}

2

- 1104 kJ {mol}^{- 1}

3

+ 1104 kJ {mol}^{- 1}

4

+ 807 kJ {mol}^{- 1}

Explanation:

$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = 117 k J m o l^{- 1} (1)$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 k J m o l^{- 1} (2)$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 k J m o l^{- 1} (3)$
$2 \times$ Eqn. (iii) + Eqn.(ii) - Eqn.(i) gives
$Δ H = 2 (- 297) + (- 393) - 117 = - 1104 k J$

CHXI06:THERMODYNAMICS

369497 A sample of gas changes from $p_{1},$ $V_{1}$ and $T_{1}$ to $p_{2},$ $V_{2}$ and $T_{2}$ by one path and then back to $p_{1},$ $V_{1}$ and $T_{1}$ by another path. How many of the following must be zero for the gas in this cycle?
$Δ T, Δ p, Δ V, q . W$ and $Δ E,$

1 1

2 2

3 3

4 4

Explanation:

supporting img

Original states are restored.
Thus, $Δ p, Δ V, Δ T a n d Δ E = 0.$

CHXI06:THERMODYNAMICS

369493 The heat of formation of $C_{2} H_{5} {OH}_{(l)}$ is $- 66 k c a l /$ mol. The heat of combustion of ${CH}_{3} {OCH}_{3 (g)}$ is $- 348 k c a l / m o l .$ $Δ H_{f}$ for $H_{2} O$ and ${CO}_{2}$ are $- 68$ $kcal / mol$ and $- 94 k c a l / m o l$ , respectively. Then, $Δ H$ for the isomerisation reaction ${CH}_{3} {OCH}_{3 (g)}$ is

1

18 k c a l / m o l

2

95 k c a l / m o l

3

22 k c a l / m o l

4

30 k c a l / m o l

Explanation:

$Δ H_{f}^{o} (C_{2} H_{5} {OH}_{(l)}) = - 66 kcal / mol$
$2 C + 3 H_{2} + \frac{1}{2} O_{2} \to C_{2} H_{5} O H; Δ H = - 66 k c a l (1)$
$C H_{3} - O - C H_{3} + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O; Δ H$
$= - 348 k c a l (2)$
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O; Δ H = - 68 k c a l$
$C + O_{2} \to {CO}_{2}; Δ H = - 94 kcal / mol$
$C_{2} H_{5} {OH}_{(1)} \to {CH}_{3} - O - {CH}_{3 (g)}; Δ H_{5} = ?$
$Δ H_{5} = 2 \times (4) + 3 \times (3) - (2) - (1)$
$Δ H = 22 k c a l / m o l$

CHXI06:THERMODYNAMICS

369494 Given,
$N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g);$
$- Δ H_{1}$
$N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g);$
$- Δ H_{2}$
$H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g);$
$Δ H_{3}$
The heat of formation of ${NCl}_{3} (g)$
in terms of $Δ H_{1}, Δ H_{2}$ and $Δ H_{3}$ is

1

Δ H_{f} = - Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

2

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

3

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} + \frac{3}{2} Δ H_{3}

4 None of the above

Explanation:

$I . N H_{3} (g) + 3 C l_{2} (g)$ $⇌$ $N C l_{3} (g) + 3 H C l (g),$
$- Δ H_{1} (1)$
$I I . N_{2} (g) + 3 H_{2} (g)$ $⇌$ $2 N H_{3} (g),$
$- Δ H_{2} (2)$
$I I I . H_{2} (g) + C l_{2} (g)$ $⇌$ $2 H C l (g),$
$Δ H_{3} (3)$
$O n m u l t i p l y i n g E q . (3) b y 3, w e g e t$
$3 H_{2} (g) + 3 C l_{2} (g)$ $⇌$ $6 H C l (g);$
$3 \times Δ H_{3} (4)$
$O n i n v e r t i n g E q . (4), w e g e t$
$6 H C l (g)$ $⇌$ $3 H_{2} (g) + 3 C l_{2} (g)$
$- 3 \times Δ H_{3} (5)$
$O n M u l t i p l y i n g E q . (1) w i t h (2), w e g e t$
$2 N H_{3} + 6 C l_{2}$ $⇌$ $2 N C l_{3} + 6 H C l$
$- 2 \times Δ H_{1} (6)$
$O n a d d i n g E q . (2), (5) a n d (6), w e g e t$
$N_{2} + 3 C l_{2}$ $⇌$ $2 N C l_{3};$
$Δ H_{r} = - Δ H_{2} - 3 \times Δ H_{3} - 2 \times Δ H_{1}$
$∴ Δ H_{f} = \frac{Δ H_{r}}{2} = - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3} - Δ H_{1}$
$= - Δ H_{1} - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

CHXI06:THERMODYNAMICS

369495 In the reaction,
$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$
heat of formation of ${SO}_{2}$ is

1

x - y

2

2 x + y

3

x + y

4

y - 2 x

Explanation:

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J (1)$
$S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J (2)$
$S + O_{2} \to S O_{2} + q k J (3)$
Equation (3) can be obtained by subtracting equation (2) from (1)
$\Rightarrow q = 2 x - ykJ$
Enthalpy of formation $(Δ H_{f}) = - q = - (2 x - y)$
$= y - 2 x$

CHXI06:THERMODYNAMICS

369496 Given
$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = + 117 . 0 k J m o l^{- 1}$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 . 0 k J m o l^{- 1}$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 . 0 k J m o l^{- 1}$
The heat of combustion of
${CS}_{2} + 3 O_{2} \to {CO}_{2} + 2 {SO}_{2}$ is

1

- 807 kJ {mol}^{- 1}

2

- 1104 kJ {mol}^{- 1}

3

+ 1104 kJ {mol}^{- 1}

4

+ 807 kJ {mol}^{- 1}

Explanation:

$C + 2 S \to C S_{2}, Δ_{f} H^{\circ} = 117 k J m o l^{- 1} (1)$
$C + O_{2} \to C O_{2}, Δ_{f} H^{\circ} = - 393 k J m o l^{- 1} (2)$
$S + O_{2} \to S O_{2}, Δ_{f} H^{\circ} = - 297 k J m o l^{- 1} (3)$
$2 \times$ Eqn. (iii) + Eqn.(ii) - Eqn.(i) gives
$Δ H = 2 (- 297) + (- 393) - 117 = - 1104 k J$

CHXI06:THERMODYNAMICS

369497 A sample of gas changes from $p_{1},$ $V_{1}$ and $T_{1}$ to $p_{2},$ $V_{2}$ and $T_{2}$ by one path and then back to $p_{1},$ $V_{1}$ and $T_{1}$ by another path. How many of the following must be zero for the gas in this cycle?
$Δ T, Δ p, Δ V, q . W$ and $Δ E,$

1 1

2 2

3 3

4 4

Explanation:

supporting img

Original states are restored.
Thus, $Δ p, Δ V, Δ T a n d Δ E = 0.$

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