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CHXI06:THERMODYNAMICS

369186 One gram sample of ${NH}_{4} {NO}_{3}$ is decomposed in a bomb calorimeter. The temperature of the calorimeter increased by 6.12 K . The heat capacity of the system is $1.23 kJ / g^{- 1} K^{- 1}$ . What is the molar heat of decomposition for ${NH}_{4} {NO}_{3}$ .

1

7.53 kJ / mol

2

398.1 kJ / mol

3

16.1 kJ / mol

4

602 kJ / mol

Explanation:

The heat evolved, $q = mS Δ T$
$m = \lg; S = 1.23 kJ . g^{-} K^{-}$ ;
$Δ T = 6.12 K$
When 1 g of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 kJ /$ gram
When 1 mole of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 \times 80 kJ /$ $mol = 602.21 kJ / mol$
$({NH}_{4} {NO}_{3}$ molar mass is 80 g.mol $^{-})$

CHXI06:THERMODYNAMICS

369187 $1 g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 K$ and $1 atm$ . pressure according to the equation
$C_{(graphite)} + O_{2 (g)} \to {CO}_{2 (g)}$
During the reaction, temperature rises from 298 $K$ to $299 K$ . If the heat capacity of the bomb calorimeter is $20.7 kJ / K$ , what is the enthalpy change for the above reaction at $298 K$ and 1 atm?

1

20.7 kJ {mol}^{- 1}

2

- 20.7 kJ {mol}^{- 1}

3

- 248 kJ {mol}^{- 1}

4

248 kJ {mol}^{- 1}

Explanation:

Suppose $q$ is the quantity of heat from the reaction mixture and $C_{V}$ is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
$q = C_{V} \times Δ T$
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
$q = - C_{v} \times Δ T = - 20.7 kJ K^{- 1} \times (299 - 298) K$ $= - 20.7 kJ$
(Here, negative sign indicates the exothermic nature of the reaction)
Thus, $Δ U$ for the combustion of the $1 g$ of graphite $= - 20.7 kJ K^{- 1}$
For combustion of $1 mol$ of graphite,
$= \frac{12.0 g {mol}^{- 1} \times (- 20.7 kJ)}{1 g}$
$= - 2.48 \times 10^{2} kJ {mol}^{- 1}$ , Since $Δ n_{g} = 0$ ,
$Δ H = Δ U = - 2.48 \times 10^{2} kJ {mol}^{- 1}$
$= - 248 k J m o l^{- 1}$

CHXI06:THERMODYNAMICS

369188 $Δ_{f} H^{\circ}$ of hypothetical $MX$ is $- 250 kJ {mol}^{- 1}$ and for ${MX}_{2}$ is $- 600 kJ {mol}^{- 1}$ . The enthalpy of disproportionation of $MX$ is $- 100 x k J m o l^{- 1} .$ Find the value of x.

1

-

100

2

-

200

3

+

100

4

+

200

Explanation:

(3) $2 MX \to M + MX, : Δ H^{\circ} =$ ?
$M (s) + \frac{1}{2} \to 2 \times MX : Δ_{f} H_{1}^{\circ} = - 250 kJ mo l^{- 1}$
$M (s) + X_{2} \to M X_{2}; Δ_{f} H_{2}^{-} = - 600 kJ mo l^{- 1}$
$∴ Δ H^{\circ} = Δ_{f} H_{2}^{\circ} - 2 Δ_{f} H_{1}^{\circ}$
$= - 600 - (2 \times 250) = - 100 kJ mo l^{- 1}$
Enthalpy of disproportionation is $-$ 100.

CHXI06:THERMODYNAMICS

369189 $4 g$ of graphite is burnt in a bomb calorimeter of heat capacity $30 kJ K^{- 1}$ in excess of oxygen at $1 atm$ pressure. The temperature rises from 300 $K$ to $304 K$ . What is the enthalpy of combustion of graphite $(i n kJ mo l^{- 1})$ ?

1

360

2

1440

3

- 360

4

- 1440

Explanation:

$Δ E = C \times Δ t \times \frac{M}{m} = 30 \times 4 \times \frac{12}{4} = 360$
$∴ Δ E = - 360 kJ / mol$ (as it is combustion, sign is $- ve$ )
for $\underset{(s)}{C} + \underset{(g)}{O_{2}} \to \underset{(g)}{{CO}_{2}}; \underset{(g)}{Δ n} = 1 - 1 = 0$
$∴ Δ H = Δ E = - 360 kJ / mol$

CHXI06:THERMODYNAMICS

369190 When 0.50 g of unknown carbon compound is burned in the bomb calorimeter, its temperature rises by $6.76^{\circ} C$ . How much energy (in kilojoules) is released during combustion? (Heat capacity of calorimeter $= 3.64 {kJ}^{\circ} C^{- 1} g^{- 1}$ )

1 6.15

2 12.3

3 0.68

4 26

Explanation:

Heat evolved $= m \times C_{v} \times Δ T$ $= 0.50 \times 3.64 \times 6.76 = 12.30 kJ$

CHXI06:THERMODYNAMICS

369186 One gram sample of ${NH}_{4} {NO}_{3}$ is decomposed in a bomb calorimeter. The temperature of the calorimeter increased by 6.12 K . The heat capacity of the system is $1.23 kJ / g^{- 1} K^{- 1}$ . What is the molar heat of decomposition for ${NH}_{4} {NO}_{3}$ .

1

7.53 kJ / mol

2

398.1 kJ / mol

3

16.1 kJ / mol

4

602 kJ / mol

Explanation:

The heat evolved, $q = mS Δ T$
$m = \lg; S = 1.23 kJ . g^{-} K^{-}$ ;
$Δ T = 6.12 K$
When 1 g of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 kJ /$ gram
When 1 mole of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 \times 80 kJ /$ $mol = 602.21 kJ / mol$
$({NH}_{4} {NO}_{3}$ molar mass is 80 g.mol $^{-})$

CHXI06:THERMODYNAMICS

369187 $1 g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 K$ and $1 atm$ . pressure according to the equation
$C_{(graphite)} + O_{2 (g)} \to {CO}_{2 (g)}$
During the reaction, temperature rises from 298 $K$ to $299 K$ . If the heat capacity of the bomb calorimeter is $20.7 kJ / K$ , what is the enthalpy change for the above reaction at $298 K$ and 1 atm?

1

20.7 kJ {mol}^{- 1}

2

- 20.7 kJ {mol}^{- 1}

3

- 248 kJ {mol}^{- 1}

4

248 kJ {mol}^{- 1}

Explanation:

Suppose $q$ is the quantity of heat from the reaction mixture and $C_{V}$ is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
$q = C_{V} \times Δ T$
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
$q = - C_{v} \times Δ T = - 20.7 kJ K^{- 1} \times (299 - 298) K$ $= - 20.7 kJ$
(Here, negative sign indicates the exothermic nature of the reaction)
Thus, $Δ U$ for the combustion of the $1 g$ of graphite $= - 20.7 kJ K^{- 1}$
For combustion of $1 mol$ of graphite,
$= \frac{12.0 g {mol}^{- 1} \times (- 20.7 kJ)}{1 g}$
$= - 2.48 \times 10^{2} kJ {mol}^{- 1}$ , Since $Δ n_{g} = 0$ ,
$Δ H = Δ U = - 2.48 \times 10^{2} kJ {mol}^{- 1}$
$= - 248 k J m o l^{- 1}$

CHXI06:THERMODYNAMICS

369188 $Δ_{f} H^{\circ}$ of hypothetical $MX$ is $- 250 kJ {mol}^{- 1}$ and for ${MX}_{2}$ is $- 600 kJ {mol}^{- 1}$ . The enthalpy of disproportionation of $MX$ is $- 100 x k J m o l^{- 1} .$ Find the value of x.

1

-

100

2

-

200

3

+

100

4

+

200

Explanation:

(3) $2 MX \to M + MX, : Δ H^{\circ} =$ ?
$M (s) + \frac{1}{2} \to 2 \times MX : Δ_{f} H_{1}^{\circ} = - 250 kJ mo l^{- 1}$
$M (s) + X_{2} \to M X_{2}; Δ_{f} H_{2}^{-} = - 600 kJ mo l^{- 1}$
$∴ Δ H^{\circ} = Δ_{f} H_{2}^{\circ} - 2 Δ_{f} H_{1}^{\circ}$
$= - 600 - (2 \times 250) = - 100 kJ mo l^{- 1}$
Enthalpy of disproportionation is $-$ 100.

CHXI06:THERMODYNAMICS

369189 $4 g$ of graphite is burnt in a bomb calorimeter of heat capacity $30 kJ K^{- 1}$ in excess of oxygen at $1 atm$ pressure. The temperature rises from 300 $K$ to $304 K$ . What is the enthalpy of combustion of graphite $(i n kJ mo l^{- 1})$ ?

1

360

2

1440

3

- 360

4

- 1440

Explanation:

$Δ E = C \times Δ t \times \frac{M}{m} = 30 \times 4 \times \frac{12}{4} = 360$
$∴ Δ E = - 360 kJ / mol$ (as it is combustion, sign is $- ve$ )
for $\underset{(s)}{C} + \underset{(g)}{O_{2}} \to \underset{(g)}{{CO}_{2}}; \underset{(g)}{Δ n} = 1 - 1 = 0$
$∴ Δ H = Δ E = - 360 kJ / mol$

CHXI06:THERMODYNAMICS

369190 When 0.50 g of unknown carbon compound is burned in the bomb calorimeter, its temperature rises by $6.76^{\circ} C$ . How much energy (in kilojoules) is released during combustion? (Heat capacity of calorimeter $= 3.64 {kJ}^{\circ} C^{- 1} g^{- 1}$ )

1 6.15

2 12.3

3 0.68

4 26

Explanation:

Heat evolved $= m \times C_{v} \times Δ T$ $= 0.50 \times 3.64 \times 6.76 = 12.30 kJ$

CHXI06:THERMODYNAMICS

369186 One gram sample of ${NH}_{4} {NO}_{3}$ is decomposed in a bomb calorimeter. The temperature of the calorimeter increased by 6.12 K . The heat capacity of the system is $1.23 kJ / g^{- 1} K^{- 1}$ . What is the molar heat of decomposition for ${NH}_{4} {NO}_{3}$ .

1

7.53 kJ / mol

2

398.1 kJ / mol

3

16.1 kJ / mol

4

602 kJ / mol

Explanation:

The heat evolved, $q = mS Δ T$
$m = \lg; S = 1.23 kJ . g^{-} K^{-}$ ;
$Δ T = 6.12 K$
When 1 g of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 kJ /$ gram
When 1 mole of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 \times 80 kJ /$ $mol = 602.21 kJ / mol$
$({NH}_{4} {NO}_{3}$ molar mass is 80 g.mol $^{-})$

CHXI06:THERMODYNAMICS

369187 $1 g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 K$ and $1 atm$ . pressure according to the equation
$C_{(graphite)} + O_{2 (g)} \to {CO}_{2 (g)}$
During the reaction, temperature rises from 298 $K$ to $299 K$ . If the heat capacity of the bomb calorimeter is $20.7 kJ / K$ , what is the enthalpy change for the above reaction at $298 K$ and 1 atm?

1

20.7 kJ {mol}^{- 1}

2

- 20.7 kJ {mol}^{- 1}

3

- 248 kJ {mol}^{- 1}

4

248 kJ {mol}^{- 1}

Explanation:

Suppose $q$ is the quantity of heat from the reaction mixture and $C_{V}$ is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
$q = C_{V} \times Δ T$
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
$q = - C_{v} \times Δ T = - 20.7 kJ K^{- 1} \times (299 - 298) K$ $= - 20.7 kJ$
(Here, negative sign indicates the exothermic nature of the reaction)
Thus, $Δ U$ for the combustion of the $1 g$ of graphite $= - 20.7 kJ K^{- 1}$
For combustion of $1 mol$ of graphite,
$= \frac{12.0 g {mol}^{- 1} \times (- 20.7 kJ)}{1 g}$
$= - 2.48 \times 10^{2} kJ {mol}^{- 1}$ , Since $Δ n_{g} = 0$ ,
$Δ H = Δ U = - 2.48 \times 10^{2} kJ {mol}^{- 1}$
$= - 248 k J m o l^{- 1}$

CHXI06:THERMODYNAMICS

369188 $Δ_{f} H^{\circ}$ of hypothetical $MX$ is $- 250 kJ {mol}^{- 1}$ and for ${MX}_{2}$ is $- 600 kJ {mol}^{- 1}$ . The enthalpy of disproportionation of $MX$ is $- 100 x k J m o l^{- 1} .$ Find the value of x.

1

-

100

2

-

200

3

+

100

4

+

200

Explanation:

(3) $2 MX \to M + MX, : Δ H^{\circ} =$ ?
$M (s) + \frac{1}{2} \to 2 \times MX : Δ_{f} H_{1}^{\circ} = - 250 kJ mo l^{- 1}$
$M (s) + X_{2} \to M X_{2}; Δ_{f} H_{2}^{-} = - 600 kJ mo l^{- 1}$
$∴ Δ H^{\circ} = Δ_{f} H_{2}^{\circ} - 2 Δ_{f} H_{1}^{\circ}$
$= - 600 - (2 \times 250) = - 100 kJ mo l^{- 1}$
Enthalpy of disproportionation is $-$ 100.

CHXI06:THERMODYNAMICS

369189 $4 g$ of graphite is burnt in a bomb calorimeter of heat capacity $30 kJ K^{- 1}$ in excess of oxygen at $1 atm$ pressure. The temperature rises from 300 $K$ to $304 K$ . What is the enthalpy of combustion of graphite $(i n kJ mo l^{- 1})$ ?

1

360

2

1440

3

- 360

4

- 1440

Explanation:

$Δ E = C \times Δ t \times \frac{M}{m} = 30 \times 4 \times \frac{12}{4} = 360$
$∴ Δ E = - 360 kJ / mol$ (as it is combustion, sign is $- ve$ )
for $\underset{(s)}{C} + \underset{(g)}{O_{2}} \to \underset{(g)}{{CO}_{2}}; \underset{(g)}{Δ n} = 1 - 1 = 0$
$∴ Δ H = Δ E = - 360 kJ / mol$

CHXI06:THERMODYNAMICS

369190 When 0.50 g of unknown carbon compound is burned in the bomb calorimeter, its temperature rises by $6.76^{\circ} C$ . How much energy (in kilojoules) is released during combustion? (Heat capacity of calorimeter $= 3.64 {kJ}^{\circ} C^{- 1} g^{- 1}$ )

1 6.15

2 12.3

3 0.68

4 26

Explanation:

Heat evolved $= m \times C_{v} \times Δ T$ $= 0.50 \times 3.64 \times 6.76 = 12.30 kJ$

CHXI06:THERMODYNAMICS

369186 One gram sample of ${NH}_{4} {NO}_{3}$ is decomposed in a bomb calorimeter. The temperature of the calorimeter increased by 6.12 K . The heat capacity of the system is $1.23 kJ / g^{- 1} K^{- 1}$ . What is the molar heat of decomposition for ${NH}_{4} {NO}_{3}$ .

1

7.53 kJ / mol

2

398.1 kJ / mol

3

16.1 kJ / mol

4

602 kJ / mol

Explanation:

The heat evolved, $q = mS Δ T$
$m = \lg; S = 1.23 kJ . g^{-} K^{-}$ ;
$Δ T = 6.12 K$
When 1 g of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 kJ /$ gram
When 1 mole of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 \times 80 kJ /$ $mol = 602.21 kJ / mol$
$({NH}_{4} {NO}_{3}$ molar mass is 80 g.mol $^{-})$

CHXI06:THERMODYNAMICS

369187 $1 g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 K$ and $1 atm$ . pressure according to the equation
$C_{(graphite)} + O_{2 (g)} \to {CO}_{2 (g)}$
During the reaction, temperature rises from 298 $K$ to $299 K$ . If the heat capacity of the bomb calorimeter is $20.7 kJ / K$ , what is the enthalpy change for the above reaction at $298 K$ and 1 atm?

1

20.7 kJ {mol}^{- 1}

2

- 20.7 kJ {mol}^{- 1}

3

- 248 kJ {mol}^{- 1}

4

248 kJ {mol}^{- 1}

Explanation:

Suppose $q$ is the quantity of heat from the reaction mixture and $C_{V}$ is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
$q = C_{V} \times Δ T$
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
$q = - C_{v} \times Δ T = - 20.7 kJ K^{- 1} \times (299 - 298) K$ $= - 20.7 kJ$
(Here, negative sign indicates the exothermic nature of the reaction)
Thus, $Δ U$ for the combustion of the $1 g$ of graphite $= - 20.7 kJ K^{- 1}$
For combustion of $1 mol$ of graphite,
$= \frac{12.0 g {mol}^{- 1} \times (- 20.7 kJ)}{1 g}$
$= - 2.48 \times 10^{2} kJ {mol}^{- 1}$ , Since $Δ n_{g} = 0$ ,
$Δ H = Δ U = - 2.48 \times 10^{2} kJ {mol}^{- 1}$
$= - 248 k J m o l^{- 1}$

CHXI06:THERMODYNAMICS

369188 $Δ_{f} H^{\circ}$ of hypothetical $MX$ is $- 250 kJ {mol}^{- 1}$ and for ${MX}_{2}$ is $- 600 kJ {mol}^{- 1}$ . The enthalpy of disproportionation of $MX$ is $- 100 x k J m o l^{- 1} .$ Find the value of x.

1

-

100

2

-

200

3

+

100

4

+

200

Explanation:

(3) $2 MX \to M + MX, : Δ H^{\circ} =$ ?
$M (s) + \frac{1}{2} \to 2 \times MX : Δ_{f} H_{1}^{\circ} = - 250 kJ mo l^{- 1}$
$M (s) + X_{2} \to M X_{2}; Δ_{f} H_{2}^{-} = - 600 kJ mo l^{- 1}$
$∴ Δ H^{\circ} = Δ_{f} H_{2}^{\circ} - 2 Δ_{f} H_{1}^{\circ}$
$= - 600 - (2 \times 250) = - 100 kJ mo l^{- 1}$
Enthalpy of disproportionation is $-$ 100.

CHXI06:THERMODYNAMICS

369189 $4 g$ of graphite is burnt in a bomb calorimeter of heat capacity $30 kJ K^{- 1}$ in excess of oxygen at $1 atm$ pressure. The temperature rises from 300 $K$ to $304 K$ . What is the enthalpy of combustion of graphite $(i n kJ mo l^{- 1})$ ?

1

360

2

1440

3

- 360

4

- 1440

Explanation:

$Δ E = C \times Δ t \times \frac{M}{m} = 30 \times 4 \times \frac{12}{4} = 360$
$∴ Δ E = - 360 kJ / mol$ (as it is combustion, sign is $- ve$ )
for $\underset{(s)}{C} + \underset{(g)}{O_{2}} \to \underset{(g)}{{CO}_{2}}; \underset{(g)}{Δ n} = 1 - 1 = 0$
$∴ Δ H = Δ E = - 360 kJ / mol$

CHXI06:THERMODYNAMICS

369190 When 0.50 g of unknown carbon compound is burned in the bomb calorimeter, its temperature rises by $6.76^{\circ} C$ . How much energy (in kilojoules) is released during combustion? (Heat capacity of calorimeter $= 3.64 {kJ}^{\circ} C^{- 1} g^{- 1}$ )

1 6.15

2 12.3

3 0.68

4 26

Explanation:

Heat evolved $= m \times C_{v} \times Δ T$ $= 0.50 \times 3.64 \times 6.76 = 12.30 kJ$

CHXI06:THERMODYNAMICS

369186 One gram sample of ${NH}_{4} {NO}_{3}$ is decomposed in a bomb calorimeter. The temperature of the calorimeter increased by 6.12 K . The heat capacity of the system is $1.23 kJ / g^{- 1} K^{- 1}$ . What is the molar heat of decomposition for ${NH}_{4} {NO}_{3}$ .

1

7.53 kJ / mol

2

398.1 kJ / mol

3

16.1 kJ / mol

4

602 kJ / mol

Explanation:

The heat evolved, $q = mS Δ T$
$m = \lg; S = 1.23 kJ . g^{-} K^{-}$ ;
$Δ T = 6.12 K$
When 1 g of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 kJ /$ gram
When 1 mole of ammonium nitrate decomposes, the heat evolved is $q = 1 \times 1.23 \times 6.12 \times 80 kJ /$ $mol = 602.21 kJ / mol$
$({NH}_{4} {NO}_{3}$ molar mass is 80 g.mol $^{-})$

CHXI06:THERMODYNAMICS

369187 $1 g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 K$ and $1 atm$ . pressure according to the equation
$C_{(graphite)} + O_{2 (g)} \to {CO}_{2 (g)}$
During the reaction, temperature rises from 298 $K$ to $299 K$ . If the heat capacity of the bomb calorimeter is $20.7 kJ / K$ , what is the enthalpy change for the above reaction at $298 K$ and 1 atm?

1

20.7 kJ {mol}^{- 1}

2

- 20.7 kJ {mol}^{- 1}

3

- 248 kJ {mol}^{- 1}

4

248 kJ {mol}^{- 1}

Explanation:

Suppose $q$ is the quantity of heat from the reaction mixture and $C_{V}$ is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter.
$q = C_{V} \times Δ T$
Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
$q = - C_{v} \times Δ T = - 20.7 kJ K^{- 1} \times (299 - 298) K$ $= - 20.7 kJ$
(Here, negative sign indicates the exothermic nature of the reaction)
Thus, $Δ U$ for the combustion of the $1 g$ of graphite $= - 20.7 kJ K^{- 1}$
For combustion of $1 mol$ of graphite,
$= \frac{12.0 g {mol}^{- 1} \times (- 20.7 kJ)}{1 g}$
$= - 2.48 \times 10^{2} kJ {mol}^{- 1}$ , Since $Δ n_{g} = 0$ ,
$Δ H = Δ U = - 2.48 \times 10^{2} kJ {mol}^{- 1}$
$= - 248 k J m o l^{- 1}$

CHXI06:THERMODYNAMICS

369188 $Δ_{f} H^{\circ}$ of hypothetical $MX$ is $- 250 kJ {mol}^{- 1}$ and for ${MX}_{2}$ is $- 600 kJ {mol}^{- 1}$ . The enthalpy of disproportionation of $MX$ is $- 100 x k J m o l^{- 1} .$ Find the value of x.

1

-

100

2

-

200

3

+

100

4

+

200

Explanation:

(3) $2 MX \to M + MX, : Δ H^{\circ} =$ ?
$M (s) + \frac{1}{2} \to 2 \times MX : Δ_{f} H_{1}^{\circ} = - 250 kJ mo l^{- 1}$
$M (s) + X_{2} \to M X_{2}; Δ_{f} H_{2}^{-} = - 600 kJ mo l^{- 1}$
$∴ Δ H^{\circ} = Δ_{f} H_{2}^{\circ} - 2 Δ_{f} H_{1}^{\circ}$
$= - 600 - (2 \times 250) = - 100 kJ mo l^{- 1}$
Enthalpy of disproportionation is $-$ 100.

CHXI06:THERMODYNAMICS

369189 $4 g$ of graphite is burnt in a bomb calorimeter of heat capacity $30 kJ K^{- 1}$ in excess of oxygen at $1 atm$ pressure. The temperature rises from 300 $K$ to $304 K$ . What is the enthalpy of combustion of graphite $(i n kJ mo l^{- 1})$ ?

1

360

2

1440

3

- 360

4

- 1440

Explanation:

$Δ E = C \times Δ t \times \frac{M}{m} = 30 \times 4 \times \frac{12}{4} = 360$
$∴ Δ E = - 360 kJ / mol$ (as it is combustion, sign is $- ve$ )
for $\underset{(s)}{C} + \underset{(g)}{O_{2}} \to \underset{(g)}{{CO}_{2}}; \underset{(g)}{Δ n} = 1 - 1 = 0$
$∴ Δ H = Δ E = - 360 kJ / mol$

CHXI06:THERMODYNAMICS

369190 When 0.50 g of unknown carbon compound is burned in the bomb calorimeter, its temperature rises by $6.76^{\circ} C$ . How much energy (in kilojoules) is released during combustion? (Heat capacity of calorimeter $= 3.64 {kJ}^{\circ} C^{- 1} g^{- 1}$ )

1 6.15

2 12.3

3 0.68

4 26

Explanation:

Heat evolved $= m \times C_{v} \times Δ T$ $= 0.50 \times 3.64 \times 6.76 = 12.30 kJ$