QBManager

Thermodynamics

272815 Given: $2 C (s) + 2 O_{2} (g) \to 2 {CO}_{2} (g)$ ;
$Δ H = - 787 k J$

$\begin{aligned} H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286 kJ \\ C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (l) \\ ; Δ H = - 1310 kJ \end{aligned}$
The heat of formation of acetylene is

1

- 1802 kJ

2

+ 1802 kJ

3

+ 237 kJ

4

- 800 kJ

Explanation:

We have to find
$2 C (s) + 2 H_{2} (g) ⟶ C_{2} H_{2} (g), Δ H = ?$
This is the equation for formation of acelytene
$\begin{aligned} 2 C (s) + 2 O_{2} (g) ⟶ 2 {CO}_{2} (g); Δ H = - 787 kJ \\ H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286 kJ \end{aligned}$
$\begin{aligned} C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l) \\ Δ H = - 1310 kJ \end{aligned}$
Add eq. (1) and (2), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l); \\ Δ H = & - 1073 kJ \end{aligned}$
Subtract eq. (3) form eq. (4), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) ⟶ C_{2} H_{2} (g) \\ Δ H = + 237 kJ \end{aligned}$

VITEEE- 2007

Thermodynamics

272767 The dissociation energy of ${CH}_{4}$ is $360 kcal {mol}^{-}$ 1 and that of ethane is $620 kcal {mol}^{- 1}$ . The bond energy of $C - C$ bond is :

1

260 kcal

2

980 kcal

3

90 kcal

4

80 kcal

Explanation:

${CH}_{4} (g) ⟶ C (g) + 2 H_{2} (g); Δ H = 360 kcal {mol}^{- 1}$
So, bond dissociation energy of one $C - H$ bond
$= \frac{360}{4} = 90 kcal / mol$
$C_{2} H_{6} (g) ⟶ 2 C (g) + 3 H_{2} (g); Δ H = 320 kcal / mol$
$Δ H = Σ (Bond energy)_{R} - Σ (Bond energy) p$
$620 = [6 \times (B . E)_{C - H} + 1 \times (B . E) C - C]$
$620 = 6 \times 90 + (B . E)_{C - C}$
$(B . E)_{C - C} = 620 - 540 = 80 kcal / mol$

CG PET-22.05.2022

Thermodynamics

272769 What is the amount of heat is released when $8.17 g$ of $Al$ (s) is converted to ${Al}_{2} O_{3} (s)$ at $25^{\circ} C$ and $1 atm$ via the reaction $4 Al (s) + 3 O_{2} (g) \to$ $2 {Al}_{2} O_{3} (s)$ ?
[For the reaction, $Δ H = - 1676 kJ / mol$ , atomic mass of $Al = 27$

1

134 kJ

2

254 kJ

3

268 kJ

4

1.01 \times 10^{3} kJ

Explanation:

$Δ H_{f}^{\circ} = - 1676 kJ mol$
Note that the minus sign is used here to illustrate that the heat is being given off.
This value tells us that when 1 mole of aluminium oxide is formed at standard conditions from its constituent elements in their most stable form, $1676 kJ$ of heat are being given off.
In other words, we know that standard enthalpy change for this reaction
$2 Al (s) + \frac{3}{2} O_{2} (g) \to {Al}_{2} O_{3} (s)$
is equal to
$Δ H_{f}^{\circ} = - 1676 kJ$
This implies that this reaction
$4 Al (s) + 3 O_{2} (g) \to 2 {Al}_{2} O_{3} (s)$
will have
$Δ H_{mmn}^{\circ} = 2 . Δ H_{f}^{\circ}, Δ H_{rmm}^{\circ} = 3352 kJ$
So, when 2 moles of aluminium oxide are formed under standard conditions, the reaction given off $3352 kJ$ of heat.
Use the molar mass of aluminium to calculate the number of moles present in sample
$8.17 g \cdot \frac{1 molAl}{26.98 g} = 0.30282$ moles A1
According to the balanced chemical equation, the reaction will produce
0.30282 moles $Al \cdot \frac{1 {molAl}_{2} O_{3}}{4 moles Al} = 0.15141$ moles ${Al}_{2} O_{3}$
We can say that when $8.17 g$ of aluminium undergo complete combustion, the reaction will give off
$0.15141 moles {Al}_{2} O_{3} \cdot \frac{3352 kJ}{2 moles {Al}_{2} O_{3}} = 254 kJ$

SCRA 2012

Thermodynamics

272771 The bond enthalpy of $H - H, Cl - Cl$ and $H - Cl$ are 435,243 and $431 kJ {mol}^{- 1}$ respectively. The enthalpy of formation of $HCl (g)$ would be:

1

92 kJ {mol}^{- 1}

2

- 92 kJ {mol}^{- 1}

3

184 kJ {mol}^{- 1}

4

- 184 kJ {mol}^{- 1}

Explanation:

For the reaction
$\frac{1}{2} H_{2} (g) + \frac{1}{2} {Cl}_{2} (g) \to HCl (g)$
For reactants-
Bond Energy of $H - H$ bond $= 435 {kJ}^{- 1} {mol}^{- 1}$
Bond Energy of $Cl - Cl$ bond $= 243 kJ {mol}^{- 1}$
For products-
Energy of formation of $H - Cl$ bond $= 431 kJ mol^{- 1}$
Thus, $Δ H$ of the above reaction is-
$Δ H = \frac{1}{2} (Δ H_{H_{2}} + Δ H_{{Cl}_{2}}) - Δ H_{HCl}$
$= \frac{1}{2} (435 + 243) - 431$
$= \frac{1}{2} (678) - 431$
$= 339 - 431$
$= - 92 kJ mol$

MPPET - 2012

Thermodynamics

272772 Consider the following reactions:
$C + O_{2} \to {CO}_{2}; Δ H = - 400 kJ {mol}^{- 1}$
$2 H_{2} + O_{2} \to 2 H_{2} O; Δ H = - 570 kJ {mol}^{- 1}$
${CO}_{2} + 2 H_{2} O \to {CH}_{4} + 2 O_{2} Δ H = 890 kJ {mol}^{- 1}$
The heat of formation of methane is

1

- 180 kJ {mol}^{- 1}

2

- 1060 kJ mol {kl}^{- 1}

3

- 720 kJ {mol}^{- 1}

4

- 80 kJ {mol}^{- 1}

Explanation:

Given that,
$C + O_{2} ⟶ {CO}_{2}; Δ H = - 400 kJ / mole$
$2 H_{2} + O_{2} ⟶ 2 H_{2} O; Δ H = - 570 kJ / mole$
${CO}_{2} + 2 H_{2} O ⟶ {CH}_{4} + 2 O_{2}; Δ H = 890 kJ / mole$ .(iii) Adding equation (i), (ii) and (iii) we get,
$C + 2 H_{2} \to {CH}_{4}; Δ H = - 400 + (- 570) + 890$
$Δ H = - 80 kJ / mole$ .

(A.P.EAMCET-1994)

Thermodynamics

272815 Given: $2 C (s) + 2 O_{2} (g) \to 2 {CO}_{2} (g)$ ;
$Δ H = - 787 k J$

$\begin{aligned} H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286 kJ \\ C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (l) \\ ; Δ H = - 1310 kJ \end{aligned}$
The heat of formation of acetylene is

1

- 1802 kJ

2

+ 1802 kJ

3

+ 237 kJ

4

- 800 kJ

Explanation:

We have to find
$2 C (s) + 2 H_{2} (g) ⟶ C_{2} H_{2} (g), Δ H = ?$
This is the equation for formation of acelytene
$\begin{aligned} 2 C (s) + 2 O_{2} (g) ⟶ 2 {CO}_{2} (g); Δ H = - 787 kJ \\ H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286 kJ \end{aligned}$
$\begin{aligned} C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l) \\ Δ H = - 1310 kJ \end{aligned}$
Add eq. (1) and (2), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l); \\ Δ H = & - 1073 kJ \end{aligned}$
Subtract eq. (3) form eq. (4), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) ⟶ C_{2} H_{2} (g) \\ Δ H = + 237 kJ \end{aligned}$

VITEEE- 2007

Thermodynamics

272767 The dissociation energy of ${CH}_{4}$ is $360 kcal {mol}^{-}$ 1 and that of ethane is $620 kcal {mol}^{- 1}$ . The bond energy of $C - C$ bond is :

1

260 kcal

2

980 kcal

3

90 kcal

4

80 kcal

Explanation:

${CH}_{4} (g) ⟶ C (g) + 2 H_{2} (g); Δ H = 360 kcal {mol}^{- 1}$
So, bond dissociation energy of one $C - H$ bond
$= \frac{360}{4} = 90 kcal / mol$
$C_{2} H_{6} (g) ⟶ 2 C (g) + 3 H_{2} (g); Δ H = 320 kcal / mol$
$Δ H = Σ (Bond energy)_{R} - Σ (Bond energy) p$
$620 = [6 \times (B . E)_{C - H} + 1 \times (B . E) C - C]$
$620 = 6 \times 90 + (B . E)_{C - C}$
$(B . E)_{C - C} = 620 - 540 = 80 kcal / mol$

CG PET-22.05.2022

Thermodynamics

272769 What is the amount of heat is released when $8.17 g$ of $Al$ (s) is converted to ${Al}_{2} O_{3} (s)$ at $25^{\circ} C$ and $1 atm$ via the reaction $4 Al (s) + 3 O_{2} (g) \to$ $2 {Al}_{2} O_{3} (s)$ ?
[For the reaction, $Δ H = - 1676 kJ / mol$ , atomic mass of $Al = 27$

1

134 kJ

2

254 kJ

3

268 kJ

4

1.01 \times 10^{3} kJ

Explanation:

$Δ H_{f}^{\circ} = - 1676 kJ mol$
Note that the minus sign is used here to illustrate that the heat is being given off.
This value tells us that when 1 mole of aluminium oxide is formed at standard conditions from its constituent elements in their most stable form, $1676 kJ$ of heat are being given off.
In other words, we know that standard enthalpy change for this reaction
$2 Al (s) + \frac{3}{2} O_{2} (g) \to {Al}_{2} O_{3} (s)$
is equal to
$Δ H_{f}^{\circ} = - 1676 kJ$
This implies that this reaction
$4 Al (s) + 3 O_{2} (g) \to 2 {Al}_{2} O_{3} (s)$
will have
$Δ H_{mmn}^{\circ} = 2 . Δ H_{f}^{\circ}, Δ H_{rmm}^{\circ} = 3352 kJ$
So, when 2 moles of aluminium oxide are formed under standard conditions, the reaction given off $3352 kJ$ of heat.
Use the molar mass of aluminium to calculate the number of moles present in sample
$8.17 g \cdot \frac{1 molAl}{26.98 g} = 0.30282$ moles A1
According to the balanced chemical equation, the reaction will produce
0.30282 moles $Al \cdot \frac{1 {molAl}_{2} O_{3}}{4 moles Al} = 0.15141$ moles ${Al}_{2} O_{3}$
We can say that when $8.17 g$ of aluminium undergo complete combustion, the reaction will give off
$0.15141 moles {Al}_{2} O_{3} \cdot \frac{3352 kJ}{2 moles {Al}_{2} O_{3}} = 254 kJ$

SCRA 2012

Thermodynamics

272771 The bond enthalpy of $H - H, Cl - Cl$ and $H - Cl$ are 435,243 and $431 kJ {mol}^{- 1}$ respectively. The enthalpy of formation of $HCl (g)$ would be:

1

92 kJ {mol}^{- 1}

2

- 92 kJ {mol}^{- 1}

3

184 kJ {mol}^{- 1}

4

- 184 kJ {mol}^{- 1}

Explanation:

For the reaction
$\frac{1}{2} H_{2} (g) + \frac{1}{2} {Cl}_{2} (g) \to HCl (g)$
For reactants-
Bond Energy of $H - H$ bond $= 435 {kJ}^{- 1} {mol}^{- 1}$
Bond Energy of $Cl - Cl$ bond $= 243 kJ {mol}^{- 1}$
For products-
Energy of formation of $H - Cl$ bond $= 431 kJ mol^{- 1}$
Thus, $Δ H$ of the above reaction is-
$Δ H = \frac{1}{2} (Δ H_{H_{2}} + Δ H_{{Cl}_{2}}) - Δ H_{HCl}$
$= \frac{1}{2} (435 + 243) - 431$
$= \frac{1}{2} (678) - 431$
$= 339 - 431$
$= - 92 kJ mol$

MPPET - 2012

Thermodynamics

272772 Consider the following reactions:
$C + O_{2} \to {CO}_{2}; Δ H = - 400 kJ {mol}^{- 1}$
$2 H_{2} + O_{2} \to 2 H_{2} O; Δ H = - 570 kJ {mol}^{- 1}$
${CO}_{2} + 2 H_{2} O \to {CH}_{4} + 2 O_{2} Δ H = 890 kJ {mol}^{- 1}$
The heat of formation of methane is

1

- 180 kJ {mol}^{- 1}

2

- 1060 kJ mol {kl}^{- 1}

3

- 720 kJ {mol}^{- 1}

4

- 80 kJ {mol}^{- 1}

Explanation:

Given that,
$C + O_{2} ⟶ {CO}_{2}; Δ H = - 400 kJ / mole$
$2 H_{2} + O_{2} ⟶ 2 H_{2} O; Δ H = - 570 kJ / mole$
${CO}_{2} + 2 H_{2} O ⟶ {CH}_{4} + 2 O_{2}; Δ H = 890 kJ / mole$ .(iii) Adding equation (i), (ii) and (iii) we get,
$C + 2 H_{2} \to {CH}_{4}; Δ H = - 400 + (- 570) + 890$
$Δ H = - 80 kJ / mole$ .

(A.P.EAMCET-1994)

Thermodynamics

272815 Given: $2 C (s) + 2 O_{2} (g) \to 2 {CO}_{2} (g)$ ;
$Δ H = - 787 k J$

$\begin{aligned} H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286 kJ \\ C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (l) \\ ; Δ H = - 1310 kJ \end{aligned}$
The heat of formation of acetylene is

1

- 1802 kJ

2

+ 1802 kJ

3

+ 237 kJ

4

- 800 kJ

Explanation:

We have to find
$2 C (s) + 2 H_{2} (g) ⟶ C_{2} H_{2} (g), Δ H = ?$
This is the equation for formation of acelytene
$\begin{aligned} 2 C (s) + 2 O_{2} (g) ⟶ 2 {CO}_{2} (g); Δ H = - 787 kJ \\ H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286 kJ \end{aligned}$
$\begin{aligned} C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l) \\ Δ H = - 1310 kJ \end{aligned}$
Add eq. (1) and (2), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l); \\ Δ H = & - 1073 kJ \end{aligned}$
Subtract eq. (3) form eq. (4), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) ⟶ C_{2} H_{2} (g) \\ Δ H = + 237 kJ \end{aligned}$

VITEEE- 2007

Thermodynamics

272767 The dissociation energy of ${CH}_{4}$ is $360 kcal {mol}^{-}$ 1 and that of ethane is $620 kcal {mol}^{- 1}$ . The bond energy of $C - C$ bond is :

1

260 kcal

2

980 kcal

3

90 kcal

4

80 kcal

Explanation:

${CH}_{4} (g) ⟶ C (g) + 2 H_{2} (g); Δ H = 360 kcal {mol}^{- 1}$
So, bond dissociation energy of one $C - H$ bond
$= \frac{360}{4} = 90 kcal / mol$
$C_{2} H_{6} (g) ⟶ 2 C (g) + 3 H_{2} (g); Δ H = 320 kcal / mol$
$Δ H = Σ (Bond energy)_{R} - Σ (Bond energy) p$
$620 = [6 \times (B . E)_{C - H} + 1 \times (B . E) C - C]$
$620 = 6 \times 90 + (B . E)_{C - C}$
$(B . E)_{C - C} = 620 - 540 = 80 kcal / mol$

CG PET-22.05.2022

Thermodynamics

272769 What is the amount of heat is released when $8.17 g$ of $Al$ (s) is converted to ${Al}_{2} O_{3} (s)$ at $25^{\circ} C$ and $1 atm$ via the reaction $4 Al (s) + 3 O_{2} (g) \to$ $2 {Al}_{2} O_{3} (s)$ ?
[For the reaction, $Δ H = - 1676 kJ / mol$ , atomic mass of $Al = 27$

1

134 kJ

2

254 kJ

3

268 kJ

4

1.01 \times 10^{3} kJ

Explanation:

$Δ H_{f}^{\circ} = - 1676 kJ mol$
Note that the minus sign is used here to illustrate that the heat is being given off.
This value tells us that when 1 mole of aluminium oxide is formed at standard conditions from its constituent elements in their most stable form, $1676 kJ$ of heat are being given off.
In other words, we know that standard enthalpy change for this reaction
$2 Al (s) + \frac{3}{2} O_{2} (g) \to {Al}_{2} O_{3} (s)$
is equal to
$Δ H_{f}^{\circ} = - 1676 kJ$
This implies that this reaction
$4 Al (s) + 3 O_{2} (g) \to 2 {Al}_{2} O_{3} (s)$
will have
$Δ H_{mmn}^{\circ} = 2 . Δ H_{f}^{\circ}, Δ H_{rmm}^{\circ} = 3352 kJ$
So, when 2 moles of aluminium oxide are formed under standard conditions, the reaction given off $3352 kJ$ of heat.
Use the molar mass of aluminium to calculate the number of moles present in sample
$8.17 g \cdot \frac{1 molAl}{26.98 g} = 0.30282$ moles A1
According to the balanced chemical equation, the reaction will produce
0.30282 moles $Al \cdot \frac{1 {molAl}_{2} O_{3}}{4 moles Al} = 0.15141$ moles ${Al}_{2} O_{3}$
We can say that when $8.17 g$ of aluminium undergo complete combustion, the reaction will give off
$0.15141 moles {Al}_{2} O_{3} \cdot \frac{3352 kJ}{2 moles {Al}_{2} O_{3}} = 254 kJ$

SCRA 2012

Thermodynamics

272771 The bond enthalpy of $H - H, Cl - Cl$ and $H - Cl$ are 435,243 and $431 kJ {mol}^{- 1}$ respectively. The enthalpy of formation of $HCl (g)$ would be:

1

92 kJ {mol}^{- 1}

2

- 92 kJ {mol}^{- 1}

3

184 kJ {mol}^{- 1}

4

- 184 kJ {mol}^{- 1}

Explanation:

For the reaction
$\frac{1}{2} H_{2} (g) + \frac{1}{2} {Cl}_{2} (g) \to HCl (g)$
For reactants-
Bond Energy of $H - H$ bond $= 435 {kJ}^{- 1} {mol}^{- 1}$
Bond Energy of $Cl - Cl$ bond $= 243 kJ {mol}^{- 1}$
For products-
Energy of formation of $H - Cl$ bond $= 431 kJ mol^{- 1}$
Thus, $Δ H$ of the above reaction is-
$Δ H = \frac{1}{2} (Δ H_{H_{2}} + Δ H_{{Cl}_{2}}) - Δ H_{HCl}$
$= \frac{1}{2} (435 + 243) - 431$
$= \frac{1}{2} (678) - 431$
$= 339 - 431$
$= - 92 kJ mol$

MPPET - 2012

Thermodynamics

272772 Consider the following reactions:
$C + O_{2} \to {CO}_{2}; Δ H = - 400 kJ {mol}^{- 1}$
$2 H_{2} + O_{2} \to 2 H_{2} O; Δ H = - 570 kJ {mol}^{- 1}$
${CO}_{2} + 2 H_{2} O \to {CH}_{4} + 2 O_{2} Δ H = 890 kJ {mol}^{- 1}$
The heat of formation of methane is

1

- 180 kJ {mol}^{- 1}

2

- 1060 kJ mol {kl}^{- 1}

3

- 720 kJ {mol}^{- 1}

4

- 80 kJ {mol}^{- 1}

Explanation:

Given that,
$C + O_{2} ⟶ {CO}_{2}; Δ H = - 400 kJ / mole$
$2 H_{2} + O_{2} ⟶ 2 H_{2} O; Δ H = - 570 kJ / mole$
${CO}_{2} + 2 H_{2} O ⟶ {CH}_{4} + 2 O_{2}; Δ H = 890 kJ / mole$ .(iii) Adding equation (i), (ii) and (iii) we get,
$C + 2 H_{2} \to {CH}_{4}; Δ H = - 400 + (- 570) + 890$
$Δ H = - 80 kJ / mole$ .

(A.P.EAMCET-1994)

Thermodynamics

272815 Given: $2 C (s) + 2 O_{2} (g) \to 2 {CO}_{2} (g)$ ;
$Δ H = - 787 k J$

$\begin{aligned} H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286 kJ \\ C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (l) \\ ; Δ H = - 1310 kJ \end{aligned}$
The heat of formation of acetylene is

1

- 1802 kJ

2

+ 1802 kJ

3

+ 237 kJ

4

- 800 kJ

Explanation:

We have to find
$2 C (s) + 2 H_{2} (g) ⟶ C_{2} H_{2} (g), Δ H = ?$
This is the equation for formation of acelytene
$\begin{aligned} 2 C (s) + 2 O_{2} (g) ⟶ 2 {CO}_{2} (g); Δ H = - 787 kJ \\ H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286 kJ \end{aligned}$
$\begin{aligned} C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l) \\ Δ H = - 1310 kJ \end{aligned}$
Add eq. (1) and (2), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l); \\ Δ H = & - 1073 kJ \end{aligned}$
Subtract eq. (3) form eq. (4), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) ⟶ C_{2} H_{2} (g) \\ Δ H = + 237 kJ \end{aligned}$

VITEEE- 2007

Thermodynamics

272767 The dissociation energy of ${CH}_{4}$ is $360 kcal {mol}^{-}$ 1 and that of ethane is $620 kcal {mol}^{- 1}$ . The bond energy of $C - C$ bond is :

1

260 kcal

2

980 kcal

3

90 kcal

4

80 kcal

Explanation:

${CH}_{4} (g) ⟶ C (g) + 2 H_{2} (g); Δ H = 360 kcal {mol}^{- 1}$
So, bond dissociation energy of one $C - H$ bond
$= \frac{360}{4} = 90 kcal / mol$
$C_{2} H_{6} (g) ⟶ 2 C (g) + 3 H_{2} (g); Δ H = 320 kcal / mol$
$Δ H = Σ (Bond energy)_{R} - Σ (Bond energy) p$
$620 = [6 \times (B . E)_{C - H} + 1 \times (B . E) C - C]$
$620 = 6 \times 90 + (B . E)_{C - C}$
$(B . E)_{C - C} = 620 - 540 = 80 kcal / mol$

CG PET-22.05.2022

Thermodynamics

272769 What is the amount of heat is released when $8.17 g$ of $Al$ (s) is converted to ${Al}_{2} O_{3} (s)$ at $25^{\circ} C$ and $1 atm$ via the reaction $4 Al (s) + 3 O_{2} (g) \to$ $2 {Al}_{2} O_{3} (s)$ ?
[For the reaction, $Δ H = - 1676 kJ / mol$ , atomic mass of $Al = 27$

1

134 kJ

2

254 kJ

3

268 kJ

4

1.01 \times 10^{3} kJ

Explanation:

$Δ H_{f}^{\circ} = - 1676 kJ mol$
Note that the minus sign is used here to illustrate that the heat is being given off.
This value tells us that when 1 mole of aluminium oxide is formed at standard conditions from its constituent elements in their most stable form, $1676 kJ$ of heat are being given off.
In other words, we know that standard enthalpy change for this reaction
$2 Al (s) + \frac{3}{2} O_{2} (g) \to {Al}_{2} O_{3} (s)$
is equal to
$Δ H_{f}^{\circ} = - 1676 kJ$
This implies that this reaction
$4 Al (s) + 3 O_{2} (g) \to 2 {Al}_{2} O_{3} (s)$
will have
$Δ H_{mmn}^{\circ} = 2 . Δ H_{f}^{\circ}, Δ H_{rmm}^{\circ} = 3352 kJ$
So, when 2 moles of aluminium oxide are formed under standard conditions, the reaction given off $3352 kJ$ of heat.
Use the molar mass of aluminium to calculate the number of moles present in sample
$8.17 g \cdot \frac{1 molAl}{26.98 g} = 0.30282$ moles A1
According to the balanced chemical equation, the reaction will produce
0.30282 moles $Al \cdot \frac{1 {molAl}_{2} O_{3}}{4 moles Al} = 0.15141$ moles ${Al}_{2} O_{3}$
We can say that when $8.17 g$ of aluminium undergo complete combustion, the reaction will give off
$0.15141 moles {Al}_{2} O_{3} \cdot \frac{3352 kJ}{2 moles {Al}_{2} O_{3}} = 254 kJ$

SCRA 2012

Thermodynamics

272771 The bond enthalpy of $H - H, Cl - Cl$ and $H - Cl$ are 435,243 and $431 kJ {mol}^{- 1}$ respectively. The enthalpy of formation of $HCl (g)$ would be:

1

92 kJ {mol}^{- 1}

2

- 92 kJ {mol}^{- 1}

3

184 kJ {mol}^{- 1}

4

- 184 kJ {mol}^{- 1}

Explanation:

For the reaction
$\frac{1}{2} H_{2} (g) + \frac{1}{2} {Cl}_{2} (g) \to HCl (g)$
For reactants-
Bond Energy of $H - H$ bond $= 435 {kJ}^{- 1} {mol}^{- 1}$
Bond Energy of $Cl - Cl$ bond $= 243 kJ {mol}^{- 1}$
For products-
Energy of formation of $H - Cl$ bond $= 431 kJ mol^{- 1}$
Thus, $Δ H$ of the above reaction is-
$Δ H = \frac{1}{2} (Δ H_{H_{2}} + Δ H_{{Cl}_{2}}) - Δ H_{HCl}$
$= \frac{1}{2} (435 + 243) - 431$
$= \frac{1}{2} (678) - 431$
$= 339 - 431$
$= - 92 kJ mol$

MPPET - 2012

Thermodynamics

272772 Consider the following reactions:
$C + O_{2} \to {CO}_{2}; Δ H = - 400 kJ {mol}^{- 1}$
$2 H_{2} + O_{2} \to 2 H_{2} O; Δ H = - 570 kJ {mol}^{- 1}$
${CO}_{2} + 2 H_{2} O \to {CH}_{4} + 2 O_{2} Δ H = 890 kJ {mol}^{- 1}$
The heat of formation of methane is

1

- 180 kJ {mol}^{- 1}

2

- 1060 kJ mol {kl}^{- 1}

3

- 720 kJ {mol}^{- 1}

4

- 80 kJ {mol}^{- 1}

Explanation:

Given that,
$C + O_{2} ⟶ {CO}_{2}; Δ H = - 400 kJ / mole$
$2 H_{2} + O_{2} ⟶ 2 H_{2} O; Δ H = - 570 kJ / mole$
${CO}_{2} + 2 H_{2} O ⟶ {CH}_{4} + 2 O_{2}; Δ H = 890 kJ / mole$ .(iii) Adding equation (i), (ii) and (iii) we get,
$C + 2 H_{2} \to {CH}_{4}; Δ H = - 400 + (- 570) + 890$
$Δ H = - 80 kJ / mole$ .

(A.P.EAMCET-1994)

Thermodynamics

272815 Given: $2 C (s) + 2 O_{2} (g) \to 2 {CO}_{2} (g)$ ;
$Δ H = - 787 k J$

$\begin{aligned} H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286 kJ \\ C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (l) \\ ; Δ H = - 1310 kJ \end{aligned}$
The heat of formation of acetylene is

1

- 1802 kJ

2

+ 1802 kJ

3

+ 237 kJ

4

- 800 kJ

Explanation:

We have to find
$2 C (s) + 2 H_{2} (g) ⟶ C_{2} H_{2} (g), Δ H = ?$
This is the equation for formation of acelytene
$\begin{aligned} 2 C (s) + 2 O_{2} (g) ⟶ 2 {CO}_{2} (g); Δ H = - 787 kJ \\ H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286 kJ \end{aligned}$
$\begin{aligned} C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l) \\ Δ H = - 1310 kJ \end{aligned}$
Add eq. (1) and (2), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) + 2 \frac{1}{2} O_{2} (g) ⟶ & 2 {CO}_{2} (g) + H_{2} O (l); \\ Δ H = & - 1073 kJ \end{aligned}$
Subtract eq. (3) form eq. (4), we get -
$\begin{aligned} 2 C (s) + H_{2} (g) ⟶ C_{2} H_{2} (g) \\ Δ H = + 237 kJ \end{aligned}$

VITEEE- 2007

Thermodynamics

272767 The dissociation energy of ${CH}_{4}$ is $360 kcal {mol}^{-}$ 1 and that of ethane is $620 kcal {mol}^{- 1}$ . The bond energy of $C - C$ bond is :

1

260 kcal

2

980 kcal

3

90 kcal

4

80 kcal

Explanation:

${CH}_{4} (g) ⟶ C (g) + 2 H_{2} (g); Δ H = 360 kcal {mol}^{- 1}$
So, bond dissociation energy of one $C - H$ bond
$= \frac{360}{4} = 90 kcal / mol$
$C_{2} H_{6} (g) ⟶ 2 C (g) + 3 H_{2} (g); Δ H = 320 kcal / mol$
$Δ H = Σ (Bond energy)_{R} - Σ (Bond energy) p$
$620 = [6 \times (B . E)_{C - H} + 1 \times (B . E) C - C]$
$620 = 6 \times 90 + (B . E)_{C - C}$
$(B . E)_{C - C} = 620 - 540 = 80 kcal / mol$

CG PET-22.05.2022

Thermodynamics

272769 What is the amount of heat is released when $8.17 g$ of $Al$ (s) is converted to ${Al}_{2} O_{3} (s)$ at $25^{\circ} C$ and $1 atm$ via the reaction $4 Al (s) + 3 O_{2} (g) \to$ $2 {Al}_{2} O_{3} (s)$ ?
[For the reaction, $Δ H = - 1676 kJ / mol$ , atomic mass of $Al = 27$

1

134 kJ

2

254 kJ

3

268 kJ

4

1.01 \times 10^{3} kJ

Explanation:

$Δ H_{f}^{\circ} = - 1676 kJ mol$
Note that the minus sign is used here to illustrate that the heat is being given off.
This value tells us that when 1 mole of aluminium oxide is formed at standard conditions from its constituent elements in their most stable form, $1676 kJ$ of heat are being given off.
In other words, we know that standard enthalpy change for this reaction
$2 Al (s) + \frac{3}{2} O_{2} (g) \to {Al}_{2} O_{3} (s)$
is equal to
$Δ H_{f}^{\circ} = - 1676 kJ$
This implies that this reaction
$4 Al (s) + 3 O_{2} (g) \to 2 {Al}_{2} O_{3} (s)$
will have
$Δ H_{mmn}^{\circ} = 2 . Δ H_{f}^{\circ}, Δ H_{rmm}^{\circ} = 3352 kJ$
So, when 2 moles of aluminium oxide are formed under standard conditions, the reaction given off $3352 kJ$ of heat.
Use the molar mass of aluminium to calculate the number of moles present in sample
$8.17 g \cdot \frac{1 molAl}{26.98 g} = 0.30282$ moles A1
According to the balanced chemical equation, the reaction will produce
0.30282 moles $Al \cdot \frac{1 {molAl}_{2} O_{3}}{4 moles Al} = 0.15141$ moles ${Al}_{2} O_{3}$
We can say that when $8.17 g$ of aluminium undergo complete combustion, the reaction will give off
$0.15141 moles {Al}_{2} O_{3} \cdot \frac{3352 kJ}{2 moles {Al}_{2} O_{3}} = 254 kJ$

SCRA 2012

Thermodynamics

272771 The bond enthalpy of $H - H, Cl - Cl$ and $H - Cl$ are 435,243 and $431 kJ {mol}^{- 1}$ respectively. The enthalpy of formation of $HCl (g)$ would be:

1

92 kJ {mol}^{- 1}

2

- 92 kJ {mol}^{- 1}

3

184 kJ {mol}^{- 1}

4

- 184 kJ {mol}^{- 1}

Explanation:

For the reaction
$\frac{1}{2} H_{2} (g) + \frac{1}{2} {Cl}_{2} (g) \to HCl (g)$
For reactants-
Bond Energy of $H - H$ bond $= 435 {kJ}^{- 1} {mol}^{- 1}$
Bond Energy of $Cl - Cl$ bond $= 243 kJ {mol}^{- 1}$
For products-
Energy of formation of $H - Cl$ bond $= 431 kJ mol^{- 1}$
Thus, $Δ H$ of the above reaction is-
$Δ H = \frac{1}{2} (Δ H_{H_{2}} + Δ H_{{Cl}_{2}}) - Δ H_{HCl}$
$= \frac{1}{2} (435 + 243) - 431$
$= \frac{1}{2} (678) - 431$
$= 339 - 431$
$= - 92 kJ mol$

MPPET - 2012

Thermodynamics

272772 Consider the following reactions:
$C + O_{2} \to {CO}_{2}; Δ H = - 400 kJ {mol}^{- 1}$
$2 H_{2} + O_{2} \to 2 H_{2} O; Δ H = - 570 kJ {mol}^{- 1}$
${CO}_{2} + 2 H_{2} O \to {CH}_{4} + 2 O_{2} Δ H = 890 kJ {mol}^{- 1}$
The heat of formation of methane is

1

- 180 kJ {mol}^{- 1}

2

- 1060 kJ mol {kl}^{- 1}

3

- 720 kJ {mol}^{- 1}

4

- 80 kJ {mol}^{- 1}

Explanation:

Given that,
$C + O_{2} ⟶ {CO}_{2}; Δ H = - 400 kJ / mole$
$2 H_{2} + O_{2} ⟶ 2 H_{2} O; Δ H = - 570 kJ / mole$
${CO}_{2} + 2 H_{2} O ⟶ {CH}_{4} + 2 O_{2}; Δ H = 890 kJ / mole$ .(iii) Adding equation (i), (ii) and (iii) we get,
$C + 2 H_{2} \to {CH}_{4}; Δ H = - 400 + (- 570) + 890$
$Δ H = - 80 kJ / mole$ .

(A.P.EAMCET-1994)

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: