Given these data C 2 H 4 ( g ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 1411 kJ/mol 2C 2 H 2 ( g ) + 5 O 2 ( g ) → 4 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 2599 kJ/mol H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O( l ) Δ H ° = − 285.8 kJ/mol calculate the heat of hydrogenation for acetylene: C 2 H 2 ( g ) + H 2 ( g ) → C 2 H 4 ( g )

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 24, Problem 24.44QP

Given these data

C 2 H 4 ( g ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 1411 kJ/mol 2C 2 H 2 ( g ) + 5 O 2 ( g ) → 4 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 2599 kJ/mol H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O( l ) Δ H ° = − 285.8 kJ/mol

calculate the heat of hydrogenation for acetylene:

C 2 H 2 ( g ) + H 2 ( g ) → C 2 H 4 ( g )

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

ΔHro = ΣnΔHo(products) - ΣnΔHo (reactants)where, n is Stoichiometric coefficient

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol

Equating the all the given equations as,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol H2(g) + 12O2(g) → H2O(l) ΔHo = −285.8 kJ/mol_overall:C2H2(g) + H2(g) → C2H4(g) ΔHo = -174.3kJ/mol _

The standard enthalpy change in kilojoules per mole for the given reaction is -174.3kJ/mol

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Don't used Ai solution

Benzene-toluene equilibrium is often approximated as αBT = 2.34. Generate the y-x diagram for this relative volatility. Also, generate the equilibrium data using Raoult’s law, and compare your results to these.

Given the most probable macrostate: s/k (K) Populations 300 4 200 8 100 16 0 32 Indicate how to demonstrate that the population of the levels is consistent with the Boltzmann distribution.

Answer 2

Textbook Question

Chapter 24, Problem 24.44QP

Given these data

C 2 H 4 ( g ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 1411 kJ/mol 2C 2 H 2 ( g ) + 5 O 2 ( g ) → 4 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 2599 kJ/mol H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O( l ) Δ H ° = − 285.8 kJ/mol

calculate the heat of hydrogenation for acetylene:

C 2 H 2 ( g ) + H 2 ( g ) → C 2 H 4 ( g )

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

ΔHro = ΣnΔHo(products) - ΣnΔHo (reactants)where, n is Stoichiometric coefficient

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol

Equating the all the given equations as,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol H2(g) + 12O2(g) → H2O(l) ΔHo = −285.8 kJ/mol_overall:C2H2(g) + H2(g) → C2H4(g) ΔHo = -174.3kJ/mol _

The standard enthalpy change in kilojoules per mole for the given reaction is -174.3kJ/mol

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 24, Problem 24.44QP

Given these data

C 2 H 4 ( g ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 1411 kJ/mol 2C 2 H 2 ( g ) + 5 O 2 ( g ) → 4 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 2599 kJ/mol H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O( l ) Δ H ° = − 285.8 kJ/mol

calculate the heat of hydrogenation for acetylene:

C 2 H 2 ( g ) + H 2 ( g ) → C 2 H 4 ( g )

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

ΔHro = ΣnΔHo(products) - ΣnΔHo (reactants)where, n is Stoichiometric coefficient

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol

Equating the all the given equations as,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol H2(g) + 12O2(g) → H2O(l) ΔHo = −285.8 kJ/mol_overall:C2H2(g) + H2(g) → C2H4(g) ΔHo = -174.3kJ/mol _

The standard enthalpy change in kilojoules per mole for the given reaction is -174.3kJ/mol

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 24, Problem 24.44QP

Given these data

C 2 H 4 ( g ) + 3 O 2 ( g ) → 2 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 1411 kJ/mol 2C 2 H 2 ( g ) + 5 O 2 ( g ) → 4 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = − 2599 kJ/mol H 2 ( g ) + 1 2 O 2 ( g ) → H 2 O( l ) Δ H ° = − 285.8 kJ/mol

calculate the heat of hydrogenation for acetylene:

C 2 H 2 ( g ) + H 2 ( g ) → C 2 H 4 ( g )

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

ΔHro = ΣnΔHo(products) - ΣnΔHo (reactants)where, n is Stoichiometric coefficient

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol

Equating the all the given equations as,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol H2(g) + 12O2(g) → H2O(l) ΔHo = −285.8 kJ/mol_overall:C2H2(g) + H2(g) → C2H4(g) ΔHo = -174.3kJ/mol _

The standard enthalpy change in kilojoules per mole for the given reaction is -174.3kJ/mol

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

ΔHro = ΣnΔHo(products) - ΣnΔHo (reactants)where, n is Stoichiometric coefficient

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol

Equating the all the given equations as,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol H2(g) + 12O2(g) → H2O(l) ΔHo = −285.8 kJ/mol_overall:C2H2(g) + H2(g) → C2H4(g) ΔHo = -174.3kJ/mol _

The standard enthalpy change in kilojoules per mole for the given reaction is -174.3kJ/mol

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

ΔHro = ΣnΔHo(products) - ΣnΔHo (reactants)where, n is Stoichiometric coefficient

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol

Equating the all the given equations as,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol H2(g) + 12O2(g) → H2O(l) ΔHo = −285.8 kJ/mol_overall:C2H2(g) + H2(g) → C2H4(g) ΔHo = -174.3kJ/mol _

The standard enthalpy change in kilojoules per mole for the given reaction is -174.3kJ/mol

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

ΔHro = ΣnΔHo(products) - ΣnΔHo (reactants)where, n is Stoichiometric coefficient

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Answer 12

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol

Equating the all the given equations as,

2CO2 (g) + 2H2O(l)→ C2H4(g) + 3 O2(g) ΔHo = +1411 kJ/mol C2H2(g) + 52O2(g) → 2CO2 (g) + H2O(l) ΔHo = −1299.5 kJ/mol H2(g) + 12O2(g) → H2O(l) ΔHo = −285.8 kJ/mol_overall:C2H2(g) + H2(g) → C2H4(g) ΔHo = -174.3kJ/mol _

The standard enthalpy change in kilojoules per mole for the given reaction is -174.3kJ/mol

Answer 13

Ch. 24.2 - How many structural isomers are there in the...Ch. 24.2 - Prob. 2PE Ch. 24.2 - Prob. 3PE Ch. 24.2 - Prob. 4PE Ch. 24.2 - Prob. 1RCF Ch. 24.2 - For which of the following compounds are cis-trans...Ch. 24.3 - Prob. 1RCF Ch. 24.4 - Prob. 5PE Ch. 24.4 - Prob. 1RCF Ch. 24 - Prob. 24.1QP

Videos

Explanation of Solution

Want to see more full solutions like this?

Chapter 24 Solutions